Exercise 7.5 - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$
\operatorname{Ex} 7.5
$
Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
If $\mathrm{a}_{\mathrm{ij}}=\frac{1}{2}(3 \mathrm{i}-2 \mathrm{j})$ and $A=\left[\mathrm{a}_{\mathrm{ij}}\right]_{2 \times 2}$ is
(a) $\left[\begin{array}{rr}\frac{1}{2} & 2 \\ -\frac{1}{2} & 1\end{array}\right]$
(b) $\left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{2} \\ 2 & 1\end{array}\right]$
(c) $\left[\begin{array}{rr}2 & 2 \\ \frac{1}{2} & -\frac{1}{2}\end{array}\right]$
(d) $\left[\begin{array}{rr}-\frac{1}{2} & \frac{1}{2} \\ 1 & 2\end{array}\right]$
Solution:
(b) $\left[\begin{array}{rr}\frac{1}{2} & -\frac{1}{2} \\ 2 & 1\end{array}\right]$
Hint: $a_{i j}=\frac{1}{2}(3 i-2 j)$
A.is a matrix of order $2 \times 2 \therefore \mathrm{A}=\left(\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right)$
$
\begin{aligned}
& a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=-\frac{1}{2} ; \\
& a_{21}=\frac{1}{2}(6-2)=\frac{4}{2}=2: a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \\
& \therefore A=\left(\begin{array}{cc}
\frac{1}{2} & -\frac{1}{2} \\
2 & 1
\end{array}\right)
\end{aligned}
$
Question 2.
What must be the matrix $\mathrm{X}$, if $2 \mathrm{X}+\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]=\left[\begin{array}{ll}3 & 8 \\ 7 & 2\end{array}\right]$ ?
(a) $\left[\begin{array}{rr}1 & 3 \\ 2 & -1\end{array}\right]$
(b) $\left[\begin{array}{rr}1 & -3 \\ 2 & -1\end{array}\right]$
(c) $\left[\begin{array}{rr}2 & 6 \\ 4 & -2\end{array}\right]$
$(d)\left[\begin{array}{rr}2 & -6 \\ 4 & -2\end{array}\right]$
Solution:

(a) $\left[\begin{array}{rr}1 & 3 \\ 2 & -1\end{array}\right]$

$
\begin{aligned}
& \text { Hint }: 2 X+\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)=\left(\begin{array}{ll}
3 & 8 \\
7 & 2
\end{array}\right) \\
& \Rightarrow \quad 2 X=\left(\begin{array}{ll}
3 & 8 \\
7 & 2
\end{array}\right)-\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)=\left(\begin{array}{cc}
3-1 & 8-2 \\
7-3 & 2-4
\end{array}\right)=\left(\begin{array}{cc}
2 & 6 \\
4 & -2
\end{array}\right) \\
&
\end{aligned}
$
$
\therefore \quad X=\frac{1}{2}\left(\begin{array}{cc}
2 & 6 \\
4 & -2
\end{array}\right)=\left(\begin{array}{cc}
1 & 3 \\
2 & -1
\end{array}\right)
$

Question 3.
Which one of the following is not true about the matrix $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 5\end{array}\right]$ ?
(a) a scalar matrix
(b) a diagonal matrix
(c) an upper triangular matrix
(d) A lower triangular matrix
Solution:
(b) a diagonal matrix
Question 4.
If $A$ and $B$ are two matrices such that $A+B$ and $A B$ are both defined, then
(a) $A$ and $B$ are two matrices not necessarily of same order.
(b) $A$ and $B$ are square matrices of same order.
(c) Number of columns of $a$ is equal to the number of rows of $B$.
(d) $\mathrm{A}=\mathrm{B}$.
Solution:
(b) $A$ and $B$ are square matrices of same order.
Question 5.
If $A=\left[\begin{array}{rr}\lambda & 1 \\ -1 & -\lambda\end{array}\right]$, then for what value of $\lambda, A^2=0$ ?
(a) 0
(b)$\pm 1$
(c) -1
(d) 1
Solution:
(b)$\pm 1$
$
\begin{aligned}
\text { Hint: } A^2 & =A \times A=\left(\begin{array}{cc}
\lambda & 1 \\
-1 & -\lambda
\end{array}\right)\left(\begin{array}{cc}
\lambda & 1 \\
-1 & -\lambda
\end{array}\right) \\
& =\left(\begin{array}{cc}
\lambda^2-1 & 0 \\
0 & \lambda^2-1
\end{array}\right)=0 \text { given } \\
\Rightarrow \quad \lambda^2-1 & =0 \Rightarrow \lambda^2=1 ; \lambda= \pm 1
\end{aligned}
$

