Question 12 Marks
If A is a skew-symmetric matrix of order 3, then prove that det A = 0.
AnswerAny skew symmetric matrix of order 3 is $\text{A} = \begin{bmatrix} 0 & \text{a} & \text{b} \\ \text{-a} & 0 & \text{c} \\ \text{-b} & \text{-c} & 0 \end{bmatrix}$
$\Rightarrow \text{|A}| = \text{-a(bc) + a(bc) = 0}$
Alternate Answer
Since A is a skew-symmetric matrix
$\therefore \text{A}^{\text{T}} = \text{-A}$
$\therefore |\text{A}^{\text{T}}| = \text{|-A|} = (-1)^{3}. \text{|A|}$
$\Rightarrow \text{|A|} = \text{|-A|}$
$\Rightarrow 2\text{|A}| = 0 \text{ }\text{ or } \text{ }\text{|A|} = 0.$
View full question & answer→Question 22 Marks
If A and B are square matrices of order 3 such that |A| = – 1, |B| = 3, then find the value of |2AB|.
Answer$|\text{2AB}| = 2^{3} \times \text{|A|} \times \text{|B|}$
$= 8 \times (-1) \times 3 = -24$
View full question & answer→Question 32 Marks
Show that all the diagonal elements of a skew symmetric matrix are zero.
AnswerLet $\text{A} = [\text{a}_{\text{ij}}]_{\text{n}\times\text{n}} $ be skew symmetric matrix.
A is skew symmetric
$\therefore \text{A = -A}' $
$\Rightarrow \text{a}_{\text{ij}} = \text{-a}_{\text{ji}} \text{ }\forall \text{ i, j}$
For diagonal elements i = j,
$\Rightarrow \text{2a}_{\text{ii}} = 0$
$\Rightarrow \text{a}_{\text{ii}} = 0 \Rightarrow$ diagonal elements are zero.
View full question & answer→Question 42 Marks
Given $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix},$ compute $A^{-1}$ and show that $2A^{-1} = 9I – A$.
Answer$\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
$\text{A}^{-1}=\frac{1}{|\text{A}|}\ \text{adj A}$
$=\frac{1}{14-12}\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}$
$=\frac{1}{2}\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}$
To Prove $2A^{-1} = 9I – A$
$LHS = 2A^{-1}$
$=2\times\frac{1}{2}\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}=\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}$
RHS = 9I – A
$=\begin{bmatrix}9 & 0 \\0 & 9 \end{bmatrix}-\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
$=\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}$
LHS = RHS.
View full question & answer→Question 52 Marks
Find a matrix A such that 2A - 3B + 5C = O, where $\text{B}=\begin{bmatrix}-2 & 2 & 0 \\3 & 1 & 4 \end{bmatrix}$ and $\text{C}=\begin{bmatrix}2 & 0 & -2 \\7 & 1 & 6\end{bmatrix}.$
AnswerGiven: $2\text{A}-3\text{B}+5\text{C}=0$
$\Rightarrow2\text{A}-3\begin{bmatrix}-2 & 2 & 0 \\3 & 1 & 4\end{bmatrix}+5\begin{bmatrix}2 & 0 & -2 \\7 & 1 & 6\end{bmatrix}0$
$\Rightarrow2\text{A}-\begin{bmatrix}-6 & 6 & 0 \\9 & 3 & 12\end{bmatrix}+\begin{bmatrix}10 & 0 & -10 \\35 & 5 & 30\end{bmatrix}=0$
$\Rightarrow2\text{A}+\begin{bmatrix}10+6 & 0-6 &-10-0 \\35 - 9 & 5-3 &30-12\end{bmatrix}=0$
$\Rightarrow2\text{A}+\begin{bmatrix}16 & -6 & -10 \\26 & 2 & 18\end{bmatrix}=0$
$\Rightarrow\text{A}=\begin{bmatrix}-8 & 3 & 5 \\-13 & -1 & -9\end{bmatrix}$
View full question & answer→Question 62 Marks
If $\text{A}=\begin{bmatrix}2 & 0 & 1 \\2 & 1 & 3\\1 & -1 & 0 \end{bmatrix},$ then find $(\text{A}^2-5\text{A}).$
Answer$\text{A}=\begin{bmatrix}2 & 0 & 1 \\2 & 1 & 3\\1 & -1 & 0 \end{bmatrix}$
Now $\text{A}^2=\text{A}.\text{A}$
$=\begin{bmatrix}2 & 0 & 1 \\2 & 1 & 3\\1 & -1 & 0 \end{bmatrix}\begin{bmatrix}2 & 0 & 1 \\2 & 1 & 3\\1 & -1 & 0 \end{bmatrix}$
$=\begin{bmatrix}5 & -1 & 2 \\9 & -2 & 5\\0 & -1 & -2 \end{bmatrix}$
Now value of $A^2-5\text{A}$
$=\begin{bmatrix}5 & -1 & 2 \\9 & -2 & 5\\0 & -1 & -2 \end{bmatrix}-5\begin{bmatrix}2 & 0 & 1 \\2 & 1 & 3\\1 & -1 & 0 \end{bmatrix}$
$=\begin{bmatrix}5 & -1 & 2 \\9 & -2 & 5\\0 & -1 & -2 \end{bmatrix}-\begin{bmatrix}10 & 0 & 5 \\10 & 5 & 15\\5 & -5 & 0 \end{bmatrix}$
So, $\text{A}^2-5\text{A}=\begin{bmatrix}-5 & -1 & -3 \\-1 & -7 & -10\\-5 & 4 & -2 \end{bmatrix}$
View full question & answer→Question 72 Marks
If $A = [a_{ij}]$ is a $2 \times 2$ matrix such that $a_{ij} = i + 2j$, write A.
AnswerHere,
$a_{ij} = i + 2j$
$\text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}$
$=\begin{bmatrix}1+2(1)&1+2(2)\\2+2(1)&2+2(2)\end{bmatrix}$
$=\begin{bmatrix}3&5\\4&6\end{bmatrix}$
Hence,
$\text{A}=\begin{bmatrix}3&5\\4&6\end{bmatrix}$
View full question & answer→Question 82 Marks
f $A$ is a matrix of order $3 \times 4$ and $B$ is a matrix of order $4 \times 3$, find the order of the matrix of $AB$.
AnswerOrder of $A=3 \times 4$
Order of $B=4 \times 3$
Order of $\mathrm{A}_{3 \times 4} \times \mathrm{B}_{4 \times 3}=3 \times 3$
So,
Order of $A B=3 \times 3$
View full question & answer→Question 92 Marks
Write the number of all possible matrices of order $2 \times 2$ with each entry $1, 2$ or $3$.
AnswerAs matrices is of order $2 \times 2$, so there are 4 entries possible.
Each entry has 3 choices that are 1, 2 or 3
So, number of ways to make up such matrices are $3 \times 3 \times 3 \times 3$ i.e, $3^4$ times or 81 times
View full question & answer→Question 102 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{AB})^{\text{T}}=\text{B}^{\text{T}}\text{A}^{\text{T}}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$\text{AB}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$
$=\begin{bmatrix}4+2&0+10\\-4+3&0+15\end{bmatrix}=\begin{bmatrix}6&10\\-1&15\end{bmatrix}$
$\therefore\ (\text{AB})^{\text{T}}=\begin{bmatrix}6&-1\\10&15\end{bmatrix}$
Now, $\text{B}^{\text{T}}\text{A}^{\text{T}}=\begin{bmatrix}4&1\\0&5\end{bmatrix}\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ $=\begin{bmatrix}6&-1\\10&15\end{bmatrix}$ $=(\text{AB})^{\text{T}}$ Hence proved.
View full question & answer→Question 112 Marks
If $\text{A}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}-4\\3\end{bmatrix},$ write AB.
Answer$\text{AB}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}\begin{bmatrix}-4\\3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-16+9\\-4+6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix} -7\\2\end{bmatrix}$
View full question & answer→Question 122 Marks
If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},$ write $A^2$.
AnswerGiven: $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2+0&0+0\\0+0&0+\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2&0\\0&\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ $(\because\ \text{i} ^2=-1)$
View full question & answer→Question 132 Marks
Find the values of x, y and z from the following equations:
$\begin{bmatrix}\text{x + y+ z}\\ \text{x + z}\\\ \text{y + z} \end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix} $
AnswerWe are given that
$\begin{bmatrix}\text{x + y + z}\\ \text{x+ z}\\ \text{y + z}\end{bmatrix}=\begin{bmatrix}9\\5\\7 \end{bmatrix}$
By defination of equality of matrices.
x + y + z = 9 ...(1)
x + z = 5 ...(2)
y + z = 7 ...(3)
Subtracting (2) from (1), y = 4
Subtracting (3) from (1), x = 2
$\therefore$ from(2), 2 + z = 5, ⇒ z = 3
$\therefore$ x = 2, y = 4, z = 3
View full question & answer→Question 142 Marks
For any square matrix write whether $AA^T$ is symmetric or skew-symmetric.
Answer$\big(\text{AA}^\text{T}\big)^\text{T}=\big(\text{A}^\text{T}\big)^\text{T}\times\text{A}^\text{T}$ $\big\{\text{since, (AB)}^\text{T}=\text{B}^\text{T}\text{A}^\text{T}\big\}$
$\therefore\ \big(\text{AA}^\text{T}\big)^\text{T}=\big(\text{AA}^\text{T}\big)\ \dots(\text{i})$ $\big\{\text{since, }(\text{A}^\text{T})^\text{T}=\text{A}\big\}$
We know that, a square matrix A is symmetric if $A^T = A$
So, from equation (i)
$(AA^T)$ is a symmetric matric.
View full question & answer→Question 152 Marks
Solve the equation for x, y, z and t if:$2\begin{bmatrix}\text{x} & \text{z} \\\text{y} & \text{t} \end{bmatrix} + 3\begin{bmatrix}1 & -1\\0 & 2 \end{bmatrix} = 3\begin{bmatrix}3 & 5 \\4 & 6 \end{bmatrix}.$
AnswerGiven: $2\begin{bmatrix}\text{x} & \text{z} \\\text{y} & \text{t} \end{bmatrix} + 3\begin{bmatrix}1 & -1\\0 & 2 \end{bmatrix} = 3\begin{bmatrix}3 & 5 \\4 & 6 \end{bmatrix}.$
$\Rightarrow \begin{bmatrix}2\text{x}&2\text{z}\\2\text{y}&2\text{t}\end{bmatrix}+\begin{bmatrix}3&-3\\0&6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2\text{x} + 3&2\text{z}-3\\2\text{y} + 0&2\text{t} + 6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix} $
Equation corresponding entries, we have
2x + 3 = 9 ⇒ 2x = 9 - 3 ⇒ 2x = 6 ⇒ x = 3
And 2z - 3 = 15
⇒ 2z = 15 + 3
⇒ 2z = 18 ⇒ z = 9
And 2y = 12 ⇒ y = 6
And 2t + 6 = 18 ⇒ 2t = 18 - 6
⇒ 2t = 12 ⇒ t = 6
$\therefore$ x = 3, y = 6, z = 9, t = 6
View full question & answer→Question 162 Marks
Find the values of x, y and z from the following equations:$\begin{bmatrix}4 & 3 \\ \text x & 5 \end{bmatrix}=\begin{bmatrix}\text y& \text z\\1 & 5 \end{bmatrix} $
AnswerWe are given that$\begin{bmatrix}4 & 3 \\ \text x & 5 \end{bmatrix}=\begin{bmatrix}\text y& \text z \\1 & 5 \end{bmatrix} $
By defination of equality of matrices$4 = \text y,\ 3 = \text z,\ \text x = 1$
$\therefore\ \text{ x = 1},\ \text{ y = 4},\text{ z = 3 } $
View full question & answer→Question 172 Marks
Give example of matrices:
A and B such that AB = O but A ≠ 0, B ≠ 0.
AnswerLet $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Thus, AB = 0 while A ≠ 0 and B ≠ 0
View full question & answer→Question 182 Marks
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&2-2&-2+2\\4+1&\text{x}+0&6-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&0&0\\5&\text{x}&5\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\text{x}-\text{y}+3=6$
$\Rightarrow\text{x}-\text{y}=6-3$
$\Rightarrow\text{x}-\text{y}=3\ \dots(1)$
Also,
$\text{x}=2\text{x}+\text{y}$
$\Rightarrow-\text{x}=\text{y}\ \dots(2)$
Putting the value of y in eq. (1), we get
$\text{x}-(-\text{x})=3$
$\Rightarrow2\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{2}$
Putting the value of x in eq. (2), we get
$-\Big(\frac{3}{2}\Big)=\text{y}$
$\Rightarrow\text{y}=-\frac{3}{2}$
View full question & answer→Question 192 Marks
Find the values of a, b, c and d from the equation:
$\begin{bmatrix}a-b&2a+c\\2a-b&3c+d\end{bmatrix}=\begin{bmatrix}-1&5\\0&13\end{bmatrix}.$
AnswerEquating corresponding entries,
a - b = -1 ...(i)
2a - b = 0 ...(ii)
2a + c = 5 ...(iii)
3c + d = 13 ...(iv)
Eq. (i) - Eq. (ii) = -a = -1 ⇒ a = 1
Putting a = 1 in eq. (i), 1 - b = -1 ⇒ -b = -2 ⇒ b = 2
Putting a = 1 in eq. (iii), 2 + c = 5 ⇒ c = 5 - 2 ⇒ c = 3
Putting c = 3 in eq. (iv), 9 + d = 13 ⇒ d = 13 - 9 ⇒ d = 4
$\therefore$ a = 1, b = 2, c = 3, d = 4
View full question & answer→Question 202 Marks
Find x, y, z and t, if.
$2\begin{bmatrix}\text{x}&5\\7&\text{y}-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
Answer$2\begin{bmatrix}\text{x}&5\\7&\text{y}-3\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}&10\\14&2\text{y}-6\end{bmatrix}+\begin{bmatrix}3&4\\1&2\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}+3&14\\15&2\text{y}-4\end{bmatrix}=\begin{bmatrix}7&14\\15&14\end{bmatrix}$
Comparing the corresponding elements from both sides,
$2\text{x}+3=7\Rightarrow2\text{x}=4\Rightarrow\text{x}=2$
$2\text{y}-4=14\Rightarrow2\text{y}=18\Rightarrow\text{y}=9$
Hence, x = 2, y = 9
View full question & answer→Question 212 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$\text{a}_\text{ij}=\frac{(\text{i}+\text{j})^2}{2}$
AnswerHere,
$\text{a}_\text{ij}=\frac{(\text{i}+\text{j})^2}{2}$
$\text{a}_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2,$ $\text{a}_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{13}=\frac{(1+3)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$\text{a}_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$ and $\text{a}_{23}=\frac{(2+3)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2}$
Required matrix = $\text{A}=\begin{bmatrix}2&\frac{9}{2}&8\\\frac{9}{2}&8&\frac{25}{2}\end{bmatrix}$
View full question & answer→Question 222 Marks
If $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix},$ show that $A - A^T$ is a skew symmetric matrix.
