2 Marks

Question 12 Marks

If A is a skew-symmetric matrix of order 3, then prove that det A = 0.

Answer

Any 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

$\therefore \text{A}^{\text{T}} = \text{-A}$

$\therefore |\text{A}^{\text{T}}| = \text{|-A|} = (-1)^{3}. \text{|A|}$

$\Rightarrow \text{|A|} = \text{|-A|}$

$\Rightarrow 2\text{|A}| = 0 \text{ }\text{ or } \text{ }\text{|A|} = 0.$

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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

$|\text{2AB}| = 2^{3} \times \text{|A|} \times \text{|B|}$
$= 8 \times (-1) \times 3 = -24$

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Question 32 Marks

Show that all the diagonal elements of a skew symmetric matrix are zero.

Answer

Let $\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.

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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.

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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 $\text{C}=\begin{bmatrix}2 & 0 & -2 \\7 & 1 & 6\end{bmatrix}.$

Answer

Given: $2\text{A}-3\text{B}+5\text{C}=0$
$\Rightarrow2\text{A}-3\begin{bmatrix}-2 & 2 & 0 \\3 & 1 & 4\end{bmatrix}+5\begin{bmatrix}2 & 0 & -2 \\7 & 1 & 6\end{bmatrix}0$
$\Rightarrow2\text{A}-\begin{bmatrix}-6 & 6 & 0 \\9 & 3 & 12\end{bmatrix}+\begin{bmatrix}10 & 0 & -10 \\35 & 5 & 30\end{bmatrix}=0$
$\Rightarrow2\text{A}+\begin{bmatrix}10+6 & 0-6 &-10-0 \\35 - 9 & 5-3 &30-12\end{bmatrix}=0$
$\Rightarrow2\text{A}+\begin{bmatrix}16 & -6 & -10 \\26 & 2 & 18\end{bmatrix}=0$
$\Rightarrow\text{A}=\begin{bmatrix}-8 & 3 & 5 \\-13 & -1 & -9\end{bmatrix}$

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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, $\text{A}^2-5\text{A}=\begin{bmatrix}-5 & -1 & -3 \\-1 & -7 & -10\\-5 & 4 & -2 \end{bmatrix}$

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Question 72 Marks

Find the value of x from the following: $\begin{bmatrix}2\text{x}-\text{y}&5\\3&\text{y} \end{bmatrix}=\begin{bmatrix}6&5\\3&-2 \end{bmatrix}$

Answer

The 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

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Question 82 Marks

Construct a 2 × 2 matrix A = [a_ij] whose elements a_ij are give by:

$\frac{(\text{i}+\text{j})^2}{2}$

Answer

Here,
$\text{a}_{11}=\frac{(1+1)^2}{2}=\frac{(2)^2}{2}=\frac{4}{2}=2,$ $\text{a}_{12}=\frac{(1+2)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2}$
$\text{a}_{21}=\frac{(2+1)^2}{2}=\frac{(3)^2}{2}=\frac{9}{2},$ $\text{a}_{22}=\frac{(2+2)^2}{2}=\frac{(4)^2}{2}=\frac{16}{2}=8$
So, the required matrix is $\begin{bmatrix}2\frac{9}{2}\\\frac{9}{2}8\end{bmatrix}.$

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Question 92 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

$[\text{a},\text{b}]\begin{bmatrix}\text{c}\\\text{d} \end{bmatrix}+\big[\text{a},\text{b},\text{c},\text{d}\big]\begin{bmatrix}\text{a}\\\text{b}\\\text{c}\\\text{d}\end{bmatrix}$
$\Rightarrow\big[\text{ac}+\text{bd}\big]+\big[\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2\big]$
$\big[\text{a}^2+\text{b}^2+\text{c}^2+\text{d}^2+\text{ac}+\text{bd}\big]$

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Question 102 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?

Answer

According 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

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Question 112 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

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Question 122 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.