Question 6.
$
\mathrm{A}=\left[\begin{array}{ll}
1 & -1 \\
2 & -1
\end{array}\right] \mathrm{B}=\left[\begin{array}{cc}
a & 1 \\
b & -1
\end{array}\right]
$
If and $(A+B)^2=A^2+B^2$, then the values of $a$ and $b$ are
(a) $a=4, b=1$
(b) $a=1, b=4$
(c) $\mathrm{a}=0, \mathrm{~b}=4$
(d) $\mathrm{a}=2, \mathrm{~b}=4$
Solution:
(b) $a=1, b=4$
$
\begin{aligned}
\text { Hint: } \mathrm{A} & =\left(\begin{array}{ll}
1 & -1 \\
2 & -1
\end{array}\right) ; \mathrm{B}=\left(\begin{array}{cc}
a & 1 \\
b & -1
\end{array}\right) \\
\mathrm{A}+\mathrm{B} & =\left(\begin{array}{cc}
1+a & 0 \\
2+b & -2
\end{array}\right) \\
(\mathrm{A}+\mathrm{B})^2 & =\left(\begin{array}{cc}
1+a & 0 \\
2+b & -2
\end{array}\right)\left(\begin{array}{cc}
1+a & 0 \\
2+b & -2
\end{array}\right) \\
& =\left(\begin{array}{cc}
(1+a)^2 & 0 \\
(1+a)(2+b) & 4 \\
-2(2+b) & 4
\end{array}\right) \\
\mathrm{A}^2 & =\mathrm{A} \times \mathrm{A}=\left(\begin{array}{cc}
1 & -1 \\
2 & -1
\end{array}\right)\left(\begin{array}{cc}
1 & -1 \\
2 & -1
\end{array}\right)=\left(\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right) \\
\mathrm{B}^2 & =\mathrm{B} \times \mathrm{B}=\left(\begin{array}{cc}
a & 1 \\
b & -1
\end{array}\right)\left(\begin{array}{cc}
a & 1 \\
b & -1
\end{array}\right)=\left(\begin{array}{cc}
a^2+b & a-1 \\
a b-b & b+1
\end{array}\right) \\
\text { Now } \mathrm{A}^2+\mathrm{B}^2 & =\left(\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right)+\left(\begin{array}{cc}
a^2+b & a-1 \\
a b-b & b+1
\end{array}\right) \\
& =\left(\begin{array}{cc}
a^2+b-1 & a-1 \\
a b-b & b
\end{array}\right)
\end{aligned}
$

Given
$
\begin{aligned}
(A+B)^2 & =A^2+B^2 \\
(1) & =(2)
\end{aligned}
$
$
\Rightarrow \quad \text { (1) }=\text { (2) }
$
$
\begin{aligned}
& \left(\begin{array}{cc}
(1+a)^2 & 0 \\
(1+a)(2+b) & 4 \\
-2(2+b) &
\end{array}\right)=\left(\begin{array}{cc}
a^2+b-1 & a-1 \\
a b-b & b
\end{array}\right) \\
& \Rightarrow \quad a-1=0 \\
& \Rightarrow a=1 \\
& \therefore a=1, b=4 \\
&
\end{aligned}
$