AnswerGiven:$\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$
$$$\text{A}-\text{A}^{\text{T}}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}^{\text{T}}$
$=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}-\begin{bmatrix}3 &1 \\-4 & -1 \end{bmatrix}$
$=\begin{bmatrix}3-3 & -4-1 \\1+4 & -1+1 \end{bmatrix}$
$\text{A}-\text{A}^{\text{T}}=\begin{bmatrix}0 & -5 \\5 & 0 \end{bmatrix}\ \dots( \text{i})$
$$$-(\text{A}-\text{A}^{\text{T}})^\text{T}=-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}^{\text{T}}$
$=-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}$
$$$-(\text{A}-\text{A}^{\text{T}})^{\text{T}}-\begin{bmatrix}0 & 5 \\-5 & 0 \end{bmatrix}\ \dots(\text{ii})$
From equation (i) and (ii)
$(\text{A}-\text{A}^{\text{T}})^{\text{T}}=-(\text{A}-\text{A}^{\text{T}})^{\text{T}}$
We know that, x is skewsym metric matrix if $x = -x^T$
So, $(A - A^T)$ is skewsym metric matrix.
View full question & answer→Question 232 Marks
Construct a $2 \times 3$ matrix $A=\left[a_{i j}\right]$ whose elements $a_{i j}$ are give by:
$a_{i j}=i . j$
AnswerHere,
$a_{i j}=\mathrm{i} . \mathrm{j} .1 \leq \mathrm{i} \leq 2$ and $1 \leq \mathrm{j} \leq 3$
$a_{11}=1 \times 1=1, a_{12}=1 \times 2=2, a_{13}=1 \times 3=3$
$a_{21}=2 \times 1=2, a_{22}=2 \times 2=4$ and $a_{23}=2 \times 3=6$
View full question & answer→Question 242 Marks
Let $A$ and $B$ be square matrices of the order $3 × 3.$ Is $(AB)^2 = A^2B^2?$ Give reasons.
AnswerWe are given that, A and B are square matrices of order $3 × 3.$
Consider, $(AB)^2= AB.AB$
$= ABAB$
$= AABB [\because AB = BA ]$
$= A^2B^2$
Thus, $AB^2 = A^2B^2$ is true if and only if $AB = BA.$
View full question & answer→Question 252 Marks
If a matrix has 5 elements, write all possible orders it can have.
AnswerWe know that if a matrix is of order m×n,then it has mn elements.If the matrix has 5 elements, then the number of elements will be 1×5 or 5×1, i.e. there will be 2 possible orders of the matrix.
View full question & answer→Question 262 Marks
Give example of matrices:
A, B and C such that AB = AC but B ≠ C, A ≠ 0
AnswerLet $\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&0\\-1&0\end{bmatrix},\ \text{C}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
Here,
$\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\-1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$
$\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$\begin{bmatrix}0&0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
LHS = RHS
So,
for A ≠ 0, BC ≠ 0 but AB = AC
We have,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}0&0\\-1&0\end{bmatrix},\ \text{C}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
View full question & answer→Question 272 Marks
If $\text{X}-\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}$ and $\text{X}+\text{Y}=\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix},$ find X and Y.
AnswerHere,
$\text{X}-\text{Y}+\text{X}+\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}1+3&1+5&1+1\\1-1&1+1&0+4\\1+11&0+8&0+0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$
Now,
$(\text{X}-\text{Y})-(\text{X}+\text{Y})=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}-\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow\text{X}-\text{Y}-\text{X}-\text{Y}=\begin{bmatrix}1-3&1-5&1-1\\1+1&1-1&0-4\\1-11&0-8&0-0\end{bmatrix}$
$\Rightarrow-2\text{Y}=\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=-\frac{1}{2}\begin{bmatrix}-2&-4&0\\2&0&-4\\-10&-8&0\end{bmatrix}$
$\Rightarrow\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$ and $\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$
View full question & answer→Question 282 Marks
If $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix},$ find $AA^T$.
AnswerGiven: $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos^2\text{x}+\sin^2\text{x}&\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}\\\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}&\sin^2\text{x}+\cos^2\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→Question 292 Marks
Find x, y satisfying the matrix equation.
$\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0$
View full question & answer→Question 302 Marks
If $\text{A}=\begin{bmatrix}9&1\\7&8\end{bmatrix},\text{ B}=\begin{bmatrix}1&5\\7&12\end{bmatrix},$ find matrix C such that 5A + 3B + 2C is a null matrix.
AnswerGiven, $5\text{A}+3\text{B}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow5\begin{bmatrix}9&1\\7&8\end{bmatrix}+3\begin{bmatrix}1&5\\7&12\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45&5\\35&40\end{bmatrix}+\begin{bmatrix}3&15\\21&36\end{bmatrix}+2\text{C}\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45+3&5+15\\35+21&40+36\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}48&20\\56&76\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}48&20\\56&76\end{bmatrix}$
$\Rightarrow\text{C}=\frac{1}{2}\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$
View full question & answer→Question 312 Marks
If $A$ is $2\times 3$ matrix and $B$ is a matrix such that $A^T B$ and $BA^T$ both are defined, then what is the order of $B?$
AnswerOrder of $A = 2 \times 3$
Order of $A^T= 3 \times 2$
Let Order of $B = m \times n$
Given: $A^TB$ and $BA^T $ are defined
If ${A^T}_{3\times 2}B_{m\times n}$ _exists, then the number of columns in $A^T $ must be equal to number of rows in $B.$
$\Rightarrow m = 2$
If_$B_{m\times n} {A^T}_{3\times 2}$ _exists, then the number of columns in B must be equal to number of rows in $A^T$
$\Rightarrow n = 3$
$\therefore$ Order of $B = 2 \times 3$
View full question & answer→Question 322 Marks
$\text{If}\ \text{A}'=\begin{bmatrix}-2&3\\1&2\end{bmatrix}, \text{and}\ \text{B}=\begin{bmatrix}-1&0\\1&2\end{bmatrix}\ \text{then find}\ (\text{A} + 2\text{B})'$
AnswerWe know that A = (A')'
$\therefore\ \text{A}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}$
$\therefore \text{A}+2\text{B}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+2\begin{bmatrix}-1&0\\1&2\end{bmatrix}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+\begin{bmatrix}-2&0\\2&4\end{bmatrix}=\begin{bmatrix}-4&1\\5&6\end{bmatrix}$
$\therefore(\text{A} + 2\text{B})'=\begin{bmatrix}-4&5\\1&6\end{bmatrix}$
View full question & answer→Question 332 Marks
A trust fund has Rs. 30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs. 30000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
- Rs. 1800
- Rs. 2000
AnswerIf Rs. x are invested in the first type of bond and Rs. (30000 - x) are invested in the second type of bond,
Then the matrix $\text{A}=\begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}$ represents investment and the matrix $\text{B}=\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}$ represents rate of interest.
- $\begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\frac{5\text{x}}{100}+\frac{7(30000-\text{x})}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}$
$\Rightarrow\frac{5\text{x}+210000-7\text{x}}{100}=1800$
$\Rightarrow210000-2\text{x}=180000$
$\Rightarrow2\text{x}=30000$
$\Rightarrow\text{x}=15000$
Thus,
Amount invested in the first bond = Rs. 15000
Amount invested in the second bond = Rs. (30000 - 15000)
= Rs. 15000
- $ \begin{bmatrix}\text{x}&30000-\text{x}\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\frac{5\text{x}}{100}+\frac{7(30000-\text{x})}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}$
$\Rightarrow\frac{5\text{x}+210000-7\text{x}}{100}=2000$
$\Rightarrow210000-2\text{x}=200000$
$\Rightarrow2\text{x}=10000$
$\Rightarrow\text{x}=5000$
Thus,
Amount invested in the first bond = Rs. 5000
Amount invested in the second bond = Rs. (30000 - 5000)
= Rs. 25000 View full question & answer→Question 342 Marks
Verify that $A^2 = I$, when $\text{A}=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}.$
AnswerWe have, $\text{A}=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$
$\therefore\ \text{A}^2=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}.\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}$ $[\because\ \text{A}^2=\text{A}.\text{A}]$$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\text{I}$
Hence proved.
View full question & answer→Question 352 Marks
If $A$ and $B$ are symmetric matrices, then write the condition for which $AB$ is also symmetric.
AnswerGiven that,
$A$ and $B$ are symmetric matrices, so
$\Rightarrow A^T = A$ and $B^T = B$
Now,
$\big(\text{AB}\big)^\text{T}=\text{B}^\text{T}\times\text{A}^\text{T}$ $\big\{\text{since, (AB)}^\text{T}=\text{B}^\text{T}\text{A}^\text{T}\big\}$
$\big(\text{AB}\big)^\text{T}=\text{BA}\ \dots(\text{i})$ $\big\{\text{since, B}^\text{T}=\text{B},\text{A}^\text{T}=\text{A}\big\}$
For AB to be symmetric matrix
$(AB)^T = AB$
From equation (i) and (ii),
$AB = BA$
So,
For $AB$ to be symmetric matrix we must have $AB = BA.$
View full question & answer→Question 362 Marks
If $\text A = \begin{bmatrix}\frac{2}{3}&1&\frac{5}{3}\\ \frac{1}{3}&\frac{2}{3}&\frac{4}{3}\\ \frac{7}{3}&2&\frac{2}{3}\end{bmatrix} \text{and}\ \text{B} = \begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\ \frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\ \frac{7}{5}&\frac{6}{5}&\frac{2}{3}\end{bmatrix}, $ then compute 3A - 5B.
Answer$3\text{A - 5B}=3\begin{bmatrix}\frac{2}{3}&1&\frac{5}{3}\\ \frac{1}{3}& \frac{2}{3} &\frac{4}{3}\\ \frac{7}{3}&2&\frac{2}{3}\end{bmatrix}-5\begin{bmatrix}\frac{2}{5}&\frac{3}{5}&1\\ \frac{1}{5}&\frac{2}{5}&\frac{4}{5}\\ \frac{7}{5}&\frac{6}{5}&\frac{2}{5}\end{bmatrix}$
$=\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}-\begin{bmatrix}2&3&5\\1&2&4\\7&6&2\end{bmatrix}$
$=\begin{bmatrix}2-2&3-3&5-5\\1-1&2-2&4-4\\7-7&6-6&2-2\end{bmatrix} = \begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$
View full question & answer→Question 372 Marks
Find the values of x and y, if $2\begin{bmatrix}1&3\\0&\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
AnswerGiven,
$2\begin{bmatrix}1&3\\0&\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&6\\0&2\text{x} \end{bmatrix}+\begin{bmatrix}\text{y}&0\\1&2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y}&6+0\\0+1&2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}2+\text{y}&6\\1&2\text{x}+2 \end{bmatrix}=\begin{bmatrix}5&6\\1&8 \end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
$2+\text{y}=5$
$\Rightarrow\text{y}=5-2$
$\Rightarrow\text{y}=3$
And $2\text{x}+2=8$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}=3$
Hence,
$\text{x}=3,\text{y}=3$
View full question & answer→Question 382 Marks
If I is the identity matrix and A is a square matrix such that $A^2 = A$, then what is the value of $(I + A)^2 = 3A$?
AnswerGiven,
$A$ is a square matrix such that $A^2=A$
Now,
$(I+A)^2-3 A=(I+A)(I+A)-3 A$
$\Rightarrow(I+A)^2-3 A=I \times I+I \times A+A \times I+A \times A-3 A \text { \{using distributive property\} }$
$\Rightarrow(I+A)^2-3 A=1+A+A+A^2-3 A\{\text { using } I \times I=I, I A=A I=A\}$
$\Rightarrow(I+A)^2-3 A=1+2 A+A-3 A\left\{\text { since, } A^2=A\right\}$
$\Rightarrow(I+A)^2-3 A=1$
View full question & answer→Question 392 Marks
For what value of $x$, is the matrix $\text{A}=\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$ a skew-symmetric matrix?
AnswerSince, A is a skew symmetric matrix.
$\therefore a^T= -A$
$\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}^{\text{T}}=-\begin{bmatrix}0&1&-2\\-1&0&3\\\text{x}&-3&0 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}0&-1&\text{x}\\1&0&-3\\-2&3&0\\ \end{bmatrix}=\begin{bmatrix}0&-1&2\\1&0&-3\\-\text{x}&3&0 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
⇒ x = 2
Hence, the value of x is 2.
View full question & answer→Question 402 Marks
If $\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6 \end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}+2\\8&\text{a}-8\text{b} \end{bmatrix},$ write the value of a - 2b.
Answer$\begin{bmatrix}\text{a}+4&3\text{b}\\8&-6 \end{bmatrix}=\begin{bmatrix}2\text{a}+2&\text{b}+2\\8&\text{a}-8\text{b} \end{bmatrix}$
Form the above equation,
$\therefore$ a + 4 = 2a + 2
⇒ a = 2
$\therefore$ 3b = b + 2
⇒ 2b = 2
⇒ b = 1
a - 2b
= 2 - 2 × 1
= 0
View full question & answer→Question 412 Marks
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix},$ then find the values of $x, y, z$ and $w$.
Answer$\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y}\end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6\end{bmatrix}$
The corresponding entries of the two equal matrices are equal,
$\Rightarrow x y=8 \ldots(1)$
$w=4 \ldots(2)$
$z+6=0 \ldots(3)$
And $x+y=6 \ldots$ (4)
From equation (2) and equation (3) we get $z=-6$ and $w=4$.
From equation (4) we have,
$x+y=6$
$\Rightarrow x=6-y$,
Subsituting value of $x$ in equation (1) we get,
$\Rightarrow(6-y) y=8$
$\Rightarrow y^2-6 y+8=0$
$\Rightarrow(y-2)(y-4)=0$,
$\Rightarrow y=2,4$
Subsituting the value of $y$ in equation (1) we get,
$\Rightarrow x=4,2$
Therefore, value of $x, y, z, w$ are $2,4,-6,4$ or $4,2,-6,4$.
View full question & answer→Question 422 Marks
Let A and B be matrices of orders 3×2 and 2×4 respectively. Write the order of matrix AB.
AnswerSince, the order of matrix A is 3×2 and order of matrix B is 2×4
So, the order of AB will be the "number of rows of A × number of columns of B" = 3×4
View full question & answer→Question 432 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{a}+\text{b})\text{B}=\text{aB}+\text{bB}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{a}+\text{b})\text{B}=\begin{bmatrix}4&-2\end{bmatrix}\begin{bmatrix}4&0\\1&5\end{bmatrix}$ $[\because$ a = 4, b = -2$]$
$=\begin{bmatrix}8&0\\2&10\end{bmatrix}$
and $\text{aB}+\text{bB}=4\text{B}-2\text{B}$ $=\begin{bmatrix}16&0\\4&20\end{bmatrix}-\begin{bmatrix}8&0\\2&10\end{bmatrix}$$=\begin{bmatrix}8&0\\2&10\end{bmatrix}$
$=(\text{a}+\text{b})\text{B}$
Hence proved.
