Exercise 7.2 - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$
\text { Ex } 7.2
$
Question 1.
$
\left|\begin{array}{ccc}
s & a^2 & b^2+c^2 \\
s & b^2 & c^2+a^2 \\
s & c^2 & a^2+b^2
\end{array}\right|=0
$
Solution:
$
\left|\begin{array}{lll}
s & a^2 & b^2+c^2 \\
s & b^2 & c^2+a^2 \\
s & c^2 & a^2+b^2
\end{array}\right|=\left|\begin{array}{ccc}
s & a^2 & a^2+b^2+c^2 \\
s & b^2 & a^2+b^2+c^2 \\
s & c^2 & a^2+b^2+c^2
\end{array}\right|\left(\mathrm{C}_3 \longrightarrow \mathrm{C}_3+\mathrm{C}_2\right)
$
Taking $s$ from $\mathrm{C}_1$ and $a^2+b^2+c^2$ from $\mathrm{C}_3$ as common factors are get
$(s)\left(a^2+b^2+c^2\right)\left|\begin{array}{lll}1 & a^2 & 1 \\ 1 & b^2 & 1 \\ 1 & c^2 & 1\end{array}\right|=0$
$
\left(\because C_1=\mathrm{C}_3\right)
$
Question 2.
$
\left|\begin{array}{lll}
b+c & b c & b^2 c^2 \\
c+a & c a & c^2 a^2 \\
a+b & a b & a^2 b^2
\end{array}\right|=0
$
Solution:

$
\left|\begin{array}{ccc}
b+c & b c & b^2 c^2 \\
c+a & c a & c^2 a^2 \\
a+b & a b & a^2 b^2
\end{array}\right|
$
Multiplying $\mathrm{R}_1$ by $a, \mathrm{R}_2$ by $b, \mathrm{R}_3$ by $c$ and dividing by $a b c$ we get
$
\begin{aligned}
& \frac{1}{a b c}\left|\begin{array}{lll}
a(b+c) & a b c & a b^2 c^2 \\
b(c+a) & a b c & a^2 b c^2 \\
c(a+b) & a b c & a^2 b^2 c
\end{array}\right| \\
& =\frac{(a b c)^2}{a b c}\left|\begin{array}{lll}
a b+a c & 1 & b c \\
b c+a b & 1 & c a \\
a c+b c & 1 & a b
\end{array}\right|=(a b c)\left|\begin{array}{lll}
a b+b c+c a & 1 & b c \\
a b+b c+c a & 1 & c a \\
a b+b c+c a & 1 & a b
\end{array}\right|\left(\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_3\right) \\
& =(a b+b c+c a)(a b c)\left|\begin{array}{lll}
1 & 1 & b c \\
1 & 1 & c a \\
1 & 1 & a b
\end{array}\right|=0 \\
&
\end{aligned}
$

Question 3.

Prove that
$
\left|\begin{array}{ccc}
a^2 & b c & a c+c^2 \\
a^2+a b & b^2 & a c \\
a b & b^2+b c & c^2
\end{array}\right|=4 a^2 b^2 c^2 .
$
LHS
Taking a from $\mathrm{C}_1, \mathrm{~b}$ from $\mathrm{C}_2$ and $\mathrm{c}$ from $\mathrm{C}_3$ we get

$
\begin{aligned}
& (a b c)\left|\begin{array}{ccc}
a & c & a+c \\
a+b & b & a \\
b & b+c & c
\end{array}\right| \\
& \left.=(a b c)\left|\begin{array}{ccc}
-b & c-b & c \\
a & -c & a-c \\
b & b+c & c
\end{array}\right|\left(\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2\right) \rightarrow \mathrm{R}_2-\mathrm{R}_3\right) \\
& =(a b c)\left|\begin{array}{ccc}
0 & 2 c & 2 c \\
a & -c & a-c \\
b & b+c & c
\end{array}\right|\left(\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_3\right)
\end{aligned}
$
Taking $2 c$ as a common factor from $\mathrm{R}_1$ we get
$
\begin{aligned}
& =(a b c)(2 c)\left|\begin{array}{ccc}
0 & 1 & 1 \\
a & -c & a-c \\
b & b+c & c
\end{array}\right| \\
& =(2 c)(a b c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
a & -a & a-c \\
b & b & c
\end{array}\right|\left(\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_3\right)
\end{aligned}
$