Answer

Given: $\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+2\begin{bmatrix}\text{x}&0\\1&-1\end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}1&0\\\text{y}&5\end{bmatrix}+\begin{bmatrix}2\text{x}&0\\2&-4\end{bmatrix}=\begin{bmatrix}1 &0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+2\text{x}&0+0\\\text{y}+2&5-4\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}1+2\text{x}&0\\\text{y}+2&1\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\therefore1+2\text{x}=1$ and $\text{y}+2=0$
$\Rightarrow2\text{x}=1-1$ and $\text{y}=-2$
$\Rightarrow2\text{x}=0$
$\Rightarrow\text{x}=\frac{0}{2}=0$

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Question 132 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 $\lambda.$

Answer

Given,
A is skew symmetric matrix
$\Rightarrow\text{A}^\text{T} = -\text{A}$
And
$(\text{A}^\text{n})^\text{T}=\lambda\text{A}^\text{n}$
$\Rightarrow(\text{A}^\text{T})^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow(-\text{A})^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow(-1)^\text{n}\text{A}^\text{n}=\lambda\text{A}^\text{n}$
$\Rightarrow\lambda=(-1)^\text{n}$

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Question 142 Marks

Find x, y, a and b if $\begin{bmatrix}2\text{x}-3\text{y}&\text{a}-\text{b}&3\\1&\text{x}+4\text{y}&3\text{a}+4\text{b}\end{bmatrix}=\begin{bmatrix}1&-2&3\\1&6&29\end{bmatrix}$

Answer

Since 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

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Question 152 Marks

If $\begin{bmatrix}\text{x}&2\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}=2,$ find x.

Answer

Given: $\begin{bmatrix}\text{x}&2\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}=2$
$\Rightarrow3\text{x}+8=2$
$\Rightarrow3\text{x}=2-8$
$\Rightarrow3\text{x}=-6$
$\Rightarrow\text{x}=\frac{-6}{3}$
$\Rightarrow\text{x}=-2$

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Question 162 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

Answer

Given, $\text{A}=\text{diag}\begin{pmatrix}2&-5&9\end{pmatrix},\text{B}\begin{pmatrix}1&1&-4\end{pmatrix}$ 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, $\text{A}-2\text{B}=\text{diag}\begin{pmatrix}0&-7&-17\end{pmatrix}$

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Question 172 Marks

If A is a skew-symmetric matrix and n is an odd natural number, write whether Aⁿ is symmetric or skew-symmetric or neither of the two.

Answer

Given,
n is odd natural number and A is kew symmetric matrix.
⇒ A^T= -A
Now,
(Aⁿ)^T= (A^T)ⁿ
⇒ (Aⁿ)^T= (-A)ⁿ{since, a^T= -A}
⇒ (Aⁿ)^T = (-1)ⁿ Aⁿ
⇒ (Aⁿ)^T = -Aⁿ {since, n is odd natural number}
We know that, a square matrix A is skew symmetric if A^T = -A
So,
Aⁿ is a skew symmetric matrix.

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Question 182 Marks

f A is a matrix of order 3×4 and B is a matrix of order 4×3, find the order of the matrix of AB.

Answer

Order of A = 3×4
Order of B = 4×3
Order of A_3×4× B_4×3= 3×3
So,
Order of AB = 3×3

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Question 192 Marks

If $\text{A}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}$ and $\text{B}=\begin{bmatrix}-4\\3\end{bmatrix},$ write AB.

Answer

$\text{AB}=\begin{bmatrix}4&3\\1&2 \end{bmatrix}\begin{bmatrix}-4\\3\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}-16+9\\-4+6\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix} -7\\2\end{bmatrix}$

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Question 202 Marks

If $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix},$ write A².

Answer

Given: $\text{A}=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\text{A}^2=\text{AA}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}\begin{bmatrix}\text{i}&0\\0&\text{i}\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2+0&0+0\\0+0&0+\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}\text{i}^2&0\\0&\text{i}^2\end{bmatrix}$
$\Rightarrow\text{A}^2=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$ $(\because\ \text{i} ^2=-1)$

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Question 212 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} $

Answer

We 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

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Question 222 Marks

Give example of matrices:
A and B such that AB = O but A ≠ 0, B ≠ 0.