View full question & answer→Question 442 Marks
Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
AnswerGiven: $\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+2+4&\text{z}-3+5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+6&\text{z}+2\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\therefore\ \text{x}+\text{y}=4\ \dots(1)$
Also,
$\text{y}+6=9$
$\Rightarrow\text{y}=3$
$\text{z}+2=12$
$\Rightarrow\text{z}=10$
Putting the value of y in eq. (1), we get
$\text{x}+3=4$
$\Rightarrow\text{x}=4-3$
$\Rightarrow\text{x}=1$
$\therefore\ \text{x}=1,\text{ y}=3$ and $\text{z}=10$
View full question & answer→Question 452 Marks
If $\text{A}=\begin{bmatrix}2&1&4\\4&1&5 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}3&-1\\2&2\\1&3\end{bmatrix}.$ Write the order of AB and BA.
AnswerOrder of $A=2 \times 3$
Order of $B=3 \times 2$
So,
$\mathrm{A}_{2 \times 3} \times \mathrm{B}_{3 \times 2}$ has order $=2 \times 2$
$\mathrm{B}_{3 \times 2} \times \mathrm{A}_{2 \times 3}$ has order $=3 \times 3$
Hence,
Order of $A B=2 \times 2$
Order of $B A=3 \times 3$
View full question & answer→Question 462 Marks
Construct a 2 × 2 matrix, A = $[\text a_{\text {ij}}]$, whose elements are given by:$\text a_{\text{ij}}=\frac{(\text{i}+\text{j})^2} {2} $
AnswerA = $[\text a_{\text{ ij}}]$ is 2 × 2 matrix where, $\text a_{\text{ij}}=\frac{(\text{i}+\text{j})^2} {2} $$\therefore\ \ \text{a}_{11}=\frac{(1+1)^2}2=\frac{4}{2}=2$, $\text a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2} $
$\text a_{21}=\frac{(2+1)^{2}}{2}=\frac{9}{2} $, $\text a_{23}=\frac{(2+2)^{2}}{2}=\frac{16}{2}=8 $
$\therefore\ \text A= \begin{bmatrix}2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix} $
View full question & answer→Question 472 Marks
If $x\begin{bmatrix}2\\3\end{bmatrix} +y\begin{bmatrix}-1\\1\end{bmatrix} = \begin{bmatrix}10\\5\end{bmatrix}, $find the values of x and y.
AnswerGiven: $x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2x\\3x\end{bmatrix}+\begin{bmatrix}-y\\ y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
Equating corresponding entries, we have
2x - y = 10 ...(i)
3x + y = 5 ...(ii)
Adding eq. (i) and (ii), we have 5x = 15 ⇒ x = 3
Putting x = 3 in eq. (ii), 9 + y = 5 ⇒ y = -4
View full question & answer→Question 482 Marks
If A is a skew-symmetric matrix and n is an even natural number, write whether $A^n$ is symmetric or skew-symmetric or neither of these two.
AnswerIf $A$ is a skew-symmetric matrix, then $A^{\top}=-A$.
$\left(A^n\right)^{\top}=\left(A^T\right)^n[$ For all $n \in N]$
$\Rightarrow\left(A^n\right)^{\top}=(-A)^n\left[\because A^{\top}=-A\right]$
$\Rightarrow\left(A^n\right)^{\top}=(-1)^n A^n$
$\Rightarrow\left(A^n\right)^{\top}=A^n$, if $n$ is even or $-A^n$, if $n$ is odd.
Hence, $A^n$ is a symmetric when n is an even natural number.
View full question & answer→Question 492 Marks
If $\begin{bmatrix}\text{x}+3&\text{z}+4&2\text{y}-7\\4\text{x}+6&\text{a}-1&0\\\text{b}-3&3\text{b}&\text{z}+2\text{c}\end{bmatrix}=\begin{bmatrix}0&6&3\text{y}-2\\2\text{x}&-3&2\text{c}-2\\2\text{b}+4&-21&0\end{bmatrix}$ Obtain the values of a, b, c, x, y and z.
AnswerSince all the corresponding elements of a matrix are equal, x + 3 = 0 ⇒ x = -3 Also, 2y - 7 = 3y - 2 ⇒ 2y - 3y = -2 + 7 ⇒ -y = 5 ⇒ y = -5⇒ z + 4 = 6
⇒ z = 6 - 4 ⇒ z = 2⇒ a - 1 = -3
⇒ a = -3 + 1 ⇒ a = -2 3b = -21 ⇒ b = -7⇒ z + 2c = 0
⇒ 2 = -2c ⇒ c = -1 Thus, x = -3, y = -5, a = -2, b = -7 and c = -1
View full question & answer→Question 502 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$a_{ij} = i + j$
AnswerHere,
$a_{ij} = i + j$
$a_{11} = 1 + 1 = 2, a_{12} = 1 + 2 = 3, a_{13} = 1 + 3 = 4$
$a_{21} = 2 + 1 = 3, a_{22} = 2 + 2 = 4$ and $a_{23} = 2 + 3 = 5$
Required matrix = $\text{A}=\begin{bmatrix}2&3&4\\3&4&5\end{bmatrix}$
View full question & answer→Question 512 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix},$ prove that $A - A^T$ is a skew symmetric matrix.
AnswerGiven: $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
Now,
$(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}-\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})=\begin{bmatrix}2-2 & 3-4 \\4-3 & 5-5 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^\text{T})=\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}\ ...(\text{i})$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}^\text{T}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}0 & 1 \\-1 & 0 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})^{\text{T}}=-\begin{bmatrix}0 & -1 \\1 & 0 \end{bmatrix}$
$\Rightarrow(\text{A}-\text{A}^{\text{T}})=(\text{A}-\text{A}^{\text{T}})^{\text{T}} $ [Using eq.(i)
Thus, $(A - A^T)$ is a skew-symmetric matrix.
View full question & answer→Question 522 Marks
If $\text{A}=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ satisfies $\text{A}^4=\lambda\text{A},$ then write the value of $\lambda.$
Answer$\text{A}^2=\text{A}\cdot\text{A}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&1\\1&1\end{bmatrix}\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+1&1+1\\1+1&1+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}2&2\\2&2\end{bmatrix}$
Now,
$\text{A}^4=\text{A}^2\cdot\text{A}^2$
$\Rightarrow\text{A}^4=\begin{bmatrix}2&2\\2&2\end{bmatrix}\begin{bmatrix}2&2\\2&2\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}4+4&4+4\\4+4&4+4\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}8&8\\8&8\end{bmatrix}$
Also,
$\text{A}^4=\lambda\text{A}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\lambda\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\begin{bmatrix}\lambda&\lambda\\\lambda&\lambda\end{bmatrix}$
$\therefore\ \lambda=8$
View full question & answer→Question 532 Marks
If $\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}$ is identity matrix, then write the value of a.
AnswerHere,
$\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
The corresponding elements of equal matrices are equal.
$\therefore \cos\text{a}=1$
$\Rightarrow\text{a}=0^{\circ}$
View full question & answer→Question 542 Marks
If A and B are square matrices of the same order, explain, why in general:
$(A + B)(A - B) \neq A^2 - B^2.$
AnswerLHS $= (A + B)(A - B)$
$= A(A - B) + B(A - B)$
$= A^2 - AB + BA - B^2$
We know that a matrix does not have commutative property. So,
$AB ≠ BA$
Thus,
$(A + B)(A - B) \neq A^2 - B^2$
View full question & answer→Question 552 Marks
If F(x) =$\begin{bmatrix}\cos x& -\sin x& 0\\ \sin x& \cos x&0\\0&0&1\end{bmatrix}, $ show that F(x) F(y) = F(x + y).
AnswerGiven: F(x) = $\text{F(x)}=\begin{bmatrix} \cos x &-\sin x&0\\ \sin x&\cos x&0\\0&0&1\end{bmatrix}.....(1)$
Changing x to y in eq. (i), $\text{F(y)}=\begin{bmatrix} \cos y &-\sin y&0\\ \sin y&\cos y&0\\0&0&1\end{bmatrix} $
$\text{L.H.S}=\begin{bmatrix} \cos x &-\sin x&0\\ \sin x&\cos x&0\\0&0&1\end{bmatrix}.\begin{bmatrix} \cos y &-\sin y&0\\ \sin y&\cos y&0\\0&0&1\end{bmatrix}$
$=\begin{bmatrix}\cos x \cos y- \sin x \sin y+0&- \cos x \sin y- \sin x \cos y+0&0-0+0\ &\\\sin x \cos y+ \cos x \sin y +0&- \sin x \sin y+ \cos x \cos y+0&0+0+0\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$=\begin{bmatrix} \cos(x+y)&-\sin(x+y)&0\\ \sin(x+y)&\ \cos(x+y)&0\\0&0&1\end{bmatrix}=\text F(x+y)= \text{R.H.S}$ [changing x to to (x + y) in eq. (i)]
View full question & answer→Question 562 Marks
If $A$ is a square matrix such that $A^2=A$, then write the value of $7 A-(I+A)^3$, where $I$ is the identity matrix.
Answer$A^2=A$
$A^3=A^2=A$
$7 A-(I+A)^3$
$=7 A-\left(I^3+A^3+3 A^2 I+3 A I^2\right)$
$=7 A-(I+A+3 A+3 A)$
$=7 A-(I+7 A)$
$=-1$
View full question & answer→Question 572 Marks
If $B$ is a symmetric matrix, write whether the matrix $AB A^T$ is symmetric or skew-symmetric.
AnswerIf B is a skew-symmetric matrix, then $B^T - B$.
$(\text{ABA}^\text{T})^\text{T}=(\text{A}^\text{T})^\text{T}\text{B}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{ABC})^\text{T}=\text{C}^\text{T}\text{B}^\text{T}\text{A}^\text{T}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{AB}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{A}^\text{T})^\text{T}=\text{A}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=-\text{ABA}^\text{T}$ $\big[\because\ \text{B}^\text{T}=\text{B}\big]$
Hence, $ABA^T$ is a skew-symmetric matrix.
View full question & answer→Question 582 Marks
If $\begin{bmatrix}\text{x}+\text{y}\\\text{x}-\text{y} \end{bmatrix}=\begin{bmatrix}2&1\\4&3 \end{bmatrix}\begin{bmatrix}1\\-2\end{bmatrix},$ then write the value of (x, y).
AnswerAfter doing the matrix multiplication we get,
$\begin{bmatrix}\text{x}+\text{y}\\\text{x}-\text{y} \end{bmatrix}=\begin{bmatrix}0\\-2\end{bmatrix}$
As corresponding entries of two equal matrices are equal so,
x + y = 0,
x - y = -2
Solving simultanecus linear equation gives the value of x = -1 and y = 1
or (x, y) = (-1, 1).
View full question & answer→Question 592 Marks
For a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $\text{a}_{\text{ij}}=\frac{\text{i}}{\text{j}}$
, write the value of $a_{12}$. AnswerGiven that a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $\text{a}_{\text{ij}}=\frac{\text{i}}{\text{j}}.$
We need to find the value of $a_{12}$.
Thus, $\text{a}_{12}=\frac{1}{2}.$
View full question & answer→Question 602 Marks
Show by an example that for $\text{A}\neq0,\ \text{B}\neq0,\ \text{AB}=0.$
AnswerLet $\text{A}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\neq0$ and $\text{B}=\begin{bmatrix}-1&1\\0&0\end{bmatrix}\neq0$
$\therefore\ \text{AB}=\begin{bmatrix}0&1\\0&2\end{bmatrix}\begin{bmatrix}-1&1\\0&0\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Hence proved.
View full question & answer→Question 612 Marks
The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are Rs. 8.30, Rs. 3.45 and Rs. 4.50 each respectively. Find the total amount the store will receive from selling all the items.
AnswerMatrix representation of stock of various types of book in the store is given by,
$\text{X}=\begin{bmatrix}\text{Physics}&\text{Chemistry}&\text{Mathematics}\\120&96&60\end{bmatrix}$
Matrix representation of selling price (Rs.) of each book is given by,
$\text{Y}=\begin{bmatrix}8.30&\text{Physics}\\3.45&\text{Chemistry}\\4.50&\text{Mathematics}\end{bmatrix}$
So, total amount recieved by the store from selling all the items is given by,
$\text{XY}=\begin{bmatrix}120&96&60\end{bmatrix}\begin{bmatrix}8.30\\3.45\\4.50\end{bmatrix}$
$\big[(120)(8.30)+(96)(3.45)+(60)(4.50)\big]$
$=\big[996+331.20+270\big]$
$=\big[1597.20\big]$
Required amount = Rs. 1597.20
View full question & answer→Question 622 Marks
If $\text{A}=\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix},$ find the matrix C such that A + B + C is zeor matrix.
AnswerGiven,
$\text{A}=\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix},\text{B}=\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix},$
And
$\text{A}+\text{B}+\text{C}=0$
$\Rightarrow\text{C}=-\text{A}-\text{B}+0$
$\Rightarrow\text{C}=-\text{A}-\text{B}$
$\Rightarrow\text{C}=-\begin{bmatrix}1&-3&2\\2&0&2\end{bmatrix}-\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-1&3&-2\\-2&0&-2\end{bmatrix}-\begin{bmatrix}2&-1&-1\\1&0&-1\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-1-2&3+1&-2+1\\-2-1&0-0&-2+1\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix}$
Hence,
$\text{C}=\begin{bmatrix}-3&4&-1\\-3&0&-1\end{bmatrix}$
View full question & answer→Question 632 Marks
If $\text{A}=\begin{pmatrix}3&5\\7&9 \end{pmatrix}$ is written as A = P + Q, where as A = P + Q, where P is a symmetric matrix and Qis skew symmetric matrix, then write the matrix P.
Answer$\text{A}=\begin{bmatrix}3&5\\7&9 \end{bmatrix}$
P is symmetric matrix. so,
$\text{P}=\frac{1}{2}(\text{A}+\text{A}^{\text{T}})$
Q is skew symmetric matrix. so, $\text{Q}=\frac{1}{2}(\text{A}-\text{A}^{\text{T}})$
$\text{A}^{\text{T}}=\begin{bmatrix}3&7\\5&9 \end{bmatrix}$
$\text{P}=\frac{1}{2}\begin{bmatrix}6&12\\12&18 \end{bmatrix}=\begin{bmatrix}3&6\\6&9 \end{bmatrix}$
View full question & answer→Question 642 Marks
If $\text{A}=\begin{bmatrix}-3&0\\0&-3\end{bmatrix}$, find $A^4$.
AnswerHere,
$\text{A}^2 = \text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-3&0\\0&-3\end{bmatrix}\begin{bmatrix}-3&0\\0&-3\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9+0&0+0\\0+0&0+9\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}9&0\\0&9\end{bmatrix}$
Now,
$\text{A}^4 = \text{A}^2\cdot\text{A}^2$
$\Rightarrow\text{A}^4=\begin{bmatrix}9&0\\0&9\end{bmatrix}\begin{bmatrix}9&0\\0&9\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}81+0&0+0\\0+0&0+81\end{bmatrix}$
$\Rightarrow\text{A}^4=\begin{bmatrix}81&0\\0&81\end{bmatrix}$
View full question & answer→Question 652 Marks
What is the total number of $2 \times 2$ matrices with each entry $0$ or $1$?
AnswerIn a $2 \times 2$ matrix
Total number of elements are 4 and each entry can be writte in 2 ways.
So, Number of ways in which 4 entries can be written
$=4^2$
$=16$
So,
Total number of $2 \times 2$ matrices with each entry $0$ or $1=16$
View full question & answer→Question 662 Marks
If matrix $A = [1 2 3]$, write $AA^T$.