Question 7.
$
A=\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]
$
is a matrix satisfying the equation $\mathrm{AA}^{\mathrm{T}}=9 \mathrm{I}$, where $\mathrm{I}$ is $3 \times 3$ identity matrix, then the ordered pair $(\mathrm{a}, \mathrm{b})$ is equal to ..........
(a) $(2,-1)$
(b) $(-2,1)$
(c) $(2,1)$
(d) $(-2,-1)$
Solution:
(d) $(-2,-1)$
So, $\begin{aligned} \text { Hint: } \mathrm{A} & =\left(\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{array}\right) \\ A^{\mathrm{T}} & =\left(\begin{array}{ccc}1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b\end{array}\right)\end{aligned}$
Given
$
\begin{aligned}
\mathrm{AA}^{\mathrm{T}} & =9 \mathrm{I} \\
& =9\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right)=\left(\begin{array}{lll}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right)
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow\left(\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right)\left(\begin{array}{ccc}
1 & 2 & a \\
2 & 1 & 2 \\
2 & -2 & b
\end{array}\right)=\left(\begin{array}{ccc}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right) \\
& \Rightarrow\left(\begin{array}{ccc}
9 & 0 & a+4+2 b \\
0 & 9 & 2 a+2-2 b \\
a+4+2 b & 2 a+2-2 b & a^2+4+b^2
\end{array}\right)=\left(\begin{array}{ccc}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right) \\
& \Rightarrow \quad a+2 b=-4 \\
& \text { and } 2 a-2 b=-2 \\
& (1)+(2) \Rightarrow \\
& 3 a=-6 \Rightarrow a=-6 / 3=-2 \\
& \text { Substituting } a=-2 \text { in (1) we get } \\
& -2+2 b=-4 \\
& 2 b=-4+2=-2 \\
&
\end{aligned}
$

$
\begin{aligned}
& b=-2 / 2=-1 \\
& \therefore a=2,1 \\
& b=-1 \\
&
\end{aligned}
$
Question 8.
If $A$ is a square matrix, then which of the following is not symmetric?
(a) $\mathrm{A}+\mathrm{A}^{\mathrm{T}}$
(b) $\mathrm{AA}^{\mathrm{T}}$
(c) $\mathrm{A}^{\mathrm{T}} \mathrm{A}$
(d) $\mathrm{A}-\mathrm{A}^{\mathrm{T}}$
Solution:
(b) $\mathrm{AA}^{\mathrm{T}}$
Question 9.
If $A$ and $B$ are symmetric matrices of order $n$, where $(A \neq B)$, then
(a) $\mathrm{A}+\mathrm{B}$ is skew-symmetric
(b) $A+B$ is symmetric
(c) $\mathrm{A}+\mathrm{B}$ is a diagonal matrix
(d) $\mathrm{A}+\mathrm{B}$ is a zero matrix
Solution:
(b) $A+B$ is symmetric

Question 10 .
If

$
A=\left[\begin{array}{ll}
a & \boldsymbol{x} \\
y & a
\end{array}\right]
$
and if $x y=1$, then $\operatorname{det}\left(\mathrm{AA}^{\mathrm{T}}\right)$ is equal to
(a) $(a-1)^2$
(b) $\left(a^2+1\right)^2$
(c) $\mathrm{a}^2-1$
(d) $\left(a^2-1\right)^2$
Solution:
$
\begin{aligned}
& \mathrm{A}=\left(\begin{array}{ll}
a & x \\
y & a
\end{array}\right) \therefore \mathrm{A}^{\mathrm{T}}=\left(\begin{array}{ll}
a & y \\
x & a
\end{array}\right) \\
& |\mathrm{A}|=\left|\begin{array}{ll}
a & x \\
y & a
\end{array}\right|=a^2-x y=a^2-1 \\
& \left|\mathrm{~A}^{\mathrm{T}}\right|=\left|\begin{array}{ll}
a & y \\
x & a
\end{array}\right|=a^2-x y=a^2-1 \\
& \therefore \quad\left|\mathrm{AA}^{\mathrm{T}}\right|=\left|\mathbf{A} \| \mathbf{A}^{\mathrm{T}}\right|=\left(a^2-1\right)\left(a^2-1\right)=\left(a^2-1\right)^2 \\
&
\end{aligned}
$
(d) Hint:
Question 11.
The value of $\mathrm{x}$, for which the matrix
$
A=\left[\begin{array}{ll}
e^{x-2} & e^{7+x} \\
e^{2+x} & e^{2 x+3}
\end{array}\right]
$
is singular is
(a) 9