Answer

Let $\text{A}=\begin{bmatrix}0&2\\0&0\end{bmatrix},\ \text{B}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\therefore\ \text{AB}=\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=0$
Thus, AB = 0 while A ≠ 0 and B ≠ 0

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Question 232 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

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Question 242 Marks

Construct a 2 × 3 matrix A = [a_ij] whose elements a_ij are give by:
a_ij = i.j

Answer

Here,
a_ij = i.j. 1 ≤ i ≤ 2 and 1 ≤ j ≤ 3
a₁₁ = 1 × 1 = 1, a₁₂ = 1 × 2 = 2, a₁₃ = 1 × 3 = 3
a₂₁ = 2 × 1 = 2, a₂₂ = 2 × 2 = 4 and a₂₃ = 2 × 3 = 6

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Question 252 Marks

Let A and B be square matrices of the order 3 × 3. Is (AB)² = A²B²? Give reasons.

Answer

We are given that, A and B are square matrices of order 3 × 3.
Consider, (AB)²= AB.AB
= ABAB
= AABB $[\because$ AB = BA$]$
= A²B²
Thus, AB² = A²B² is true if and only if AB = BA.

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Question 262 Marks

If $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix},$ find AA^T.

Answer

Given: $\text{A}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{A}^\text{T}=\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos\text{x}&-\sin\text{x}\\\sin\text{x}&\cos\text{x}\end{bmatrix}\begin{bmatrix}\cos\text{x}&\sin\text{x}\\-\sin\text{x}&\cos\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}\cos^2\text{x}+\sin^2\text{x}&\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}\\\cos\text{x}\sin\text{x}-\sin\text{x}\cos\text{x}&\sin^2\text{x}+\cos^2\text{x}\end{bmatrix}$
$\Rightarrow\text{AA}^\text{T}=\begin{bmatrix}1&0\\0&1\end{bmatrix}$

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Question 272 Marks

Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$

Answer

Given: $\begin{bmatrix}\text{x}-\text{y}&2&-2\\4&\text{x}&6\end{bmatrix}+\begin{bmatrix}3&-2&2\\1&0&-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&2-2&-2+2\\4+1&\text{x}+0&6-1\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}-\text{y}+3&0&0\\5&\text{x}&5\end{bmatrix}=\begin{bmatrix}6&0&0\\5&2\text{x}+\text{y}&5\end{bmatrix}$
$\Rightarrow\text{x}-\text{y}+3=6$
$\Rightarrow\text{x}-\text{y}=6-3$
$\Rightarrow\text{x}-\text{y}=3\ \dots(1)$
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
$-\Big(\frac{3}{2}\Big)=\text{y}$
$\Rightarrow\text{y}=-\frac{3}{2}$

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Question 282 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.

Answer

Given, $5\text{A}+3\text{B}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow5\begin{bmatrix}9&1\\7&8\end{bmatrix}+3\begin{bmatrix}1&5\\7&12\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45&5\\35&40\end{bmatrix}+\begin{bmatrix}3&15\\21&36\end{bmatrix}+2\text{C}\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}45+3&5+15\\35+21&40+36\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}48&20\\56&76\end{bmatrix}+2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\Rightarrow2\text{C}=\begin{bmatrix}0&0\\0&0\end{bmatrix}-\begin{bmatrix}48&20\\56&76\end{bmatrix}$
$\Rightarrow\text{C}=\frac{1}{2}\begin{bmatrix}-48&-20\\-56&-76\end{bmatrix}$
$\Rightarrow\text{C}=\begin{bmatrix}-24&-10\\-28&-38\end{bmatrix}$

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Question 292 Marks

If A is 2×3 matrix and B is a matrix such that A^T B and BA^T both are defined, then what is the order of B?