AnswerGiven: $A = [1 2 3]$
$\text{A}^{\text{T}}=\begin{bmatrix}1\\2\\3 \end{bmatrix}$
$\text{AA}^{\text{T}}=\begin{bmatrix}1&2&3 \end{bmatrix}\begin{bmatrix}1\\2\\3 \end{bmatrix}$
$\Rightarrow AA^T= 1 + 4 + 9$
$\Rightarrow AA^T= 14$
View full question & answer→Question 672 Marks
Simplify: $$$\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} + \sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\ \cos\theta&\sin\theta\end{bmatrix}$
AnswerGiven: $\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix} + \sin\theta\begin{bmatrix}\sin\theta&-\cos\theta \\ \cos\theta&\ \sin\theta\end{bmatrix}$
$=\begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{bmatrix} + \begin{bmatrix}\sin^2\theta&-\cos\theta\sin\theta\\ \cos\theta\sin\theta&\ \sin^2\theta\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}$
View full question & answer→Question 682 Marks
Let $\text{A}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ \text{B}=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}.$ Verify that AB = AC though B ≠ C, A ≠ O.
AnswerHere,
$\text{A}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix},\ \text{B}=\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$ and $\text{C}=\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
Now,
$\text{A}\text{B}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}3&1\\5&2\\-2&4\end{bmatrix}$
$\Rightarrow\text{A}\text{B}=\begin{bmatrix}3+5-2&1+2+4\\9+15-6&3+6+12\end{bmatrix}$
$\Rightarrow\text{A}\text{B}=\begin{bmatrix}6&7\\18&21\end{bmatrix}$
$\text{AC}=\begin{bmatrix}1&1&1\\3&3&3\end{bmatrix}\begin{bmatrix}4&2\\-3&5\\5&0\end{bmatrix}$
$\Rightarrow\text{AC}=\begin{bmatrix}4-3+5&2+5+0\\12-9+15&6+15+0\end{bmatrix}$
$\Rightarrow\text{AC}=\begin{bmatrix}6&7\\18&21\end{bmatrix}$
So, AB = AC though B ≠ C, A ≠ O.
View full question & answer→Question 692 Marks
If $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix},$ find $A + A^T$.
AnswerGiven: $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}2&5\\3&7\end{bmatrix}$
Now,
$\text{A}+\text{A}^{\text{T}}=\begin{bmatrix}2&3\\5 &7\end{bmatrix}+\begin{bmatrix}2&5\\3&7\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}2+2&3+5\\5+3&7+7\end{bmatrix}$
$\Rightarrow\text{A}+\text{A}^\text{T}=\begin{bmatrix}4&8\\8&14\end{bmatrix}$
View full question & answer→Question 702 Marks
If $A = [a_{ij}]$ is a skew-symmetric matrix, then write the value of $\sum\limits_\text{i}\sum\limits_\text{j}\text{a}_\text{ij}.$
AnswerGiven: $A = [a_{ij}]$ is a skew-symmetric matrix.
$\Rightarrow a_{ij} = -a_{ij}$ [For all values of i, j]
$\Rightarrow a_{ij} = -a_{ij}$ [For all values of i]
$\Rightarrow a_{ij} = 0$
Now,
$\sum\limits_{\text{i}}\sum\limits_{\text{j}} \text{a}_{\text{ij}} = \text{a}_{11} +\text{a}_{12} +\text{a}_{13 } + ... + \text{a}_{21} +\text{a}_{22} +\text{a}_{23} + ... + \text{a}_{31} +\text{a}_{32} +\text{a}_{33} +...$
$= 0 + \text{a}_{12} +\text{a}_{13} + ... -\text{a}_{12} + 0 + \text{a}_{23} +...-\text{a}_{13} - \text{a}_{23} +0 + ...$
$=0$
Thus,
$\sum\limits_{\text{i}}\sum\limits_{\text{j}} \text{a}_{\text{ij}} =0$
View full question & answer→Question 712 Marks
If $\begin{bmatrix}2&1&3 \end{bmatrix}$ $\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}\begin{pmatrix}1\\0\\-1 \end{pmatrix}=\text{A},$ then write the order of matrix A.
AnswerConsider, $\begin{pmatrix}2&1&3 \end{pmatrix}\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}\begin{pmatrix}1\\0\\-1 \end{pmatrix}=\text{A}$
Order of matrix $\begin{pmatrix}2&1&3 \end{pmatrix}$ is 1 × 3.
Order of matrix $\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}$ is 3 × 3
Order of matrix $\begin{pmatrix}1\\0\\-1 \end{pmatrix}$ is 3 × 1
Therefore, order of $\begin{pmatrix}2&1&3 \end{pmatrix}\begin{pmatrix}-1&0&-1\\-1&1&0\\0&1&1 \end{pmatrix}\begin{pmatrix}1\\0\\-1 \end{pmatrix}$ is 1 × 1.
View full question & answer→Question 722 Marks
Construct a $3 \times 4$ matrix $A = [a_{ij}]$ whose element $a_{ij}$ are given by:
$\text{a}_\text{ij}=\frac{1}{2}|-3\text{i}+\text{j}|$
AnswerHere, $\text{A}=(\text{a}_\text{ij})_{3\times4}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}&\text{a}_{13}&\text{a}_{14}\\\text{a}_{21}&\text{a}_{22}&\text{a}_{23}&\text{a}_{24}\\\text{a}_{31}&\text{a}_{32}&\text{a}_{33}&\text{a}_{34}\end{bmatrix}\ \dots(1)$
$\text{a}_{11}|-3(1)+1|=\frac{1}{2}|-2|=1,$ $\text{a}_{12}=\frac{1}{2}|-3(1)+2|=\frac{1}{2}|-1|=\frac{1}{2},$
$\text{a}_{13}=\frac{1}{2}|-3(1)+3|=\frac{1}{2}|0|=\frac{0}{2}=0,$ $\text{a}_{14}=\frac{1}{2}|-3(1)+4|=\frac{1}{4}|1|=\frac{1}{2}$
$\text{a}_{21}=\frac{1}{2}|-3(2)+1|=\frac{1}{2}|-5|=\frac{5}{2},$ $\text{a}_{22}=\frac{1}{2}|-3(2)+2|=\frac{1}{2}|-4|=2,$
$\text{a}_{23}=\frac{1}{2}|-3(2)+3|=\frac{1}{2}|-3|=\frac{3}{2},$ $\text{a}_{24}=\frac{1}{2}|-3(2)+4|=\frac{1}{2}|-2|=1$
$\text{a}_{31}=\frac{1}{2}|-3(3)+1|=\frac{1}{2}|-8|=4,$ $\text{a}_{32}=\frac{1}{2}|-3(3)+2|=\frac{1}{2}|-7|=\frac{7}{2},$
$\text{a}_{33}=\frac{1}{2}|-3(3)+3|=\frac{1}{2}|-6|=3$ and $\text{a}_{34}=\frac{1}{2}|-3(3)+4|=\frac{1}{2}|-5|=\frac{5}{2}$
So, the required matrix is $\begin{bmatrix}1&\frac{1}{2}&0&\frac{1}{2}\\\frac{5}{2}&2&\frac{3}{2}&1\\4&\frac{7}{2}&3&\frac{5}{2}\end{bmatrix}.$
View full question & answer→Question 732 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are given by:
$\text{a}_\text{ij}=\text{e}^{2\text{ix}}\sin(\text{xj})$
AnswerHere,
$\text{a}_{11}=\text{e}^{2\times1\times\text{x}}\sin(\text{x}\times1)=\text{e}^{2\text{x}}\sin(\text{x}),$ $\text{a}_{12}=\text{e}^{2\times1\times\text{x}}\sin(\text{x}\times2)=\text{e}^{2\text{x}}\sin(2\text{x})$
$\text{a}_{21}=\text{e}^{2\times2\times\text{x}}\sin(\text{x}\times1)=\text{e}^{4\text{x}}\sin(\text{x}),$ $\text{a}_{22}=\text{e}^{2\times2\times\text{x}}\sin(\text{x}\times2)=\text{e}^{4\text{x}}\sin(2\text{x})$
So, the required matrix is $\begin{bmatrix}\text{e}^{2\text{x}}\sin(\text{x})&\text{e}^{2\text{x}}\sin(2\text{x})\\\text{e}^{4\text{x}}\sin(\text{x})&\text{e}^{4\text{x}}\sin(2\text{x})\end{bmatrix}.$
View full question & answer→Question 742 Marks
If a matrix has 24 elements, what are the possible orders it can have? What, if has 13 elements?
AnswerWe know that a matrix of order m × n has mn elements. Therefore, for finding all possible orders of a matrix with 24 elements, we will find all ordered pairs with products of elements as 24.$\therefore$ all possible ordered pairs are
(1, 24), (24, 1), (2, 12), (12, 2), (3, 8), (8, 3), (4, 6), (6, 4)
$\therefore$ possible orders are
1 × 24, 24 × 1, 2 × 12, 12 × 2, 3 × 8, 8 × 3, 4 × 6, 6 × 4
if number of elements = 13, then possible orders are 1 × 13, 13 × 1.
View full question & answer→Question 752 Marks
For what valuse of x and y are the following matrices equal?
$\text{A}=\begin{bmatrix}2\text{x}+1&2\text{y}\\0&\text{y}^2-5\text{y}\end{bmatrix}\text{B}=\begin{bmatrix}\text{x}+3&\text{y}^2+2\\0&-6\end{bmatrix}$
AnswerGiven,
$\text{A}=\begin{bmatrix}2\text{x}+1&2\text{y}\\0&\text{y}^2-5\text{y}\end{bmatrix}\text{B}=\begin{bmatrix}\text{x}+3&\text{y}^2+2\\0&-6\end{bmatrix}$
Since equal matrics has all corresponding elements equal,
So,
$2x + 1 = x + 3 ...(i)$
$2y = y^2 + 2 ...(ii)$
$y^2 - 5y = -6 ...(iii)$
Solving equation (i),
$2x + 1 = x + 3$
$2x - x = 3 - 1$
$x = 2$
Solving equation (ii),
$2y = y^2 + 2$
$y^2 - 2y + 2 = 0$
$D = b^2 - 4ac$
$= (-2)^2 - 4$
$= 4 - 8$
$= -2$
So, There is no real value of y from equation (ii),
Solving equation (iii),
$y^2 - 5y = -6$
$y^2 - 5y + 6 = 0$
$y^2 - 3y - 2y + 6 = 0$
$y(y - 3) - 2(y - 3) = 0$
$(y - 3)(y - 2) = 0$
$y = 3$ or $y = 2$
From solution of equation (i), (ii) and (iii), we can say that A and B can not be equal for any value of y.
View full question & answer→Question 762 Marks
If A and B are square matrices of the same order such that $AB = BA$, then show that $(A + B)^2 = A^2 + 2AB + B^2$.
AnswerGiven,
$A$ and $B$ two square matrices of same order such that $A B=B A$
To prove: $(A+B)^2=A^2+2 A B+B^2$
Now, solving LHS gives,
$(A+B)^2=(A+B)(A+B)$
$=A(A+B)+B(A+B)$[ by dist, of matrix multiplication over addition]
$=A^2+A B+B A+B^2$[ by dist, of matrix multiplication over addition ]
$=A^2+2 A B+B^2[A s, A B=B A]$
$=R H S$
Hence proved.
View full question & answer→Question 772 Marks
If matrix $\text{A}=\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$ and $A^2= pA$, then write the value of p.
AnswerGiven: $\text{A}=\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$
$\text{A}^{2}=\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$
$=\begin{bmatrix}4+4&-4-4\\-4-4&4+4 \end{bmatrix}$
$=\begin{bmatrix}8&-8\\-8&8 \end{bmatrix}$
$=4\begin{bmatrix}2&-2\\-2&2 \end{bmatrix}$
$=4\text{A}$
Hence, p = 4.
View full question & answer→Question 782 Marks
The sales figure of two car dealers during January 2013 showed that dealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7 deluxe, 2 premium and 3 standard cars. Total sales over the 2 month period of January-February revealed that dealer A sold 8 deluxe 7 premium and 6 standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premium and 7 standard cars. Write 2 × 3 matrices summarizing sales data for January and 2-month period for each dealer.
AnswerAccording to the data, dealer A sold 5 deluxe cars, 3 premium cars and 4 standard cars in January.
Also, dealer B sold 7 deluxe cars, 2 premium cars and 3 standard cars in January.
The above information can be given by,
Deluxe Premium Standard $\begin{matrix}\text{Dealer A} \\\text{Dealer B} \end{matrix}\begin{bmatrix}5&3&4\\7&2&3\end{bmatrix}$
Total sales over the period of January-February reveal that dealer A sold 8 deluxe cars,7 premium cars and 6 standard cars, while dealer B sold 10 deluxe cars, 5 premium cars and 7 standard cars.
This information can be given by,
Deluxe Premium Standard $\begin{matrix}\text{Dealer A} \\\text{Dealer B} \end{matrix}\begin{bmatrix}8&7&6\\10&5&7\end{bmatrix}$
View full question & answer→Question 792 Marks
If A and B are square matrices of the same order, explain, why in general:
$(A − B)^2 \neq A^2 − 2AB + B^2$
Answer$(A-B)^2-(A-B)(A-B)$
$=A(A-B)-B(A-B)\{\text { using distributive property }\}$
$=A \times A-A B-B A+B \times B$
$=A^2-A B-B A+B^2$
$\neq A^2-2 A B+B^2$
Since, in general matrix multiplication is not commutative $(A B \neq B A)$,
So, $(A-B)^2 \neq A^2-2 A B+B^2$
View full question & answer→Question 802 Marks
Compute the products AB and BA whichever exists the following cases:
$\text{A}=\begin{bmatrix}1&-2\\2&3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$
Answer$\text{AB}=\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\2&3&1\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}1-4&2-6&3-2\\2+6&4+9&6+3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-3&-4&1\\8&13&9\end{bmatrix}$
Since the number of columns in B is greater then the number of rows in A, BA does not exists.
View full question & answer→Question 812 Marks
Find a matrix X such that 2A + B + X = 0, where.
$\text{A}=\begin{bmatrix}-1&2\\3&4\end{bmatrix},\text{B}=\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
AnswerGiven: $\text{A}=\begin{bmatrix}-1&2\\3&4\end{bmatrix},\text{B}=\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
and
$2\text{A}+\text{B}+\text{x}=0$
$\Rightarrow\text{x}=-2\text{A}-\text{B}$
$\Rightarrow\text{x}=-2\begin{bmatrix}-1&2\\3&4\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
$\Rightarrow\text{x}=\begin{bmatrix}2&-4\\-6&-8\end{bmatrix}-\begin{bmatrix}3&-2\\1&5\end{bmatrix}$
$\Rightarrow\text{x}=\begin{bmatrix}2-3&-4+2\\-6-1&-8-5\end{bmatrix}$
$\Rightarrow\text{x}=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}$
Hence,
$\text{x}=\begin{bmatrix}-1&-2\\-7&-13\end{bmatrix}$
View full question & answer→Question 822 Marks
If $\begin{bmatrix}1&-1\\-1&1\end{bmatrix},$ satisfies the matrix equation $A^2 = kA$, write the value of $k$.