Expanding along $R_1$ we get
$
\begin{aligned}
& \text { (2c) }(a b c)(1)[a b+a b]=a b c(2 c)(2 a b) \\
& 1=(a b c)(4 a b c)=4 a^2 b^2 c^2 \\
& =\text { RHS }
\end{aligned}
$
Question 4.
$
\text { Prove that }\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \text {. }
$

Solution:
$
\mathrm{LHS}=\left|\begin{array}{ccc}
1+a & 1 & 1 \\
1 & 1+b & 1 \\
1 & 1 & 1+c
\end{array}\right|=\left|\begin{array}{ccc}
a & -b & 0 \\
0 & b & -c \\
1 & 1 & 1+c
\end{array}\right|\left(\begin{array}{l}
\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\end{array}\right)
$
Expanding along $R_1$
$
\begin{aligned}
& a\left|\begin{array}{cc}
b & -c \\
1 & 1+c
\end{array}\right|+b\left|\begin{array}{cc}
0 & -c \\
1 & 1+c
\end{array}\right|+0 \\
& =a[b(1+c)+c]+b[0+c] \\
& =a(b+b c+c)+b c=a b+a b c+a c+b c \\
& =a b c+a b+b c+a c=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \\
& =\mathrm{RHS} \\
&
\end{aligned}
$

Question 5.
$\text { Prove that }\left|\begin{array}{ccc}
\sec ^2 \theta & \tan ^2 \theta & 1 \\
\tan ^2 \theta & \sec ^2 \theta & -1 \\
38 & 36 & 2
\end{array}\right|=0$

$\begin{aligned}
& \text { LHS }=\left|\begin{array}{ccc}
\sec ^2 \theta & \tan ^2 \theta & 1 \\
\tan ^2 \theta & \sec ^2 \theta & -1 \\
38 & 36 & 2
\end{array}\right| \\
& =\left|\begin{array}{ccc}
\sec ^2 \theta-\tan ^2 \theta & \tan ^2 \theta & 1 \\
\tan ^2 \theta-\sec ^2 \theta & \sec ^2 \theta & -1 \\
38-36 & 36 & 2
\end{array}\right| C_1 \rightarrow C_1-C_2=\left|\begin{array}{ccc}
1 & \tan ^2 \theta & 1 \\
-1 & \sec ^2 \theta & -1 \\
2 & 36 & 2
\end{array}\right| \\
& =0\left(\because \mathrm{C}_1=\mathrm{C}_3\right) \\
& =\text { RHS } \\
&
\end{aligned}$
Question 6.
$
\text { Show that }\left|\begin{array}{ccc}
x+2 a & y+2 b & z+2 c \\
x & y & z \\
a & b & c
\end{array}\right|=0 .
$
Solution:

$\begin{aligned}
\text { LHS } & =\left|\begin{array}{ccc}
x+2 a & y+2 b & 2+2 c \\
x & y & z \\
a & b & c
\end{array}\right| \\
& =\left|\begin{array}{ccc}
x & y & z \\
x & y & z \\
a & b & c
\end{array}\right|+\left|\begin{array}{ccc}
2 a & 2 b & 2 c \\
x & y & z \\
a & b & c
\end{array}\right| \\
& =\left(\because \mathrm{R}_1=\mathrm{R}_2\right)+2\left|\begin{array}{ccc}
a & b & c \\
x & y & z \\
a & b & c
\end{array}\right| \\
& =0+2(0)\left(\because \mathrm{R}_1=\mathrm{R}_3\right) \\
& =0=\mathrm{RHS}
\end{aligned}$

Question 7.
Write the general form of a $3 \times 3$ skew-symmetric matrix and prove that its determinant is 0 .

Solution:
$
\begin{aligned}
\text { Let } \mathrm{A} & =\left(\begin{array}{ccc}
0 & a & b \\
-a & 0 & -c \\
-b & c & 0
\end{array}\right) \\
\text { Here } \mathrm{A}^{\mathrm{T}} & =\left(\begin{array}{ccc}
0 & -a & -b \\
a & 0 & c \\
b & -c & 0
\end{array}\right)=-\mathrm{A}
\end{aligned}
$
$\Rightarrow \mathrm{A}$ is a skew symmetric matrix
Now
Question 8.
If $\left|\begin{array}{ccc}a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0\end{array}\right|=0$
Solution:

Let $\Delta=\left|\begin{array}{ccc}a & b & a \alpha+b \\ b & c & b \alpha+c \\ a \alpha+b & b \alpha+c & 0\end{array}\right|=\left|\begin{array}{ccc}a & b & a \alpha \\ b & c & b \alpha \\ a \alpha+b & b \alpha+c & -(b \alpha+c)\end{array}\right|\left(\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2\right)$ $=\left|\begin{array}{ccc}a & b & 0 \\ b & c & 0 \\ a \alpha+b & b \alpha+c & -(b \alpha+c) \\ & & -\left(a \alpha^2+b \alpha\right)\end{array}\right|\left(\mathrm{C}_3 \rightarrow \mathrm{C}_3-\alpha \mathrm{C}_1\right)$ $=\left|\begin{array}{ccc}a & b & 0 \\ b & c & 0 \\ a \alpha+b & b \alpha+c & -\left(a \alpha^2+2 b \alpha+c\right)\end{array}\right|$
expanding along $\mathrm{C}_3$
we get $-\left(a \alpha^2+2 b \alpha+c\right)\left[a c-b^2\right]$
So $\Delta=0 \Rightarrow\left(a \alpha^2+2 \mathrm{~b} \alpha+\mathrm{c}\right)\left(\mathrm{ac}-\mathrm{b}^2\right)=-0=0$
$\Rightarrow \mathrm{a} \alpha^2+2 \mathrm{~b} \alpha+\mathrm{c}=0$ or $\mathrm{ac}-\mathrm{b}^2=0$
(i.e.) $a$ is a root of $\mathrm{ax}^2+2 b x+c=0$
or $\mathrm{ac}=\mathrm{b}^2$
$\Rightarrow \mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P.
Question 9.

Prove that:
$\left|\begin{array}{lll}1 & a & a^2-b c \\ 1 & b & b^2-c a \\ 1 & c & c^2-a b\end{array}\right|=0$
Solution:

$
\begin{aligned}
& \text { LHS }=\left|\begin{array}{lll}
1 & a & a^2-b c \\
1 & b & b^2-a c \\
1 & c & c^2-a b
\end{array}\right| \\
& =\left|\begin{array}{ccc}
0 & a-b & a^2-b c-b^2+a c \\
0 & b-c & b^2-a c-c^2+a b \\
1 & c & c^2-a b
\end{array}\right| \begin{array}{l}
\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\end{array} \\
& =\left|\begin{array}{ccc}
0 & a-b & \left(a^2-b^2\right)+(a c-b c) \\
0 & b-c & \left(b^2-c^2\right)+(a b-a c) \\
1 & c & c^2-a b
\end{array}\right| \\
& =\left|\begin{array}{ccc}
0 & a-b & (a+b)(a-b)+c(a-b) \\
0 & b-c & (b+c)(b-c)+a(b-c) \\
1 & c & c^2-a b
\end{array}\right| \\
& =\left|\begin{array}{ccc}
0 & a-b & (a-b)(a+b+c) \\
0 & b-c & (b-c)(a+b+c) \\
1 & c & c^2-a b
\end{array}\right| \\
&
\end{aligned}
$
Taking $(a-b)$ and $(b-c)$ from $\mathrm{R}_1$ and $\mathrm{R}_2$ respectively
we get $(a-b)(b-c)\left|\begin{array}{ccc}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^2-a b\end{array}\right|$
expanding along $\mathrm{C}_1$
$
(a-b)(b-c)\{0-0+1[(a+b+c)-(a+b+c]\}=0=\text { RHS }
$

Question 10.
If $a, b, c$ are $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A.P., find the value of
$
\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
1 & 1 & 1
\end{array}\right|
$
Solution:
We are given $\mathrm{a}=\mathrm{t}_{\mathrm{p}}, \mathrm{b}=\mathrm{t}_{\mathrm{q}}$ and $\mathrm{c}=\mathrm{t}_{\mathrm{r}}$