Answer

Order of A = 2 × 3
Order of A^T= 3 × 2
Let Order of B = m × n
Given: A^TB and BA^T are defined
If A^T_3×2B_m×nexists, then the number of columns in A^T must be equal to number of rows in B.
⇒ m = 2
IfB_m×n A^T_3×2exists, then the number of columns in B must be equal to number of rows in A^T
⇒ n = 3
$\therefore$ Order of B = 2 × 3

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Question 302 Marks

$\text{If}\ \text{A}'=\begin{bmatrix}-2&3\\1&2\end{bmatrix}, \text{and}\ \text{B}=\begin{bmatrix}-1&0\\1&2\end{bmatrix}\ \text{then find}\ (\text{A} + 2\text{B})'$

Answer

We know that A = (A')'
$\therefore\ \text{A}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}$
$\therefore \text{A}+2\text{B}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+2\begin{bmatrix}-1&0\\1&2\end{bmatrix}=\begin{bmatrix}-2&1\\3&2\end{bmatrix}+\begin{bmatrix}-2&0\\2&4\end{bmatrix}=\begin{bmatrix}-4&1\\5&6\end{bmatrix}$
$\therefore(\text{A} + 2\text{B})'=\begin{bmatrix}-4&5\\1&6\end{bmatrix}$

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Question 312 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

Answer

If 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

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Question 322 Marks

If I is the identity matrix and A is a square matrix such that A² = A, then what is the value of (I + A)² = 3A?

Answer

Given,
A is a square matrix such that A² = A
Now,
(I + A)²- 3A = (I + A)(I + A) - 3A
⇒ (I + A)²- 3A = I × I + I × A + A × I + A × A - 3A {using distributive property}
⇒ (I + A)²- 3A = I + A + A + A²- 3A {using I × I = I, IA = AI = A}
⇒ (I + A)²- 3A = I + 2A + A - 3A {since, A²= A}
⇒ (I + A)²- 3A = I

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Question 332 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?

Answer

Since, 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.

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Question 342 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

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Question 352 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.

Answer

Here,
$\text{X}-\text{Y}+\text{X}+\text{Y}=\begin{bmatrix}1&1&1\\1&1&0\\1&0&0\end{bmatrix}+\begin{bmatrix}3&5&1\\-1&1&4\\11&8&0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}1+3&1+5&1+1\\1-1&1+1&0+4\\1+11&0+8&0+0\end{bmatrix}$
$\Rightarrow2\text{X}=\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\frac{1}{2}\begin{bmatrix}4&6&2\\0&2&4\\12&8&0\end{bmatrix}$
$\Rightarrow\text{X}=\begin{bmatrix}2&3&1\\0&1&2\\6&4&0\end{bmatrix}$
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 $\text{Y}=\begin{bmatrix}1&2&0\\-1&0&2\\5&4&0\end{bmatrix}$

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Question 362 Marks

Verify that A² = I, when $\text{A}=\begin{bmatrix}0&1&-1\\4&-3&4\\3&-3&4\end{bmatrix}.$

Answer

We 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.

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Question 372 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,
⇒ xy = 8 ...(1),
w = 4 ...(2),
z + 6 = 0 ...(3),
And x + y = 6 ...(4)
From equation (2) and equation (3) we get z = -6 and w = 4.
From equation (4) we have,
x + y = 6
⇒ x = 6 - y,
Subsituting value of x in equation (1) we get,
⇒ (6 - y)y = 8
⇒ y² - 6y + 8 = 0
⇒ (y - 2)(y - 4) = 0,
⇒ y = 2, 4
Subsituting the value of y in equation (1) we get,
⇒ x = 4, 2
Therefore, value of x, y, z, w are 2, 4, -6, 4 or 4, 2, -6, 4.

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Question 382 Marks

Let A and B be matrices of orders 3×2 and 2×4 respectively. Write the order of matrix AB.

Answer

Since, 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

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Question 392 Marks

Find x, y satisfying the matrix equation.
$\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$

Answer

Given: $\begin{bmatrix}\text{x}&\text{y}+2&\text{z}-3\end{bmatrix}+\begin{bmatrix}\text{y}&4&5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+2+4&\text{z}-3+5\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\Rightarrow\begin{bmatrix}\text{x}+\text{y}&\text{y}+6&\text{z}+2\end{bmatrix}=\begin{bmatrix}4&9&12\end{bmatrix}$
$\therefore\ \text{x}+\text{y}=4\ \dots(1)$
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 $\text{z}=10$

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Question 402 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.

Answer

Order of A = 2×3
Order of B = 3×2
So,
A_2×3 × B_3×2 has order = 2×2
B_3×2 × A_2×3 has order = 3×3
Hence,
Order of AB = 2×2
Order of BA = 3×3

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Question 412 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.