AnswerGiven,
$\text{A}=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$
and
$\text{A}^2=\text{kA}$
$\Rightarrow\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}=\text{k}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+1&-1-1\\-1-1&1+1\end{bmatrix}=\begin{bmatrix}\text{k}&-\text{k}\\-1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2&-2\\-2&2\end{bmatrix}=\begin{bmatrix}\text{k}&-\text{k}\\-\text{k}&\text{k}\end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
k = 2
View full question & answer→Question 832 Marks
If $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{ B}=\text{diag}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-6&3&4\end{pmatrix},$ find.
A - 2B
AnswerGiven, $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{B}\begin{pmatrix}1&1&-4\end{pmatrix}$ and $\text{C}=\text{diag}\begin{pmatrix}-\text{b}&3&4\end{pmatrix}$
$\text{A}-2\text{B}$
$=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix}-2\text{diag}\begin{pmatrix}1&1&-4\end{pmatrix}$
$=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix}-\text{diag}\begin{pmatrix}2&2&-8\end{pmatrix}$
$=\text{diag}\begin{pmatrix}2-2&-5-2&9+8\end{pmatrix}$
$=\text{diag}\begin{pmatrix}0&-7&-17\end{pmatrix}$
So, $\text{A}-2\text{B}=\text{diag}\begin{pmatrix}0&-7&-17\end{pmatrix}$
View full question & answer→Question 842 Marks
If A is an m × n matrix and B is n × p matrix does AB exist? If yes, write its order.
AnswerGiven: Order of A = m × n
Order of B = n × p
Since the number of columns in A are equal to the number of rows in B, i.e. n, AB exists.
Order of AB = Number of rows in A × Number of columns in B = m × p
View full question & answer→Question 852 Marks
Write matrix satisfying $\text{A}+\begin{bmatrix}2&3\\-1&4\end{bmatrix}=\begin{bmatrix}3&-6\\-3&8\end{bmatrix}.$
AnswerGiven: $\text{A}+\begin{bmatrix}2&3\\-1&4\end{bmatrix}=\begin{bmatrix}3&-6\\-3&8\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}3&-6\\-3&8\end{bmatrix}-\begin{bmatrix}2&3\\-1&4\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}3-2&-6-3\\-3+1&8-4\end{bmatrix}$
$\Rightarrow\text{A}=\begin{bmatrix}1&-9\\-2&4\end{bmatrix}$
View full question & answer→Question 862 Marks
If A and B are symmetric matrices of the same order, write whether $AB − BA$ is symmetric or skew-symmetric or neither of the two.
AnswerSince $A$ and $B$ are symmetric matrices, $A^{\top}=A$ and $B^{\top}=B$.
Here,
$(A B-B A)^{\top}=(A B)^{\top}-(B A)^{\top}$
$\Rightarrow(A B-B A)^{\top}=B^{\top} A^{\top}-A^{\top} B^{\top}\left[\because(A B)^{\top}=B^{\top} A^{\top}\right]$
$\Rightarrow(A B-B A)^{\top}=B A-A B\left[\because B^{\top}=B \text { and } A^{\top}=A\right]$
$\Rightarrow(A B-B A)^{\top}=-(A B-B A)$
Therefore, $A B$ - $B A$ is skew - symmetric.
View full question & answer→Question 872 Marks
If $\begin{bmatrix}\text{a+b}&2\\5&\text{b} \end{bmatrix}=\begin{bmatrix}6&5\\2&2 \end{bmatrix}$, then find a.
AnswerThe corresponding elements of two equal matrices are equal.
$\Rightarrow\begin{bmatrix}\text{a+b}&2\\5&\text{b} \end{bmatrix}=\begin{bmatrix}6&5\\2&2 \end{bmatrix}$
⇒ a + b = 6 ...(1)
$\therefore$ b = 2
Putting the value of b in eq.(1)
a + b = 6
⇒ a = 6 - 2
$\therefore$ a = 4
View full question & answer→Question 882 Marks
Write a 2×2 matrix which is both symmetric and skew-symmetric.
AnswerA matrix which is both symmetric and skew-symmetric is a null matrix.
Hence, the required matrix is $\begin{bmatrix}0&0\\0&0 \end{bmatrix}$.
View full question & answer→Question 892 Marks
If $\begin{bmatrix}1&2\\3&4 \end{bmatrix}\begin{bmatrix}3&1\\2&5 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix},$ then write the value of k.
AnswerGiven: $\begin{bmatrix}1&2\\3&4 \end{bmatrix}\begin{bmatrix}3&1\\2&5 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}3+4&1+10\\9+8&3+20 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}7&11\\17&23 \end{bmatrix}=\begin{bmatrix}7&11\\\text{k}&23 \end{bmatrix}$
The corresponding elements of two equal matrices are equal.
$\therefore$ k = 17
View full question & answer→Question 902 Marks
Write a square matrix which is both symmetric as well as skew-symmetric.
AnswerLet $\text{A}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
$\text{A}^{\text{T}{}}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
Since $A^T = A, A$ is a symmetric matrix.
Now,
$-\text{A}=-\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
$\Rightarrow-\text{A}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$
Since $A^T = -A, A$ is a skew-symmetric matrix.
Thus, $\text{A}=\begin{bmatrix}0&0\\0&0 \end{bmatrix}$is an exampal of a matrix that is both symmetric and skew-symmetric.
View full question & answer→Question 912 Marks
If $\begin{bmatrix}\text{a }-\text{b}&2\text{a}+\text{c}\\2\text{a}-\text{b}&3\text{c}+\text{d} \end{bmatrix}=\begin{bmatrix}-1&5\\0&13 \end{bmatrix},$ fine the value of b.
Answer $\begin{bmatrix}\text{a }-\text{b}&2\text{a}+\text{c}\\2\text{a}-\text{b}&3\text{c}+\text{d} \end{bmatrix}=\begin{bmatrix}-1&5\\0&13 \end{bmatrix}$
From the above matrices,
a - b = -1 ...(1)
2a - b = 0 ...(2)
Solving (1) and (2),
a = 1, b = 2
$\therefore$ b = 2
View full question & answer→Question 922 Marks
Find the value of a, b, c and d from the following equations:
$\begin{bmatrix}2\text{a}+\text{b}&\text{a}-2\text{b}\\5\text{c}-\text{d}&4\text{c}+3\text{d}\end{bmatrix}=\begin{bmatrix}4&-3\\11&24\end{bmatrix}$
AnswerAs the given m atrices are equal, therefore their corresponding elements must be equal.
Comparing the corresponding elements, we get
2a + b = 4 ...(i)
a - 2b = -3 ...(ii)
5c - d = 11 ...(iii)
4c + 3d = 24 ...(iv)
Multiplying (i) by 2 and adding to (ii)
5a = 5 ⇒ a = 1
(i) ⇒ b = 4 - 2 × 1 = 2
Multiplying (iii) by 3 and adding to (iv)
19c = 57 ⇒ c = 3
(iii) ⇒ d = 5 × 3 - 11 = 4
Hence, a = 1, b = 2, c = 3, d = 4
View full question & answer→Question 932 Marks
If $\text{A}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix},$ find $A^3$.
AnswerGiven,
$\text{A}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$\text{A}^2=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$=\begin{bmatrix}1+0+0&0+0+0&0+0+0\\0+0+0&0+1+0&0+0+0\\0+0+0&0+0+0&0+1+0\end{bmatrix}$
$=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
$\text{A}^3=\text{A}^2\times\text{A}$
$=\begin{bmatrix}1&0&0\\1&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$=\begin{bmatrix}-1+0+0&0+0+0&0+0+0\\0+0+0&0-1+0&0+0+0\\0+0+0&0+0+0&0+0-1\end{bmatrix}$
$=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$\text{A}^3=\text{A}$
Hence,
$\text{A}^3=\text{A}$
View full question & answer→Question 942 Marks
If $\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+2\begin{bmatrix}\text{x}&0\\1&-2\end{bmatrix}=\text{I},$ where I is 2×2 unit matrix. Find x and y.
AnswerGiven: $\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+2\begin{bmatrix}\text{x}&0\\1&-1\end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+\begin{bmatrix}2\text{x}&0\\2&-4\end{bmatrix}=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+2\text{x}&0+0\\\text{y}+2&5-4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+2\text{x}&0\\\text{y}+2&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\therefore1+2\text{x}=1$ and $\text{y}+2=0$
$\Rightarrow2\text{x}=1-1$ and $\text{y}=-2$
$\Rightarrow2\text{x}=0$
$\Rightarrow\text{x}=\frac{0}{2}=0$
View full question & answer→Question 952 Marks
Construct a $2 \times 3$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$a_{ij} = 2i - j$
AnswerHere,
$a_{11} = 2(1) -1 = 1, a_{12} = 2(1) -2 = 0, a_{13} = 2(1) -3 = -1$
$a_{21} = 2(2) -1 = 3, a_{22} = 2(2) -2 = 2, a_{23} = 2(2) -3 = 1$
Using equation (i)
$\text{A}=\begin{bmatrix}1 &0&-1\\3&2&1\end{bmatrix}$
View full question & answer→Question 962 Marks
In a parliament election, a political party hired a public relations firm to promote its candidates in three ways - telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as
$\text{A}=\begin{bmatrix}140&\text{Telephone}\\200&\text{House calls}\\150&\text{Letters}\end{bmatrix}$
The number of contacts of each type made in two cities X and Yis given in the matrix B as
$\begin{matrix}\text{Telephone}&\text{House calls}&\text{Letters}\end{matrix}\\\text{B}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{matrix}\text{City X}\\\text{City Y}\end{matrix}$
Find the total amount spent by the party in the two cities.
What should one consider before casting his/ her vote - party's promotional activity of their social activities?
AnswerAccording to the question,
Let A be the matrix showing the cost per contact (in paisa).
$\text{A}=\begin{bmatrix}140&\text{Telephone}\\200&\text{House calls}\\150&\text{Letters}\end{bmatrix}$
And, B be a matrix showing the number of contacts of each type made in two cities X and Y.
$\begin{matrix}\text{Telephone}&\text{House calls}&\text{Letters}\end{matrix}\\\text{B}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{matrix}\text{City X}\\\text{City Y}\end{matrix}$
Now, The total amount spent by the party in the two cities will shown by BA.
$\text{BA}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{bmatrix}140\\200\\150\end{bmatrix}$
$=\begin{bmatrix}140000+100000+750000\\420000+200000+1500000\end{bmatrix}$
$=\begin{bmatrix}990000\\2120000\end{bmatrix}$
Hence, the total amount spent by the party in the two cities is
X: Rs. 9900
Y: Rs. 21200
One should consider social activities of a party before casting his/ her vote.
View full question & answer→Question 972 Marks
Compute the products AB and BA whichever exists the following cases:
$[\text{a},\text{b}]\begin{bmatrix}\text{c}\\\text{d} \end{bmatrix}+\big[\text{a},\text{b},\text{c},\text{d}\big]\begin{bmatrix}\text{a}\\\text{b}\\\text{c}\\\text{d}\end{bmatrix}$
Answer$[\text{a},\text{b}]\begin{bmatrix}\text{c}\\\text{d} \end{bmatrix}+\big[\text{a},\text{b},\text{c},\text{d}\big]\begin{bmatrix}\text{a}\\\text{b}\\\text{c}\\\text{d}\end{bmatrix}$
$\Rightarrow\big[\text{ac}+\text{bd}\big]+\big[\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2\big]$
$\big[\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2+\text{ac}+\text{bd}\big]$
View full question & answer→Question 982 Marks
There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommend daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrix. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. What awareness can you create among people about the planned diet from this question?
AnswerAccording to the question, Let X be the matrix showing number of family members in family A and B. $\text{X}=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}$And, Y be a matrix showing the recommend daily amount of calories.
$\text{Y}=\begin{bmatrix}2400\\1900\\1800\end{bmatrix}$And, Z be a matrix showing the recommend daily amount of proteins.
$\text{Z}=\begin{bmatrix}45\\55\\33\end{bmatrix}$Now, the total requirement of calories of the two families will be shown by XY.
$\text{XY}=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{bmatrix}2400\\1900\\1800\end{bmatrix}$ $=\begin{bmatrix}9600+11400+3600\\4800+3800+7200\end{bmatrix}$ $=\begin{bmatrix}24600\\15800\end{bmatrix}$ Also, the total requirement of proteins of the two families will be shown by XZ. $\text{XZ}=\begin{bmatrix}4&6&2\\2&2&4\end{bmatrix}\begin{bmatrix}45\\55\\33\end{bmatrix}$ $=\begin{bmatrix}180+330+66\\90+110+132\end{bmatrix}$ $=\begin{bmatrix}576\\332\end{bmatrix}$Hence, the total requirement of calories and proteins for each of the two families is shown as:
| |
Calories |
Proteins |
| Family A: |
24600 |
576 |
| Family B: |
15800 |
332 |
View full question & answer→Question 992 Marks
If A is a skew-symmetric and n ∈ N such that $(\text{A}^\text{n})^\text{T}=\lambda\text{A}^\text{n},$ write the value of $\lambda.$
AnswerGiven,
A is skew symmetric matrix
$\Rightarrow\text{A}^\text{T} = -\text{A}$
And
$(\text{A}^\text{n})^\text{T}=\lambda\text{A}^\text{n}$
$\Rightarrow(\text{A}^\text{T})^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow(-\text{A})^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow(-1)^\text{n}\text{A}^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow\lambda=(-1)^\text{n}$
View full question & answer→Question 1002 Marks
Find x, y, a and b if $\begin{bmatrix}2\text{x}-3\text{y}&\text{a}-\text{b}&3\\1&\text{x}+4\text{y}&3\text{a}+4\text{b}\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}$
AnswerSince the corresponding elements of two equal matrices are equal,
$\begin{bmatrix}2\text{x}-3\text{y}&\text{a}-\text{b}&3\\1&\text{x}+4\text{y}&3\text{a}+4\text{b}\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}$
⇒ 2x - 3y = 1 ...(1)
⇒ x + 4y = 6
⇒ x = 6 - 4y ...(2)
Putting the value of x in eq. (1), we get
2(6 - 4y) - 3y = 1
⇒ 12 - 8y - 3y = 1
⇒ 12 + 11y = 1
⇒ -11y = -11
$\Rightarrow\text{y}=\frac{-11}{-11}=1$
Putting the value of y in eq. (2), we get
x = 6 - 4(1)
⇒ x = 6 - 4
⇒ x = 2
Now,
a - b = -2
⇒ a = -2 + b ...(3)
3a + 4b = 29 ...(4)
Putting the value of a in eq. (4), we get
3(-2 + b) + 4b = 29
⇒ -6 + 3b + 4b = 29
⇒ -6 + 7b = 29
⇒ 7b = 29 + 6
⇒ 7b = 35
$\Rightarrow\text{b}=\frac{35}{7}=5$
Putting the value of b in eq. (3), we get
a = -2 + 5
⇒ a = 3
$\therefore$ a = 3, b = 5, x = 2 and y = 1
View full question & answer→Question 1012 Marks
Find a matrix X such that 2A + B + X = 0, where.