Let a be the first term and $\mathrm{d}$ be the common difference
So $t_p=a+(p-1) d=a$
$
\text { and } \begin{aligned}
t_q & =a+(q-1) d=b \\
t_r & =a+(r-1) d=c
\end{aligned}
$
To find the value of $\left|\begin{array}{lll}a & b & c \\ p & q & r \\ 1 & 1 & 1\end{array}\right|$ :
$
\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
t_p & t_q & t_r \\
p & q & r \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
t_p-t_q & t_q-t_r & t_r \\
p-q & q-r & r \\
0 & 0 & 1
\end{array}\right| C_1 \rightarrow C_1-C_1
$
expanding along $\mathrm{R}_3$
$
\begin{aligned}
& 1\left\{\left(t_p-t_q\right)(q-r)-(p-q)\left(t_q-t_r\right)\right\} \\
& =q t_p-r t_p-q t_q+r t_q-p t_q+p t_r-q t_q-q t_r \\
& =(p-q) t r+(q-r) t p+(r-p) t q \\
& =(p-q)[a+(r-1) d+(q-r)[a+(p-1) d]+(r-p)[(a+(q-1) d] \\
& =a[p-q+q-r+r-p]+d[(r-1)(p-q)+(p-1)(q-r)+(q-1)(r-p)] \\
& =a(0)+d(0)=0
\end{aligned}
$
Question 11.
Show that
$
\left|\begin{array}{ccc}
a^2+x^2 & a b & a c \\
a b & b^2+x^2 & b c \\
a c & b c & c^2+x^2
\end{array}\right|
$
is divisible by $x^4$
$
\text { LHS }=\left|\begin{array}{ccc}
a^2+x^2 & a b & a c \\
a b & b^2+x^2 & b c \\
a c & b c & c^2+x^2
\end{array}\right|
$
Multiplying $R_1$ by a, $R_2$ by b and $R_3$ by c and

taking out a from $\mathrm{C}_1$ b from $\mathrm{C}_2$ and $\mathrm{c}$ from $\mathrm{C}_3$ we get
$
\begin{aligned}
& \frac{a b c}{a b c}\left|\begin{array}{ccc}
a^2+x^2 & a^2 & a^2 \\
b^2 & b^2+x^2 & b^2 \\
c^2 & c^2 & c^2+x^2
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a^2+b^2+c^2+x^2 & a^2+b^2+c^2+x^2 & a^2+b^2+c^2+x^2 \\
b^2 & b^2+x^2 & b^2 \\
c^2 & c^2 & c^2+x^2
\end{array}\right| R_1 \rightarrow R_1+R_2+R_3 \\
& =\left(a^2+b^2+c^2+x^2\right)\left|\begin{array}{ccc}
1 & 1 & 1 \\
b^2 & b^2+x^2 & b^2 \\
c^2 & c^2 & c^2+x^2
\end{array}\right| \\
& =\left(a^2+b^2+c^2+x^2\right)\left|\begin{array}{ccc}
0 & 0 & 1 \\
-x^2 & x^2 & b^2 \\
0 & -x^2 & c^2+x^2
\end{array}\right| \begin{array}{l}
C_1 \rightarrow C_1-C_2 \\
C_2 \rightarrow C_2-C_3
\end{array} \\
&
\end{aligned}
$
expanding along $R_1$ we get
$
\begin{aligned}
=\left(a^2+b^2+c^2+x^2\right)\left[1\left(-x^2\right)\left(-x^2\right)\right. & -0] \\
& =x^4\left(a^2+b^2+c^2+x^2\right)
\end{aligned}
$
$\Rightarrow$ The given determinant is divisible by $x^4$.
Question 12.
If a, b, c are all positive, and are ${ }^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a G.P., show that $\mathrm{p}\left|\begin{array}{lll}\log a & p & 1 \\ \log b & q & 1 \\ \log c & r & 1\end{array}\right|=0$.
Solution:

Let the G.P. be $k, k x, k x^2, \ldots .$.
expanding along $\mathrm{C}_3$
$
\begin{aligned}
& {[(\log a-\log b)(q-r)-(\log b-\log c)(p-q)] } \\
&=\left(\log \frac{a}{b}\right)(q-r)-\left(\log \frac{b}{c}\right)(p-q)
\end{aligned}
$
$
\begin{aligned}
& \left\{\begin{array}{c}
\text { Here } \frac{a}{b}=\frac{k x^{p-1}}{k x^{q-1}}=x^{p-1-q+1}=x^{p-q} \\
\text { and } \frac{b}{c}=\frac{k x^{q-1}}{k x^{r-1}}=x^{q-1-r+1}=x^{q-r}
\end{array}\right\} \\
& =(q-r)\left[\log x^{p-q}\right]-(p-q)\left[\log x^{q-r}\right] \\
& =(q-r)(p-q) \log x-(p-q)(q-r) \log x \\
& =0=\text { RHS } \\
&
\end{aligned}
$

