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
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 82 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:
$3\text{A}-2\text{B}+3\text{C}$
Answer Given, $\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 92 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 102 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 112 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 122 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}.$
Answer Equating 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 132 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 142 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 152 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 162 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 172 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 182 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 192 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 202 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 212 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 222 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 232 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 242 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 252 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 262 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 272 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 282 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 292 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 302 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 312 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 322 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 332 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 342 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 352 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 362 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 $$\text a_{31}=\frac{1}{2}\left|-9+1\right|=\frac{1}{2}(8)=4, $
$\text a_{32}=\frac{1}{2}\left|-9+2\right|=\frac{1}{2}(7)=\frac{7}{2} $
$\text a_{33}=\frac{1}{2}\left|-9+3\right|=\frac{1}{2}(6)=3,$ $\text a_{34}=\frac{1}{2}\left|-9+4\right|=\frac {1}{2}(5)=\frac{5}{2} $ $\therefore\ \text{A}=\begin{bmatrix}1 & \frac{1}{2}&0&\frac{1}{2}\\ \frac{5}{2}&2 &\frac{3}{2} & 1\\ 4&\frac{7}{2}&3& \frac{5}{2}\end{bmatrix}$
View full question & answer→Question 372 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $(\text{bA})^{\text{T}}=\text{bA}^{\text{T}}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{bA})^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}\ [\because\ \text{b}=-2]$
$=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}$
and $\text{A}^{\text{T}}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ $\therefore\ \text{bA}^{\text{T}}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}=(\text{bA})^{\text{T}}$ Hence proved.
View full question & answer→Question 382 Marks
Show that if $A$ and $B$ are square matrices such that $AB = BA,$ then $(A + B)^2 = A^2 + 2AB + B^2.$
AnswerSince, $A$ and $B$ are square matrices such that $AB = BA$
$\therefore (A + B)^2 = (A + B).(A + B)$
$= A^2 + AB + BA + B^2$
$= A^2 + AB + AB + B^2 [\because AB = BA]$
$= A^2 + 2AB + B^2$
Hence proved.
View full question & answer→Question 392 Marks
Construct a 2 × 2 matrix, A = $[\text{a}_{\text{ ij}}]$, whose elements are given by:$\text {a}_\text {ij}=\frac {(\text{i}+2 \text{j})^{2}}{2} $
Answer$\text A=\left[\text {a}_{\text {ij}} \right]\text{is}\ 2\times2\ {\text {matrix where}}\ \text {a}_{\text {ij}}=\frac{(\text{i}+2 \text{j})^{2}}{2}$$\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 402 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 412 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 422 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 432 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$]$
$\Rightarrow x - 3x = 4 - 10$
$\Rightarrow -2x = -6$
$\therefore x = 3$
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 442 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 452 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 462 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}$$=\begin{bmatrix}\text{xa}&0&0\\0&\text{yb}&0\\0&0&\text{zc}\end{bmatrix}\ ....(\text{i})$
and $\text{QP}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix}\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$
$=\begin{bmatrix}\text{ax}&0&0\\0&\text{by}&0\\0&0&\text{cz}\end{bmatrix}\ ....(\text{ii})$
Thus, we see that, PQ = QP [using Eq. (i) and (ii)]
Hence proved.
View full question & answer→Question 472 Marks
Let $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ a = 4, b = -2, then show that $\text{a}(\text{C}-\text{A})=\text{aC}-\text{aA}.$
AnswerWe have, $\text{A}=\begin{bmatrix}1&2\\-1&3\end{bmatrix},\ \text{B}=\begin{bmatrix}4&0\\1&5\end{bmatrix},$ $\text{C}=\begin{bmatrix}2&0\\1&-2\end{bmatrix},$ and a = 4, b = -2$(\text{C}-\text{A})=\begin{bmatrix}2-1&0-2\\1+1&-2-3\end{bmatrix}=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}$
and $\text{a}(\text{C}-\text{A})=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}\ [\because\ \text{a}=4]$ Also, $\text{aC}-\text{aA}=\begin{bmatrix}8&0\\4&-8\end{bmatrix}-\begin{bmatrix}4&8\\-4&12\end{bmatrix}$$=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}$
$=\text{a}(\text{C}-\text{A})$
Hence proved.
View full question & answer→Question 482 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 492 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 502 Marks
$A$ 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 $A_{3\times 4 }\times B_{4\times 3 }= 3\times 3$
So,
Order of $AB = 3\times 3$
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