If $\text{A}=\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix},$ then find the matrix X of order 3 × 2 such that 2A + 3X = 5B.
Answer$2\text{A}+3\text{X}=5\text{B}$
$\Rightarrow2\begin{bmatrix}8&0\\4&-2\\3&6\end{bmatrix}+3\text{X}=5\begin{bmatrix}2&-2\\4&2\\-5&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}16&0\\8&-4\\6&12\end{bmatrix}+3\text{X}=\begin{bmatrix}10&-10\\20&10\\-25&5\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}10&-10\\20&10\\-25&5\end{bmatrix}-\begin{bmatrix}16&0\\8&-4\\6&12\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}10-16&-10-0\\20-8&10+4\\-25-6&5-12\end{bmatrix}$
$\Rightarrow3\text{X}=\begin{bmatrix}-6&-10\\12&14\\-31&-7\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{3}\begin{bmatrix}-6&-10\\12&14\\-31&-7\end{bmatrix}$
$\therefore\ \text{X}=\begin{bmatrix}-2&\frac{-10}{3}\\4&\frac{14}{3}\\\frac{-31}{3}&\frac{-7}{3}\end{bmatrix}$
View full question & answer→Question 1022 Marks
If $\begin{bmatrix}\text{x}&2\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}=2,$ find x.
AnswerGiven: $\begin{bmatrix}\text{x}&2\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}=2$
$\Rightarrow3\text{x}+8=2$
$\Rightarrow3\text{x}=2-8$
$\Rightarrow3\text{x}=-6$
$\Rightarrow\text{x}=\frac{-6}{3}$
$\Rightarrow\text{x}=-2$
View full question & answer→Question 1032 Marks
If $\begin{bmatrix}\text{x}-\text{y}&\text{z}\\2\text{x}-\text{y}&\text{w}\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix},$ find x, y, z, w.
AnswerGiven, $\begin{bmatrix}\text{x}-\text{y}&\text{z}\\2\text{x}-\text{y}&\text{w}\end{bmatrix}=\begin{bmatrix}-1&4\\0&5\end{bmatrix}$ Since corresponding entries of equal matrices are equal, So x - y = -1 ...(i) z = 4 ...(ii) 2x - y = 0 ...(iii) w = 5 ...(iv) Solving equation (i) and (iii)
x = 1 Put x = 1 in equation (i), x - y = -1 1 - y = -1 -y = -1 - 1 -y = -2 y = 2 Equation (ii) and (iv) given the values of z and w respectively, So z = 4, w = 5 Hence, x = 1, y = 2, z = 4, w = 5 View full question & answer→Question 1042 Marks
If A is a skew-symmetric matrix and n is an odd natural number, write whether $A^n$ is symmetric or skew-symmetric or neither of the two.
AnswerGiven,
n is odd natural number and A is kew symmetric matrix.
$\Rightarrow A^T= -A$
Now,
$(A^n)^T= (A^T)^n$
$\Rightarrow (A^n)^T= (-A)^n$ since, $a^T= -A$
$\Rightarrow (A^n)^T = (-1)^n A^n$
$\Rightarrow (A^n)^T = -A^n$ {since, n is odd natural number}
We know that, a square matrix A is skew symmetric if $A^T = -A$
So,
$A^n$ is a skew symmetric matrix.
View full question & answer→Question 1052 Marks
Matrix $\text{A}=\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}$ is given to be symmetric, find values of a and b.
AnswerWe have
$\text{A}=\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}$
$\text{A}'=\begin{bmatrix}0&3&3\text{a}\\2\text{b}&1&3\\-2&3&-1\end{bmatrix}$
We know thet a matrix is symmetric if A = A'.
Thus,
$\begin{bmatrix}0&2\text{b}&-2\\3&1&3\\3\text{a}&3&-1 \end{bmatrix}=\begin{bmatrix}0&3&3\text{a}\\2\text{b}&1&3\\-2&3&-1\end{bmatrix}$
Now,
2b = 3
$\Rightarrow\text{b}=\frac{3}{2}$
Also,
3a = -2
$\Rightarrow\text{a}=\frac{-2}{3}$
Therefore,
$\text{a}=\frac{-2}{3}$ and $\text{b}=\frac{3}{2}$
View full question & answer→Question 1062 Marks
Find the value of $\lambda,$ non-zero scalar, if $\lambda\begin{bmatrix}1&0&2\\3&4&5\end{bmatrix}+2\begin{bmatrix}1&2&3\\-1&-3&2\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
AnswerGiven: $\lambda\begin{bmatrix}1&0&2\\3&4&5\end{bmatrix}+2\begin{bmatrix}1&2&3\\-1&-3&2\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\lambda&0&2\lambda\\3\lambda&4\lambda&5\lambda\end{bmatrix}+\begin{bmatrix}2&4&6\\-2&-6&4\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\lambda+2&0+4&2\lambda+6\\3\lambda-2&4\lambda-6&5\lambda+4\end{bmatrix}=\begin{bmatrix}4&4&10\\4&2&14\end{bmatrix}$
$\Rightarrow\lambda+2=4$
$\Rightarrow\lambda=4-2$
$\therefore\ \lambda=2$
View full question & answer→Question 1072 Marks
Let $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix}$ and $\text{C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}.$ Find each of the following:
$3\text{A}-2\text{B}+3\text{C}$
AnswerGiven, $\text{A}=\begin{bmatrix}2&4\\3&2\end{bmatrix},\text{ B}=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\text{ C}=\begin{bmatrix}-2&5\\3&4\end{bmatrix}$$3\text{A}-2\text{B}+3\text{C}$
$=3\begin{bmatrix}2&4\\3&2\end{bmatrix}-2\begin{bmatrix}1&3\\-2&5 \end{bmatrix}+3\begin{bmatrix}-2&5\\3&4 \end{bmatrix}$
$=\begin{bmatrix}6&12\\9&6\end{bmatrix}-\begin{bmatrix}2&6\\-4&10\end{bmatrix}+\begin{bmatrix}-6&15\\9&12\end{bmatrix}$
$=\begin{bmatrix}6-2-6&12-6+15\\9+4+9&6-10+12\end{bmatrix}$
$=\begin{bmatrix}-2&21\\22&8\end{bmatrix}$
Hence,
$3\text{A}-2\text{B}+3\text{C}=\begin{bmatrix}-2&21\\22&8\end{bmatrix}$
View full question & answer→Question 1082 Marks
Construct a 3 × 4 matrix, whose elements are given by:$\text{a}_{\text{ij}}=\frac{1}{2} \left|-3{\text{i}+\text{j}}\right| $
Answer$\text{Let A}=\left[\text {a}_{\text {ij}}\ \text {be required}\ 3\times4\ {\text {matrix where}}\ {\text a_{\text {ij} }}={\frac{1}{2}}\left|-3{\text{i+j}}\right|\right] $ $\therefore\ \text a_{11}=\frac{1}{2}\left|{-3+1}\right|=\frac{1}{2}(2)=1, $ $\text a_{12}=\frac{1}{2}\left|{-3+2}\right|=\frac{1}{2}(1)=\frac{1}{2} $$\text a_{13}=\frac{1}{2}\left|{-3+3}\right|=\frac{1}{2}(0)=0, $
$\text a_{14}=\frac{1}{2}\left|{-3+4}\right|=\frac{1}{2}(1)=\frac {1}{2} $ $\text a_{21}=\frac{1}{2}\left|{-6+1}\right|=\frac{1}{2}(5)=\frac {5}{2}, $$\text a_{22}=\frac{1}{2}\left|-6+2 \right|=\frac {1}{2}(4)=2 $
$\text a_{23}=\frac{1}{2}\left|-6+3\right|=\frac {1}{2}(3)=\frac {3}{2}, $ $\text a_{24}=\frac{1}{2}\left|-6+4\right|=\frac{1}{2}|-2|=\frac{1}{2}\times2=1 $$\text a_{31}=\frac{1}{2}\left|-9+1\right|=\frac{1}{2}(8)=4, $
$\text a_{32}=\frac{1}{2}\left|-9+2\right|=\frac{1}{2}(7)=\frac{7}{2} $
$\text a_{33}=\frac{1}{2}\left|-9+3\right|=\frac{1}{2}(6)=3,$ $\text a_{34}=\frac{1}{2}\left|-9+4\right|=\frac {1}{2}(5)=\frac{5}{2} $ $\therefore\ \text{A}=\begin{bmatrix}1 & \frac{1}{2}&0&\frac{1}{2}\\ \frac{5}{2}&2 &\frac{3}{2} & 1\\ 4&\frac{7}{2}&3& \frac{5}{2}\end{bmatrix}$
View full question & answer→Question 1092 Marks
construct a 3 × 4 matrix, whose elements are given by:
$\text a_{\text {ij}}=2{\text{i}}-{\text{j}} $
Answer$\text{Let A}=\left[\text a_{\text{ij}}\right]\text {be required}\ 3\times4\ \text{matrix where}\ {\text a_{\text {ij}}} =2{\text{i - j}}$ $\therefore\ \text a_{11}=2-1=1,\ \text a_{12}=2-2=0,$ $ \text a_{13}=2-3=-1,\ \text a_{14}=2-4=-2 $ $\text a_{21}=4-1=3,\ \text{a}_{22}=4-2=2,$ $ \text{a}_{23}=4-3=1,\ \ \text{a}_{24}=4-4=0 $ $\text a_{31}=6-1=5,\ \ \text a_{32}=6-2=4,$ $\text a_{33}=6-3=3,\ \ \text a_{34}=6-4=2 $$\therefore\ \text A = \begin{bmatrix}1 & 0 &-1 & -2 \\3 & 2&1&0\\\ 5&4&3&2 \end{bmatrix} $
View full question & answer→Question 1102 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are given by:
$\text{a}_\text{ij}=\frac{(2\text{i}-\text{j})^2}{2}$
AnswerHere,
$\text{a}_{11}=\frac{[2(1)+1]^2}{2}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{12}=\frac{[2(1)+2]^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
$\text{a}_{21}=\frac{[2(2)+1]^2}{2}=\frac{(4+1)^2}{2}=\frac{(5)^2}{2}=\frac{25}{2},$ $\text{a}_{22}=\frac{[2(2)+2]^2}{2}=\frac{(4+2)^2}{2}=\frac{(6)^2}{2}=\frac{36}{2}=18$
So, the required matrix is $\begin{bmatrix}\frac{9}{2}&8\\\frac{25}{2}&18\end{bmatrix}.$
View full question & answer→Question 1112 Marks
Given an example of two non-zero 2×2 matrices A and B such that AB = 0.
AnswerLet,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq0$
$\text{B}=\begin{bmatrix}0&0\\0&1\end{bmatrix}\neq0$
$\text{AB}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&0\\0&1\end{bmatrix}$
$=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$=0$
So,
$\text{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\text{B}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$
View full question & answer→Question 1122 Marks
A matrix $X$ has $a + b$ rows and $a + 2$ columns while the matrix $Y$ has $b + 1$ rows and $a + 3$ columns. Both matrices $XY$ and $YX$ exist. Find $a$ and $b$. Can you say $XY$ and $YX$ are of the same type$?$ Are they equal.
AnswerHere,
$[X]_{(a+b) \times (a+2)}$
$[Y]_{(b+1) \times (a+3)}$_
Since XY exists, the number of columns in $X$ is equal to the number of rows in$ Y.$
$\Rightarrow a + 2 = b + 1 ...(1)$
Similarly, since YX exists, the number of columns in $Y$ is equal to the number of rows in $X.$
$\Rightarrow a + b = a + 3$
$\Rightarrow b = 3$
Putting the value of b in $(1)$, we get
$a + 2 ≈ 3 + 1$
$\Rightarrow a = 2$
Since the order of the matrices $XY$ and $YX$ is not same, $XY$ and $YX$ are not of the same type and they are unequal.
View full question & answer→Question 1132 Marks
Find x, y and z so that A = B, where.
$\text{A}=\begin{bmatrix}\text{x}-2&3&2\text{z}\\18\text{z}&\text{y}+2&6\text{z}\end{bmatrix},\text{B}=\begin{bmatrix}\text{y}&\text{z}&6\\6\text{y}&\text{x}&2\text{y}\end{bmatrix}$
AnswerSince all the corresponding elements of a matrix are equal,
$\text{A}=\begin{bmatrix}\text{x}-2&3&2\text{z}\\18\text{z}&\text{y}+2&6\text{z}\end{bmatrix},\text{B}=\begin{bmatrix}\text{y}&\text{z}&6\\6\text{y}&\text{x}&2\text{y}\end{bmatrix}$
Here,
x - 2 = y ...(1)
z = 3 ...(2)
18z = 6y ...(3)
Putting the value of z in eq. (3), we get
18(3) = 6y
⇒ 54 = 6y
$\Rightarrow\text{y}=\frac{54}{6}=9$
Putting the value of y in eq. (1), we get
x - 2 = 9
⇒ x = 9 + 2
⇒ x = 11
$\therefore$ x = 11, y = 9 and z = 3
View full question & answer→Question 1142 Marks
In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrix A as.
$\ \ \ \ \ \ \ \ \ \ \ \ \text{Cost per contact}\\\text{A}=\begin{bmatrix}40&\text{Telephone}\\100&\text{House call}\\50&\text{Letter}\end{bmatrix}$
The number of contacts of each type made in two cities X and Y is given in matrix B as
$\text{BA}=\begin{bmatrix}\text{Telephone}&\text{House call}&\text{Letter}\\1000&500&5000\\3000&1000&10000\end{bmatrix} \begin{matrix}\rightarrow\text{X}\\\rightarrow\text{Y}\end{matrix}$
Find the total amount spent by the group in the two cities X and Y.
AnswerThe cost per contact (in paise) is given by,
$\text{A}=\begin{bmatrix}40&\text{Telephone}\\100&\text{House call}\\50&\text{Letter}\end{bmatrix}$
The number of contacts of each type made in the two cities X and Y is given by,
$\text{BA}=\begin{bmatrix}\text{Telephone}&\text{House call}&\text{Letter}\\1000&500&5000\\3000&1000&10000\end{bmatrix} \begin{matrix}\rightarrow\text{X}\\\rightarrow\text{Y}\end{matrix}$
Total amount spent by the group in the two cities X and Y is given by,
$\text{BA}=\begin{bmatrix}1000&500&5000\\3000&1000&10000\end{bmatrix}\begin{bmatrix}40\\100\\50\end{bmatrix}$
$=\begin{bmatrix}40000+50000+250000\\120000+100000+500000\end{bmatrix}$
$=\begin{bmatrix}340000\\720000\end{bmatrix}\begin{matrix}\text{X}\\\text{Y}\end{matrix}$
Thus, Amount spent on X = Rs. 3400
Amount spent on Y = Rs. 7200
View full question & answer→Question 1152 Marks
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves Rs. 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
AnswerLet the monthly incomes of Aryan and Babban be 3x and 4x, respectively.