(b) 8
(c) 7
(d) 6
Solution:
(b) Hint: Given $A$ is a singular matrix $\Rightarrow|A|=0$
$
\begin{aligned}
& \text { (i.e.) }\left|\begin{array}{cc}
e^{x-2} & e^{7+x} \\
e^{2+x} & e^{2 x+3}
\end{array}\right|=0 \\
& \Rightarrow \mathrm{e}^{\mathrm{x}-2} \cdot \mathrm{e}^{2 \mathrm{x}+3}-\mathrm{e}^{2+\mathrm{x}} \cdot \mathrm{e}^{7+\mathrm{x}}=0 \\
& \Rightarrow \mathrm{e}^{3 \mathrm{x}+1}-\mathrm{e}^{9+2 \mathrm{x}}=0 \Rightarrow \mathrm{e}^{3 \mathrm{x}+1}=\mathrm{e}^{9+2 \mathrm{x}} \\
& \Rightarrow 3 \mathrm{x}+1=9+2 \mathrm{x} \\
& 3 \mathrm{x}-2 \mathrm{x}=9-1 \Rightarrow \mathrm{x}=8
\end{aligned}
$
Question 12.
If the points $(\mathrm{x},-2),(5,2),(8,8)$ are collinear, then $\mathrm{x}$ is equal to
(a) -3
(b) $\frac{1}{3}$
(c) 1
(d) 3
Solution:
(d) Hint: Given that the points are collinear
So, area of the triangle formed by the points $=0$
$
\begin{aligned}
& \Rightarrow \quad \frac{1}{2}\left|\begin{array}{ccc}
x & -2 & 1 \\
5 & 2 & 1 \\
8 & 8 & 1
\end{array}\right|=0 \\
& \Rightarrow \quad\left|\begin{array}{ccc}
x & -2 & 1 \\
5 & 2 & 1 \\
8 & 8 & 1
\end{array}\right|=0 \\
& \Rightarrow \quad x(2-8)+2(5-8)+1(40-16)=0 \\
& -6 x-6+24=0 \\
& -6 x=-18 \Rightarrow x=3 \\
&
\end{aligned}
$

Question 13.
If $\left[\begin{array}{lll}2 a & x_1 & y_1 \\ 2 b & x_2 & y_2 \\ 2 c & x_3 & y_3\end{array}\right]=\frac{a b c}{2} \neq 0$, then the area of the triangle whose vertices are $\left(\frac{x_1}{a}, \frac{y_1}{a}\right),\left(\frac{x_2}{b}, \frac{y_2}{b}\right),\left(\frac{x_3}{c}, \frac{y_3}{c}\right)$ is
(a) $\frac{1}{4}$
(b) $\frac{1}{4} a b c$
(c) $\frac{1}{8}$
(d) $\frac{1}{8} a b c$

Solution:
Now the area of the triangle with vertices $\left(\frac{x_1}{a}, \frac{y_1}{a}\right),\left(\frac{x_2}{b}, \frac{y_2}{b}\right)$ and $\left(\frac{x_3}{c}, \frac{y_3}{c}\right)$
$
\frac{1}{2}\left|\begin{array}{lll}
\frac{x_1}{a} & \frac{y_1}{a} & 1 \\
\frac{x_2}{b} & \frac{y_2}{b} & 1 \\
\frac{x_3}{c} & \frac{y_3}{c} & 1
\end{array}\right|
$
multiplying $\mathrm{R}_1$ by $a, \mathrm{R}_2$ by $b$ and $\mathrm{R}_3$ by $c$ and dividing by $a b c$ we get
$
\frac{1}{2 a b c}\left|\begin{array}{lll}
x_1 & y_1 & a \\
x_2 & y_2 & b \\
x_3 & y_3 & c
\end{array}\right|
$
Interchanging $\mathrm{C}_1$ and $\mathrm{C}_3$, then $\mathrm{C}_2$ and $\mathrm{C}_3$
we get
$
\begin{aligned}
& \frac{1}{2 a b c}(-)(-)\left|\begin{array}{lll}
a & x_1 & y_1 \\
b & x_2 & y_2 \\
c & x_3 & y_3
\end{array}\right| \\
& =\frac{1}{2 a b c} \times \frac{a b c}{4}=\frac{1}{8}
\end{aligned}
$