Answer

Given: $x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2x\\3x\end{bmatrix}+\begin{bmatrix}-y\\ y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
$\Rightarrow \begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}$
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

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Question 422 Marks

If A is a skew-symmetric matrix and n is an even natural number, write whether Aⁿ is symmetric or skew-symmetric or neither of these two.

Answer

If A is a skew-symmetric matrix, then A^T = -A.
(Aⁿ)^T = (A^T)ⁿ [For all n ∈ N]
⇒ (Aⁿ)^T = (-A)ⁿ [$\because$ A^T = -A]
⇒ (Aⁿ)^T = (-1)ⁿ Aⁿ
⇒ (Aⁿ)^T = Aⁿ, if n is even or -Aⁿ, if n is odd.
Hence, Aⁿ is a symmetric when n is an even natural number.

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Question 432 Marks

If $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix},$ show that A - A^T is a skew symmetric matrix.

Answer

Given:

$\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$

$$$\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.

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Question 442 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.

Answer

Since 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

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Question 452 Marks

If $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix},$ prove that A - A^T is a skew symmetric matrix.

Answer

Given: $\text{A}=\begin{bmatrix}2 & 3 \\4 & 5 \end{bmatrix}$
$\text{A}^{\text{T}}=\begin{bmatrix}2 & 4 \\3 & 5 \end{bmatrix}$
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.

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Question 462 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.

Answer

Here,
$\text{A}=\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\text{I}$
$\Rightarrow\begin{bmatrix}\cos\text{a}&-\sin\text{a}\\\sin\text{a}&\cos\text{a} \end{bmatrix}=\begin{bmatrix}1&0\\0&1 \end{bmatrix}$
The corresponding elements of equal matrices are equal.
$\therefore \cos\text{a}=1$
$\Rightarrow\text{a}=0^{\circ}$

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Question 472 Marks

Find x, y satisfying the matrix equation.
$\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=0$

Answer

Given: $\text{x}\begin{bmatrix}2\\1\end{bmatrix}+\text{y}\begin{bmatrix}3\\5\end{bmatrix}+\begin{bmatrix}-8\\-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow\begin{bmatrix}2\text{x}+3\text{y}-8\\\text{x}+5\text{y}-11\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$
$\Rightarrow2\text{x}+3\text{y}-8=0$
$\Rightarrow2\text{x}+3\text{y}=8\ \dots(1)$
Also,
$\text{x}+5\text{y}-11=0$
$\Rightarrow\text{x}+5\text{y}=11$
$\Rightarrow\text{x}=11-5\text{y}\ \dots(2)$
Putting the value of x in eq. (1), we get
$2(11-5\text{y})+3\text{y}=8$
$\Rightarrow22-10\text{y}+3\text{y}=8$
$\Rightarrow-7\text{y}=8-22$
$\Rightarrow-7\text{y}=-14$
$\Rightarrow\text{y}=2$
Putting the value of y in eq. (2), we get
$\text{x}=11-5(2)$
$\Rightarrow\text{x}=11-10$
$\Rightarrow\text{x}=1$
$\therefore\ \text{x}=1$ and $\text{y}=2$

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Question 482 Marks

If A and B are square matrices of the same order, explain, why in general:
(A + B)(A - B) ≠ A² - B².

Answer

LHS = (A + B)(A - B)

= A(A - B) + B(A - B)

= A² - AB + BA - B²

We know that a matrix does not have commutative property. So,

AB ≠ BA

Thus,

(A + B)(A - B) ≠ A² - B²

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Question 492 Marks

If B is a symmetric matrix, write whether the matrix AB A^T is symmetric or skew-symmetric.

Answer

If 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.

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Question 502 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,
$\text{A}^4=\lambda\text{A}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\lambda\begin{bmatrix}1&1\\1&1\end{bmatrix}$
$\Rightarrow\begin{bmatrix}8&8\\8&8\end{bmatrix}=\begin{bmatrix}\lambda&\lambda\\\lambda&\lambda\end{bmatrix}$
$\therefore\ \lambda=8$

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