Suppose their monthly expenditures are 5y and 7y, respectively
Since each saves Rs. 15,000 per month,
Monthly saving of Aryan: 3x - 5y = 15,000
Monthly saving of Babban: 4x - 7y = 15,000
The above system of equations can be written in the matrix form as follows:
$\begin{bmatrix}3&-5\\4&-7\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y} \end{bmatrix}=\begin{bmatrix}15000\\15000 \end{bmatrix}$
Or,
AX = B, where $\text{A}=\begin{bmatrix}3&-5\\4&-7\end{bmatrix},\text{X}=\begin{bmatrix}\text{x}\\\text{y} \end{bmatrix}$ and $\text{B}=\begin{bmatrix}15000\\15000 \end{bmatrix}$
Now,
$|\text{A}|=\begin{vmatrix}3&-5\\4&-7\end{vmatrix}=-21-(-20)=-1$
$\text{Adj A}=\begin{bmatrix}-7&-4\\5&3\end{bmatrix}^\text{T}=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}$
So, $\text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=-1=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}\begin{bmatrix}15000\\15000 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}105000-75000\\60000-45000 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y} \end{bmatrix}=\begin{bmatrix}30000\\15000\end{bmatrix}$
$\Rightarrow\text{x}=30,000$ and $\text{y}=15,000$
Therefore,
Monthly income of Aryan = 3 × Rs. 30,000 = Rs. 90,000
Monthly income of Babban = 4 × Rs. 30,000 = Rs. 1,20,000
From this problem, we are encouraged to understand the power of savings.
We should save certain part of our monthly income for the future.
View full question & answer→Question 1162 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{bA})^{\text{T}}=\text{bA}^{\text{T}}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{bA})^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}\ [\because\ \text{b}=-2]$
$=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}$
and $\text{A}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ $\therefore\ \text{bA}^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}=(\text{bA})^{\text{T}}$ Hence proved.
View full question & answer→Question 1172 Marks
Find the value of x from the following: $\begin{bmatrix}2\text{x}-\text{y}&5\\3&\text{y} \end{bmatrix}=\begin{bmatrix}6&5\\3&-2 \end{bmatrix}$
AnswerThe corresponding elements of two equal matrices are equai.
Given: $\begin{bmatrix}2\text{x}-\text{y}&5\\3&\text{y} \end{bmatrix}=\begin{bmatrix}6&5\\3&-2 \end{bmatrix}$
2x - y = 6 $\dots(1)$
y = - 2
Putting the value of y in eq.(1)
2x - (-2) = 6
⇒ 2x + 2 = 6
⇒ 2x = 6 - 2
⇒ 2x = 4
$\Rightarrow\text{x}=\frac{4}{2}=2$
$\therefore$ x = 2
View full question & answer→Question 1182 Marks
Find the values of $x, y$ and $z$ from the following equations:
$\begin{bmatrix}\text{x+y} & 2 \\5+\text{z} & \text {xy} \end{bmatrix}=\begin{bmatrix}6 & 2 \\5& 8 \end{bmatrix}$
AnswerWe are given that
$\begin{bmatrix}\text{x+y} & 2 \\5+\text{z} & \text {xy} \end{bmatrix}=\begin{bmatrix}6 & 2 \\5& 8 \end{bmatrix}$
By defination of equality of matrices,
$x + y = 6 ...(1)$
$5 + z = 5 ...(2)$
$xy = 8 ...(3)$
Form $(2), z = 0$
Form $y = 6 - x ...(4)$
Putting $y = 6 - x$ in $(3)$, we get.
$x(6 - x) =$ or $6 x - x^2 - 8 = 0$
$\therefore x^2 - 6x + 8 = 0 ? (x - 2) (x - 4) = 0;$
$? x = 2.4$
$\therefore$ from $(4), y = 6 - 2, 6 - 4 = 4, 2$
$\therefore$ we have
$x =2, y = 4, z = 0; x = 4, y = 2, z = 0$
View full question & answer→Question 1192 Marks
Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements $a_{ij}$ are give by:$\frac{(\text{i}+\text{j})^2}{2}$
AnswerHere,
$\text{a}_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2,$ $\text{a}_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$\text{a}_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
So, the required matrix is $\begin{bmatrix}2\frac{9}{2}\\\frac{9}{2}8\end{bmatrix}.$
View full question & answer→Question 1202 Marks
Show that if A and B are square matrices such that $AB = BA$, then $(A + B)^2 = A^2 + 2AB + B^2$.
AnswerSince, A and B are square matrices such that $AB = BA$
$\therefore (A + B)^2 = (A + B).(A + B)$
$= A^2 + AB + BA + B^2$
$= A^2 + AB + AB + B^2$ $[\because$ AB = BA$]$
$= A^2 + 2AB + B^2$
Hence proved.
View full question & answer→Question 1212 Marks
Construct a 2 × 2 matrix, A = $[\text{a}_{\text{ ij}}]$, whose elements are given by:$\text {a}_\text {ij}=\frac {(\text{i}+2 \text{j})^{2}}{2} $
Answer$\text A=\left[\text {a}_{\text {ij}} \right]\text{is}\ 2\times2\ {\text {matrix where}}\ \text {a}_{\text {ij}}=\frac{(\text{i}+2 \text{j})^{2}}{2}$$\therefore\ \text{a}_{11}=\frac{(1+2)^2}{2}=\frac{9}{2}$, $\text{a}_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}$
$\text{a}_{21}=\frac{(2+2)^2}{2}=\frac{16}{2}=8,$ $\text{a}_{22}=\frac{(2+4)^2}{2}=\frac{36}{2}=18 $
$\therefore\ \ \text{A}=\begin{bmatrix}\frac{9}{2}& \frac{25}{2} \\8 & 18 \end{bmatrix} $
View full question & answer→Question 1222 Marks
If $2\begin{bmatrix}3&4\\5&\text{x} \end{bmatrix}+\begin{bmatrix}1&\text{y}\\0&1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix},$ find x - y.
Answer$2\begin{bmatrix}3&4\\5&\text{x} \end{bmatrix}+\begin{bmatrix}1&\text{y}\\0&1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}6+1&8+\text{y}\\10+0&2\text{x}+1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix}$
$\Rightarrow\begin{bmatrix}7&8+\text{y}\\10&2\text{x}+1 \end{bmatrix}=\begin{bmatrix}7&0\\10&5 \end{bmatrix}$
⇒ 8 + y = 0 and 2x + 1 = 5
⇒ y = - 8 and 2x = 4
⇒ y = - 8 and x = 2
Hence, x - y = 2 - (- 8) = 10.
View full question & answer→Question 1232 Marks
Let A and B be square matrices of the same order. Does $(A + B)^2 = A^2 + 2AB + B^2$ hold? If not, why?
Answer$LHS = (A + B)^2$
$= (A + B)(A + B)$
$= A(A + B) + B(A + B)$
$= A^2 + AB + BA + B^2$
We know that a matrix does not have commutative property. So,
$AB ≠ BA$
Thus,
$(A + B)^2 \neq A^2 + 2AB + B^2$
View full question & answer→Question 1242 Marks
If possible, find the sum of the matrices A and B, where $\text{A}=\begin{bmatrix}\sqrt{3}&1\\2&3\end{bmatrix}$ and $\text{B}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}$
AnswerWe have, $\text{A}=\begin{bmatrix}\sqrt{3}&1\\2&3\end{bmatrix}_{2\times2}$ and $\text{B}=\begin{bmatrix}\text{x}&\text{y}&\text{z}\\\text{a}&\text{b}&\text{c}\end{bmatrix}_{2\times3}$Here, A and B are of different Orders. Also, we know that the addition of two matrices A and B is possible only if order of both the matrices A and B should be same.
Hence, the sum of matrices A and B is not possible.
View full question & answer→Question 1252 Marks
If $A = [a_{ij}]$ is a square matrix such that $a_{ij} = i^2 - j^2$, then write whether A is symmetric or skew-symmetric.
AnswerHere,
$\text{a}_{\text{ij}}=\text{i}^2-\text{j}^2,1\leq\text{i}\leq2$ and $1\leq\text{j}\leq2$
$\therefore\ \text{a}_{11}=1^2-1^2=1-1=0,$ $\text{a}_{12}=1^2-2^2=1-4=-3$
$\text{a}_{21}=2^2-1^2=4-1=3$ and $\text{a}_{22}=2^2-2^2=4-4=0$
$\therefore\ \text{A}=\begin{bmatrix}\text{a}_{11}&\text{a}_{12}\\\text{a}_{21}&\text{a}_{22}\end{bmatrix}=\begin{bmatrix}0&-3\\3&0\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}0&3\\-3&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\begin{bmatrix}0&-3\\3&0\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=-\text{A}$
Since $A^T = -A, A$ is skew-symmetric.
View full question & answer→Question 1262 Marks
If $\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6 \end{bmatrix},$ write the value of (x + y + z).
Answer$\begin{bmatrix}\text{xy}&4\\\text{z}+6&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}8&\text{w}\\0&6 \end{bmatrix}$
Corresponding elements of equal matrices are equal.
$\therefore$ z + 6 = 0 and x + y = 6
⇒ z = -6 and x + y = 6
Therefore, x + y + z = 6 - 6 = 0.
View full question & answer→Question 1272 Marks
Give example of matrices:
A and B such that AB = 0 but BA ≠ 0
AnswerLet $\text{A}=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$ \therefore\ \text{AB}=0$
and $\text{BA}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0+0&1+0\\0+0&0+0\end{bmatrix}$
$\Rightarrow\text{BA}=\begin{bmatrix}0&1\\0&0\end{bmatrix}$
Thus, AB = 0 but BA ≠ 0
View full question & answer→Question 1282 Marks
If $B$ is a skew-symmetric matrix, write whether the matrix $AB A^T$ is symmetric or skew-symmetric.
AnswerIf $B$ is a skew-symmetric matrix, then $B^T = -B$.
$(\text{ABA}^\text{T})^\text{T}=(\text{A}^\text{T})^\text{T}\text{B}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{ABC})^\text{T}=\text{C}^\text{T}\text{B}^\text{T}\text{A}^\text{T}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{AB}^\text{T}\text{A}^\text{T}$ $\big[\because\ (\text{A}^\text{T})^\text{T}=\text{A}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=\text{A}(-\text{B})\text{A}^\text{T}$ $\big[\because\ \text{B}^\text{T}=-\text{B}\big]$
$\Rightarrow(\text{ABA}^\text{T})^\text{T}=-\text{ABA}^\text{T}$
$\therefore ABA^T$ is a skew-symmetric matrix.
View full question & answer→Question 1292 Marks
If $\begin{bmatrix}2\text{x}+1&5\text{x}\\0&\text{y}^2+1\end{bmatrix}=\begin{bmatrix}\text{x}+3&10\\0&26\end{bmatrix},$ find the value of (x + y).
AnswerAs the given matrices are equal, therefore, their corresponding elements must be equal.
Comparing the corresponding elements, we get
$\begin{bmatrix}2\text{x} + 1 = \text{x} + 3&5\text{x}=10\\0=0&\text{y}^2+1=26\end{bmatrix}$
On simplifying, we get
x = 2 and $\text{y}=\pm5$
Therefore, x + y = 2 + 5 = 7
or x + y = 2 - 5 = -3
Hence, the value of (x + y) is 7, -3.
View full question & answer→Question 1302 Marks
If $\begin{bmatrix}\text{x}+3&4\\\text{y}-4&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}5&4\\3&9 \end{bmatrix},$ find x abd y
AnswerThe corresponding elements of two equal matrices are equai.Given: $\begin{bmatrix}\text{x}+3&4\\\text{y}-4&\text{x}+\text{y} \end{bmatrix}=\begin{bmatrix}5&4\\3&9 \end{bmatrix}$
x + 3 = 5 and y - 4 = 3
⇒ x = 5 - 3 and y = 3 + 4
⇒ x = 2 and y = 7
$\therefore$ x = 2 and y = 7
View full question & answer→Question 1312 Marks
If $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix},$ then verify that $(\text{A}')'=\text{A}.$
AnswerWe have, $\text{A}=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}4&0\\1&3\\2&6\end{bmatrix}$We have to verify that, $(\text{A}')'=\text{A}$
$\therefore\ \text{A}'=\begin{bmatrix}0&4\\-1&3\\2&-4\end{bmatrix}$
and $(\text{A}')'=\begin{bmatrix}0&-1&2\\4&3&-4\end{bmatrix}=\text{A}$ Hence Proved.
View full question & answer→Question 1322 Marks
Find x, y, z and t, if.
$3\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{t}\end{bmatrix}=\begin{bmatrix}\text{x}&6\\-1&2\text{t}\end{bmatrix}+\begin{bmatrix}4&\text{x}+\text{y}\\\text{z}+\text{t}&3\end{bmatrix}$
Answer$3\begin{bmatrix}\text{x}&\text{y}\\\text{z}&\text{t}\end{bmatrix}=\begin{bmatrix}\text{x}&6\\-1&2\text{t}\end{bmatrix}+\begin{bmatrix}4&\text{x}+\text{y}\\\text{z}+\text{t}&3\end{bmatrix}$
$\Rightarrow\begin{bmatrix}3\text{x}&3\text{y}\\3\text{z}&3\text{t}\end{bmatrix}=\begin{bmatrix}\text{x}+4&6+\text{x}+\text{y}\\-1+\text{z}\text+{t}&2\text{t}+3\end{bmatrix}$
$\therefore\ 3\text{x}=\text{x}+4$
$\Rightarrow3\text{x}-\text{x}=4$
$\Rightarrow2\text{x}=4$
$\Rightarrow\text{x}=2$
Also,
$3\text{y}=6+\text{x}+\text{y}$
$\Rightarrow3\text{y}-\text{y}=6+\text{x}$
$\Rightarrow2\text{y}=6+\text{x}\ \dots(1)$
Putting the value of x in eq. (1), we get
$2\text{y}=6+2$
$\Rightarrow2\text{y}=8$
$\Rightarrow\text{y}=4$
Now,
$3\text{t}=2\text{t}+3$
$\Rightarrow3\text{t}-2\text{t}=3$
$\Rightarrow\text{t}=3$
$3\text{z}=-1+\text{z}+\text{t}$
$\Rightarrow3\text{z}-\text{z}=-1+\text{t}$
$\Rightarrow2\text{z}=-1+\text{t}\ \dots(2)$
Putting the value of t in eq. (2), we get
$2\text{z}=-1+3$
$\Rightarrow2\text{z}=2$
$\Rightarrow\text{z}=1$
$\therefore\ \text{x}=2,\text{ y}=4,\text{ z}=1$ and $\text{t}=3 $
View full question & answer→Question 1332 Marks
If $\text{A}=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix},$ find $A^2$.
AnswerHere,
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1+0+0&0+0+0&0+0+0\\0+0+0&0+1+0&0+0+0\\0+0+0&0+0+0&0+0+1\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
View full question & answer→Question 1342 Marks
If $\text{A}=\begin{bmatrix}1\\2\\3\end{bmatrix},$ write $AA^T$.