Question 14 .
If the square of the matrix
$
\left[\begin{array}{rr}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right]
$
is the unit matrix of order 2 , then $\alpha, \beta$ and $\gamma$ should satisfy the relation.
(a) $1+\alpha^2+\beta \gamma=0$
(b) $1-\alpha^2-\beta \gamma=0$
(c) $1-\alpha^2+\beta \gamma=0$
(d) $1+\alpha^2-\beta \gamma=0$

Solution:
(b) Hint: $\quad$ LetA $=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha\end{array}\right)$
Given $\mathrm{A}^2$ is a unit matrix
$
\begin{aligned}
& \Rightarrow \quad\left(\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right)\left(\begin{array}{cc}
\alpha & \beta \\
\gamma & -\alpha
\end{array}\right)=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& \Rightarrow \quad\left(\begin{array}{ll}
\alpha^2+\beta \gamma & \alpha \beta-\alpha \beta \\
\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^2
\end{array}\right)=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& \Rightarrow \quad \alpha^2+\beta \gamma=1 \Rightarrow 1-\alpha^2-\beta \gamma=0 \\
&
\end{aligned}
$

Question 15 .
If $\Delta=\left|\begin{array}{lll}a & b & c \\ x & y & z \\ p & q & r\end{array}\right|$, then $\left|\begin{array}{lll}k a & k b & k c \\ k x & k y & k z \\ k p & k q & k r\end{array}\right|$ is
(a) $\Delta$
(b) $\mathrm{k} \Delta$
(c) $3 \mathrm{k} \Delta$
(d) $\mathrm{k}^3 \Delta$
Solution:
(d) Hint: Given $\Delta=\left|\begin{array}{lll}a & b & c \\ x & y & z \\ p & q & r\end{array}\right|$
$
\begin{aligned}
\therefore \quad\left|\begin{array}{lll}
k a & k b & k c \\
k x & k y & k z \\
k p & k q & k r
\end{array}\right| & =(k)(k)(k)\left|\begin{array}{lll}
a & b & c \\
x & y & z \\
p & q & r
\end{array}\right| \\
& =k^3 \Delta
\end{aligned}
$
Question 16.
A root of the equation
$
\left|\begin{array}{ccc}
3-x & -6 & 3 \\
-6 & 3-x & 3 \\
3 & 3 & -6-x
\end{array}\right|=0
$
is
(a) 6
(b) 3
(c) 0
(d) -6

Solution:
(c) Hint
$
\begin{aligned}
\text { Let } \Delta & =\left|\begin{array}{ccc}
3-x & -6 & 3 \\
-6 & 3-x & 3 \\
3 & 3 & -6-x
\end{array}\right| \\
& =\left|\begin{array}{ccc}
-x & -x & -x \\
-6 & 3-x & 3 \\
3 & 3 & -6-x
\end{array}\right| \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3
\end{aligned}
$
Put $x=0$ we get $\Delta=0$
$\therefore x=0$ is a root

Question 17.
The value of the determinant of
(a) $-2 a b c$
(b) abc
(c) 0
(d) $a^2+b^2+c^2$
Solution:
(c) Hint:
$
|\mathrm{A}|=\left|\begin{array}{ccc}
0 & a & -b \\
-a & 0 & c \\
b & -c & 0
\end{array}\right|=0
$
( $\because$ the determinant of a skew symmetric matrix is 0 )
Question 18 .
If $\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3$ as well as $\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3$ are in geometric progression with the same common ratio, then the points $\left(\mathrm{x}_1, \mathrm{y}_1\right),\left(\mathrm{x}_2, \mathrm{y}_2\right),\left(\mathrm{x}_3, \mathrm{y}_3\right)$ are
(a) vertices of an equilateral triangle
(b) vertices of a right angled triangle
(c) vertices of a right angled isosceles triangle
(d) collinear
Solution:
(d) collinear
Question 19.
If $\lfloor\cdot\rfloor$ denotes the greatest integer less than or equal to the real number under consideration and $-1 \leq x<$

