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→