AnswerIf $\text{A}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2&3\end{bmatrix}$
Now,
$\text{AA}^\text{T}=\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}1&2&3\end{bmatrix}$
$\Rightarrow\ \text{AA}^\text{T}=\begin{bmatrix}1&2&3\\2&4&6\ \\3&6&9\end{bmatrix}$
View full question & answer→Question 1352 Marks
If $\text{A}=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix},$ then verify that $(\text{A}-\text{B})'=\text{A}'-\text{B}'.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}$ and $\text{B}=\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix}$$(\text{A}-\text{B})=\begin{bmatrix}1&2\\4&1\\5&6\end{bmatrix}-\begin{bmatrix}1&2\\6&4\\7&3\end{bmatrix}$
$=\begin{bmatrix}0&0\\-2&-3\\-2&3\end{bmatrix}$
and $(\text{A}-\text{B})'=\begin{bmatrix}0&-2&-2\\0&-3&3\end{bmatrix}$
Also, $\text{A}'-\text{B}'=\begin{bmatrix}1&4&5\\2&1&6\end{bmatrix}-\begin{bmatrix}1&6&7\\2&4&3\end{bmatrix}$
$=\begin{bmatrix}0&-2&-2\\0&-3&3\end{bmatrix}$
$=(\text{A}-\text{B})'$
Hence proved.
View full question & answer→Question 1362 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $\text{a}(\text{C}-\text{A})=\text{aC}-\text{aA}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{C}-\text{A})=\begin{bmatrix}2-1&0-2\\1+1&-2-3\end{bmatrix}=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}$
and $\text{a}(\text{C}-\text{A})=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}\ [\because\ \text{a}=4]$ Also, $\text{aC}-\text{aA}=\begin{bmatrix}8&0\\4&-8\end{bmatrix}-\begin{bmatrix}4&8\\-4&12\end{bmatrix}$$=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}$
$=\text{a}(\text{C}-\text{A})$
Hence proved.
View full question & answer→Question 1372 Marks
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ then prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{yb}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}.$
Answer$\text{PQ}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}$$=\begin{bmatrix}\text{xa}&0&0\\0&\text{yb}&0\\0&0&\text{zc}\end{bmatrix}\ ....(\text{i})$
and $\text{QP}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}&0&0\\0&\text{by}&0\\0&0&\text{cz}\end{bmatrix}\ ....(\text{ii})$
Thus, we see that, PQ = QP [using Eq. (i) and (ii)]
Hence proved.
View full question & answer→Question 1382 Marks
For the matrix $\text{A}=\begin{bmatrix}1&5\\6&7\end{bmatrix}$, verify that
- (A + A') is a symmetric matrix.
- (A - A') is a skew symmentric matrix.
Answer$\text{A}'=\begin{bmatrix}1&6\\5&7\end{bmatrix}$
- $\text{A}+\text{A}'=\begin{bmatrix}1&5\\6&7\end{bmatrix}+\begin{bmatrix}1&6\\5&7\end{bmatrix}=\begin{bmatrix}2&11\\11&14\end{bmatrix}$
$\therefore\ (\text{A}+\text{A})'=\begin{bmatrix}2&11\\11&14\end{bmatrix} = \text{A}+\text{A}'$
Hence, $(\text{A}+\text{A}')$ is a symmentric matrix.
- $\text{A} - \text{A}'=\begin{bmatrix}1&5\\6&7\end{bmatrix}-\begin{bmatrix}1&6\\5&7\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$
$(\text{A} - \text{A}')'=\begin{bmatrix}0&1\\-1&0\end{bmatrix}=-\begin{bmatrix}0&-1\\1&0\end{bmatrix}=-(\text{A} -\text{A}')$
Hence, $(\text{A} - \text{A}')$ is a skew - symmentric matrix. View full question & answer→Question 1392 Marks
To promote making of toilets for women, an organisation tried to generate awarness through (i) house calls, (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below:
- ₹ 50
- ₹ 20
- ₹ 40
The number of attempts made in three villages X, Y and Z are given below:
| |
(i) |
(ii) |
(iii) |
| X |
400 |
300 |
100 |
| Y |
300 |
250 |
75 |
| Z |
500 |
400 |
150 |
Find the total cost incurred by the organisation for three villages separately, using matrices.
AnswerThe cost for each mode per attempt is represented by 3 × 1 matrix:
$\text{A}=\begin{bmatrix}50\\20\\40\end{bmatrix}$
The number of attempts made in the three villages X, Y, and Z are represented by a 3 × 3 matrix:
$\text{B}=\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}$
The total cost incurred by the prganization for the three villages seperately is given by matrix multiplication,
$\text{BA}=\begin{bmatrix}400&300&100\\300&250&75\\500&400&150\end{bmatrix}\begin{bmatrix}50\\20\\40\end{bmatrix}$
$\text{BA}=\begin{bmatrix}400\times50+300\times20+100\times40\\300\times50+250\times20+75\times40\\500\times50+400\times20+150\times40\end{bmatrix}$
$=\begin{bmatrix}30,000\\23,000\\39,000\end{bmatrix}$
Note: The answer given in the book is incorrect.
View full question & answer→Question 1402 Marks
If a matrix has 18 element, what are the possible orders it can have? What, if it has 5 elements?
AnswerWe know that a matrix of order m × n has m n elements. Therefore, for finding all possible orders of a matrix with 18 elements, we will find all ordered pairs with products of elements as 18.$\therefore$ all possible ordered pair are all possible ordered pairs are all possible ordered pair are all possible ordered pairs are all possoble ordered pairs are all possible ordered pairs are(1, 18), (18, 1), (2, 9), (9, 2), (3, 6), (6, 3)
$\therefore $ possible orders are 1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3.
if number of elements = 5, then possible orders are 1 × 5, 5 × 1.
View full question & answer→Question 1412 Marks
Find the values of x and y if.$\begin{bmatrix}\text{x}+10&\text{y}^2+2\text{y}\\0&-4\end{bmatrix}=\begin{bmatrix}3\text{x}+4&3\\0&\text{y}^2-5\text{y}\end{bmatrix}$
AnswerHere,
$x + 10 = 3x + 4 [\because$ All the corresponding elements of the matrix are equal]
$⇒ x - 3x = 4 - 10$
$⇒ -2x = -6$
$\therefore x = 3$
Also,
$y^2 + 2y = 3$
$\Rightarrow y^2 + 2y - 3 = 0$
$\Rightarrow y^2 + 3y - y - 3 = 0$
$\Rightarrow y(y + 3) - 1(y + 3) = 0$
$\Rightarrow (y + 3)(y - 1) = 0$
$\Rightarrow y + 3 = 0$ or $y - 1 = 0$
$\Rightarrow y = -3$ or $y = 1$
Now
$-4 = y^2 - 5y$
$\Rightarrow y^2 - 5y + 4 = 0$
$\Rightarrow y^2 - 4y - y + 4 = 0$
$\Rightarrow y(y - 4) - 1(y - 4) = 0$
$\Rightarrow (y - 4)(y - 1) = 0$
$\Rightarrow y - 4 = 0$ or $y - 1 = 0$
$\Rightarrow y = 4$ or $y = 1$
Since $y^2 + 2y = 3$ and $y^2 - 5y = -4$ must hold good simultaneously, we take the common solution of these two equations.
Thus,
$y = 1, x = 3$ and $y = 1$
View full question & answer→Question 1422 Marks
The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves ₹ 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value?
AnswerLet the monthly incomes of Aryan and Babban be 3x and 4x, respectively.
Suppose their monthly expenditures are 5y and 7y, respectively.
Since each saves Rs. 15,000 per month,
Monthly saving of Aryan: 3x - 5y = 15,000
Monthly saving of Babban: 4x - 7y = 15,000
The above system of equations can be written in the matrix form as follows:
$\begin{bmatrix}3&-5\\4&-7\end{bmatrix}\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}15000\\15000\end{bmatrix}$
or,
AX = B, where $\text{A}=\begin{bmatrix}3&-5\\4&-7\end{bmatrix},\ \text{X}=\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}$ and $\text{B}=\begin{bmatrix}15000\\15000\end{bmatrix}$
Now,
$\big|\text{A}\big|=\begin{vmatrix}3&-5\\4&-7\end{vmatrix}=-21-(-20)=-1$
$\text{Adj}\ \text{A}=\begin{bmatrix}-7&-4\\5&3\end{bmatrix}^\text{T}=\begin{bmatrix}-7&5\\-4&3\end{bmatrix}$
So, $\text{A}^{-1}=\frac{1}{\big|\text{A}\big|}\text{ adj }\text{A}=-1\begin{bmatrix}-7&5\\-4&3\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}$
$\therefore\ \text{X}=\text{A}^{-1}\text{B}$
$ \Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}7&-5\\4&-3\end{bmatrix}\begin{bmatrix}15000\\15000\end{bmatrix}$
$ \Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}105000-75000\\60000-45000\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}\\\text{y}\end{bmatrix}=\begin{bmatrix}30000\\15000\end{bmatrix}$
$\Rightarrow\text{x}=30,000$ and $\text{y}=15,000$
Therefore,
Monthly income of Aryan = 3 × Rs. 30,000 = Rs. 90,000
Monthly income of Babban = 4 × Rs. 30,000 = Rs. 1,20,000
From this problem, we are encouraged to understand the power of savings. We should save certain part of our monthly income for the future.
View full question & answer→Question 1432 Marks
Construct a 3 × 2 matrix whose elements are given by $\text{a}_{\text{ij}}=\text{e}^{\text{i.x}}=\sin\text{jx}.$
AnswerWe have, $\text{A}=[\text{a}_{\text{ij}}]_{3\times2},$ Such that, $\text{a}_{\text{ij}}=\text{e}^{\text{i.x}}=\sin\text{jx};$ where $1\leq\text{i}\leq3$ and $1\leq\text{j}\leq2,\ \text{i, j}\in\text{N}$
| $\therefore\ \text{a}_{11}=\text{e}^\text{x}\sin\text{x}$ |
$\text{a}_{12}=\text{e}^{\text{x}}\sin2\text{x}$ |
| $\text{a}_{21}=\text{e}^{2\text{x}}\sin\text{x}$ |
$\text{a}_{22}=\text{e}^{2\text{x}}\sin2\text{x}$ |
| $\text{a}_{31}=\text{e}^{3\text{x}}\sin\text{x}$ |
$\text{a}_{32}=\text{e}^{3\text{x}}\sin2\text{x}$ |
$\therefore\ \text{A}=\begin{bmatrix}\text{e}^{\text{x}}\sin\text{x}&\text{e}^{\text{x}}\sin2\text{x}\\\text{e}^{2\text{x}}\sin\text{x}&\text{e}^{2\text{x}}\sin2\text{x}\\\text{e}^{3\text{x}}\sin\text{x}&\text{e}^{3\text{x}}\sin2\text{x}\end{bmatrix}$ View full question & answer→Question 1442 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{A}^{\text{T}})^{\text{T}}=\text{A}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2
$\text{A}^{\text{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$
Now, $(\text{A}^{\text{T}})^{\text{T}}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}^{\text{T}}$
$=\text{A}$
Hence proved.
View full question & answer→Question 1452 Marks
If $\begin{bmatrix}2\text{x+y}&3\text{y}\\0&4 \end{bmatrix}=\begin{bmatrix}6&0\\6&4\end{bmatrix}$, then find x.
AnswerGiven,
$\begin{bmatrix}2\text{x+y}&3\text{y}\\0&4 \end{bmatrix}=\begin{bmatrix}6&0\\6&4\end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
3y = 0
⇒ y = 0
And 2x + y = 6
⇒ 2x + 0 = 6
⇒ 2x = 6
⇒ x = 3
So,
x = 3, y = 0
View full question & answer→Question 1462 Marks
Show that:
$\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\neq\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}$
Answer$\text{L.H.S}=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}$$=\begin{bmatrix}5(2) + (-1)3&5(1) + (-1)4\\6(2) + 7(3)&6(1) + 7(4)\end{bmatrix}$$ = \begin{bmatrix}7&1\\33&34\end{bmatrix} $
$\text{R.H.S} = \begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix} $$= \begin{bmatrix}2(5) + 1(6)&2(-1) + 1(7)\\3(5) + 4(6)&3(-1) + 4(7)\end{bmatrix} $$= \begin{bmatrix}16&5\\39&25\end{bmatrix} $
$\therefore \text{L.H.S.} \neq \text{R.H.S.}$
View full question & answer→Question 1472 Marks
Three shopkeepers A, B and C go to a store to buy stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. Cpurchases 11 dozen notebooks, 13 dozen pens and 8 dozen pencils. A notebook costs 40 paise, a pen costs Rs. 1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual's bill.
Answer
| Shopkeepers |
Notebooks In dozen |
Pens In dozen |
Pencils In dozen |
| A |
12 |
5 |
6 |
| B |
10 |
6 |
7 |
| C |
11 |
3 |
8 |
Here,
Cost of notebooks per dozen = (12 × 40) paise = Rs. 4.80
Cost of pens per dozen = Rs. (12 × 1.25) = Rs. 15
Cost of Pencils per dozen = (12 × 35) paise = Rs. 4.20
$\therefore\ \begin{bmatrix}12&5&6\\10&6&7\\11&13&8\end{bmatrix}\begin{bmatrix}4.80\\15\\4.20\end{bmatrix}=\begin{bmatrix}12\times4.80+5\times15+6\times4.20\\10\times4.80+6\times15+7\times4.20\\11\times4.80+13\times15+8\times4.20\end{bmatrix}$
$=\begin{bmatrix}57.60+75+25.20\\48+90+29.40\\52.80+195+33.60\end{bmatrix}$
$=\begin{bmatrix}157.80\\167.40\\281.40\end{bmatrix}$
Thus, the bills of A, B and C are Rs. 157.80, Rs. 167.40 and 281.40, respectively. View full question & answer→Question 1482 Marks
Find the value of y, if $\begin{bmatrix}\text{x}-\text{y}&2\\\text{x}&5 \end{bmatrix}=\begin{bmatrix}2&2\\3&5 \end{bmatrix}$
AnswerGiven,
$\begin{bmatrix}\text{x}-\text{y}&2\\\text{x}&5 \end{bmatrix}=\begin{bmatrix}2&2\\3&5 \end{bmatrix}$
Since, corresponding entries of equal matrices are equal, so
x = 3
and x - y = 2
⇒ 3 - y = 2
⇒ -y = 2 - 3
⇒ -y = -1
⇒ y = 1
View full question & answer→Question 1492 Marks
Compute the indicated products:
$\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\-3&2&-1\end{bmatrix}$
Answer$\begin{bmatrix}1&-2\\2&3\end{bmatrix}\begin{bmatrix}1&2&3\\-3&2&-1\end{bmatrix}$
$=\begin{bmatrix}(1)(1)+(-2)(-3)&(1)(2)+(-2)(2)&(1)(3)+(-2)(1)\$2)(1)+(3)(-3)&(2)(2)+(3)(2)&(2)(3)+(3)(-1)\end{bmatrix}$
$=\begin{bmatrix}1+6&2-4&3+2\\2-9&4+6&6-3\end{bmatrix}$
$=\begin{bmatrix}7&-2&5\\-7&10&3\end{bmatrix}$
View full question & answer→Question 1502 Marks
If A is a symmetric matrix and $n ∈ N$, write whether $A^n$ is symmetric or skew-symmetric or neither of these two.
AnswerIf A is a symmetric matrix, then $A^T = A$.
Now,
$(A^n)^T = (A^T)^n [for all n \in N]$
$\Rightarrow (A^n)^T = (A)^n [\because A^T = A]$
Hence, $A^n$ is a symmetric matrix.
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