$0,0 \leq y<1,1 \leq z \leq 2$, then the value of the determinant
$
\left|\begin{array}{ccc}
\lfloor x\rfloor+1 & \lfloor y\rfloor & \lfloor z\rfloor \\
\lfloor x\rfloor & \lfloor y\rfloor+1 & \lfloor z\rfloor \\
\lfloor x\rfloor & \lfloor y\rfloor & \lfloor z\rfloor+1
\end{array}\right|
$
is
(a) $\lfloor z\rfloor$
(b) $\lfloor y\rfloor$
(c) $\lfloor x\rfloor$
(d) $\lfloor x\rfloor+1$
Solution:
(a) Hint: From the given values
$
\lfloor x\rfloor=-1 ;\lfloor y\rfloor=0 \text { and }\lfloor z\rfloor=1
$
$\therefore$ The given determinant is $\left|\begin{array}{ccc}0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2\end{array}\right|=1(0+1)=1=\lfloor z\rfloor$

Question 20.
If $\mathrm{a} \neq \mathrm{b}, \mathrm{b}, \mathrm{c}$ satisfy $\left|\begin{array}{rrr}a & 2 b & 2 c \\ 3 & b & c \\ 4 & a & b\end{array}\right|=0$,
(a) $a+b+c$
(b) 0
(c) $b^3$
(d) $a b+b c$
Solution:
(c) Hint: Expanding along $R_1$,
$
\begin{aligned}
& a\left(b^2-a c\right)-2 b(3 b-4 c)+2 c(3 a-4 b)=0 \\
& \left(b^2-a c\right)(a-b)=0 \\
& b^2=a c(o r) a=b \\
& \Rightarrow a b c=b\left(b^2\right)=b^3
\end{aligned}
$
Question 21.
$
A=\left|\begin{array}{rrr}
-1 & 2 & 4 \\
3 & 1 & 0 \\
-2 & 4 & 2
\end{array}\right| \text { and } B=\left|\begin{array}{rrr}
-2 & 4 & 2 \\
6 & 2 & 0 \\
-2 & 4 & 8
\end{array}\right|
$
then $B$ is given by
(a) $\mathrm{B}=4 \mathrm{~A}$
(b) $\mathrm{B}=-4 \mathrm{~A}$
(c) $B=-A$
(d) $\mathrm{B}=6 \mathrm{~A}$
Solution:

(b)
$
\text { Hint : } A=\left|\begin{array}{ccc}
-1 & 2 & 4 \\
3 & 1 & 0 \\
-2 & 4 & 2
\end{array}\right|
$
Now
$
\begin{aligned}
B & =\left|\begin{array}{ccc}
-2 & 4 & 2 \\
6 & 2 & 0 \\
-2 & 4 & 8
\end{array}\right|=(2)(2)\left|\begin{array}{ccc}
-2 & 4 & 2 \\
3 & 1 & 0 \\
-1 & 2 & 4
\end{array}\right| \\
& =4\left|\begin{array}{ccc}
-2 & 4 & 2 \\
3 & 1 & 0 \\
-1 & 2 & 4
\end{array}\right| R_1 \leftrightarrow R_3 \\
& =(-4)\left|\begin{array}{ccc}
-1 & 2 & 4 \\
3 & 1 & 0 \\
-2 & 4 & 2
\end{array}\right|=-4 A \Rightarrow B=-4 A
\end{aligned}
$

Question 22.
IfA is skew-symmetric of order $n$ and $C$ is a column matrix of order $n \times 1$, then $C^T A C$ is
(a) an identity matrix of order $n$
(b) an identity matrix of order 1
(e) a zero matrix of order I
(d) an Identity matrix of order 2
Solution:
(c) Hint: Given $A$ is of order $n \times n$
$\mathrm{C}$ is of order $n \times 1$
so, $\mathrm{CT}$ is of order $1 \times \mathrm{n}$
Now $\mathrm{C}^{\mathrm{T}} \mathrm{AC}$ is of order $\left(\begin{array}{ccc}1 \times n & & \\ & n \times n & \\ & & n \times 1\end{array}\right)=1 \times 1$
Let it be equal to (x) say
Taking transpose on either sides
$\left(C^{\mathrm{T}}, \mathrm{AC}\right)^{\mathrm{T}}(\mathrm{x})^{\mathrm{T}}$.
(i.e.) $C^{\mathrm{T}}\left(\mathrm{A}^{\mathrm{T}}\right)(\mathrm{C})=\mathrm{x}$
$\mathrm{C}^{\mathrm{T}}(-\mathrm{A})(\mathrm{C})=\mathrm{x}$
$\Rightarrow \mathrm{C}^{\mathrm{T}} \mathrm{AC}=-\mathrm{x}$
$\Rightarrow \mathrm{x}=-\mathrm{x} \Rightarrow 2 \mathrm{x}=0 \Rightarrow \mathrm{x}=0$

Question 23.
The matrix A satisfying the equation $\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \mathbf{A}=\left[\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right]$,
(a) $\left[\begin{array}{rr}1 & 4 \\ -1 & 0\end{array}\right]$
(b) $\left[\begin{array}{rr}1 & -4 \\ 1 & 0\end{array}\right]$
(c) $\left[\begin{array}{rr}1 & 4 \\ 0 & -1\end{array}\right]$
$(d)\left[\begin{array}{rr}1 & -4 \\ 1 & 1\end{array}\right]$
Solution:
(c) Hint: Let $\mathrm{A}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$
Given $\left(\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right) A=\left(\begin{array}{cc}1 & 1 \\ 0 & -1\end{array}\right)$
$
\Rightarrow \quad\left(\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right)\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)=\left(\begin{array}{cc}
1 & 1 \\
0 & -1
\end{array}\right)
$
$
\begin{aligned}
& a=1, \quad b-3=1, b=4 \\
& \therefore \quad \mathrm{A}=\left(\begin{array}{cc}
1 & 4 \\
0 & -1
\end{array}\right) \\
&
\end{aligned}
$
Question 24.
If $A+I=\left[\begin{array}{rr}3 & -2 \\ 4 & 1\end{array}\right]$ then $(A+I)(A-I)$ is equal to
(a) $\left[\begin{array}{rr}-5 & -4 \\ 8 & -9\end{array}\right]$
(b) $\left[\begin{array}{rr}-5 & 4 \\ -8 & 9\end{array}\right]$
(c) $\left[\begin{array}{ll}5 & 4 \\ 8 & 9\end{array}\right]$
(d) $\left[\begin{array}{rr}-5 & -4 \\ -8 & -9\end{array}\right]$

Solution:
(a)
$
\begin{aligned}
& \text { Hint: } A+I=\left(\begin{array}{cc}
3 & -2 \\
4 & 1
\end{array}\right) \\
& \Rightarrow A+\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{cc}
3 & -2 \\
4 & 1
\end{array}\right) \\
& \therefore \quad A=\left(\begin{array}{cc}
3 & -2 \\
4 & 1
\end{array}\right)-\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right) \\
& \text { Now } \quad A^2=A \times A=\left(\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right) \downarrow\left(\begin{array}{cc}
2 & -2 \\
4 & 0
\end{array}\right) \\
& =\left(\begin{array}{cc}
-4 & -4 \\
8 & -8
\end{array}\right) \\
& \therefore(\mathrm{A}+\mathrm{I})(\mathrm{A}-\mathrm{I})=\mathrm{A}^2-\mathrm{I}^2=\mathrm{A}^2-\mathrm{I} \\
& =\left(\begin{array}{cc}
-4 & -4 \\
8 & -8
\end{array}\right)-\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{cc}
-5 & -4 \\
8 & -9
\end{array}\right) \\
&
\end{aligned}
$
Question 25 .
Let $A$ and $\mathrm{B}$ be two symmetric matrices of same order. Then which one of the following statement is not true?
(a) $\mathrm{A}+\mathrm{B}$ is a symmetric matrix
(b) $A B$ is a symmetric matrix
(c) $\mathrm{AB}=(\mathrm{BA})^{\mathrm{T}}$
(d) $\mathrm{A}^{\mathrm{T}} \mathrm{B}=\mathrm{AB}^{\mathrm{T}}$
Solution:
(b) $A B$ is a symmetric matrix