Question 13.

find the value of 
$
\left|\begin{array}{ccc}
1 & \log _x y & \log _x z \\
\log _y x & 1 & \log _y z \\
\log _z x & \log _z y & 1
\end{array}\right|
$
if $x, y, z \neq 1$.
Solution:
Expanding the determinant along $\mathrm{R}_1$
$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & \log _x y & \log _x z \\
\log _y x & 1 & \log _y z \\
\log _z x & \log _z y & 1
\end{array}\right| \\
& =1\left[1-\log _z y \log _y z\right]-\log _x y\left[\log _y x-\log _z x \log _y z\right]+\log _x z \\
& =1[1-1]-\log _x y\left[\log _y x-\log _y x\right]+\log _x z\left[\log _z x-\log _z x\right]=0
\end{aligned}
$

Question 14.
If $A=\left[\begin{array}{cc}\frac{1}{2} & \alpha \\ 0 & \frac{1}{2}\end{array}\right]$, prove that $\sum_{k=1}^n \operatorname{det}\left(A^k\right)=\frac{1}{3}\left(1-\frac{1}{4^n}\right)$.
Solution:
$
\begin{aligned}
& \mathrm{A}=\left[\begin{array}{cc}
\frac{1}{2} & \alpha \\
0 & \frac{1}{2}
\end{array}\right] \quad|\mathrm{A}|=\left|\begin{array}{cc}
\frac{1}{2} & \alpha \\
0 & \frac{1}{2}
\end{array}\right|=\frac{1}{4}-0=\frac{1}{4} \\
& \mathrm{~A}^2=\mathrm{A} \times \mathrm{A}=\left(\begin{array}{cc}
\frac{1}{2} & \alpha \\
0 & \frac{1}{2}
\end{array}\right)\left(\begin{array}{cc}
\frac{1}{2} & \alpha \\
0 & \frac{1}{2}
\end{array}\right)=\left(\begin{array}{cc}
\frac{1}{4} & \alpha \\
0 & \frac{1}{4}
\end{array}\right) \\
& \left|A^2\right|=\left|\begin{array}{ll}
\frac{1}{4} & \alpha \\
0 & \frac{1}{4}
\end{array}\right|=\frac{1}{4} \times \frac{1}{4}-0=\left(\frac{1}{4}\right)^2=\frac{1}{4^2} \\
& \left|\mathrm{~A}^k\right|=\frac{1}{4^k} \\
&
\end{aligned}
$
So, $\sum_{k=1}^n \operatorname{det}\left(\mathrm{A}^k\right)=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\ldots .+\frac{1}{4^n}$
Which is a G.P with $a=\frac{1}{4}$ and $r=\frac{1}{4}$
$
\begin{aligned}
\therefore \quad \mathrm{S}_n & =\frac{a\left(1-r^n\right)}{1-r}=\frac{\frac{1}{4}\left[1-\left(\frac{1}{4}\right)^n\right]}{1-\frac{1}{4}} \\
& =\frac{\frac{1}{4}\left[1-\frac{1}{4^n}\right]}{\frac{3}{4}}=\frac{1}{4} \times \frac{4}{3}\left[1-\frac{1}{4^n}\right] \\
& =\frac{1}{3}\left[1-\frac{1}{4^n}\right] .
\end{aligned}
$

Question 15.
Without expanding, evaluate the following determinants:
(i) $\left|\begin{array}{ccc}2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x\end{array}\right|$
(ii) $\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$
Solution:
(i)
$
\begin{aligned}
\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 x & 9 x & 12 x
\end{array}\right| & =x\left|\begin{array}{ccc}
2 & 3 & 4 \\
5 & 6 & 8 \\
6 & 9 & 12
\end{array}\right| \\
& =(x)(3)\left|\begin{array}{lll}
2 & 3 & 4 \\
5 & 6 & 8 \\
2 & 3 & 4
\end{array}\right|=3 x(0)=0 \quad\left(\therefore \mathrm{R}_1=\mathrm{R}_3\right)
\end{aligned}
$
(ii)
$
\begin{aligned}
& \left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=\left|\begin{array}{ccc}
x+y+z & x+y+z & x+y+z \\
z & x & y \\
1 & 1 & 1
\end{array}\right| \mathrm{R}_1=\mathrm{R}_1+\mathrm{R}_2 \\
& =(x+y+z)\left|\begin{array}{lll}
1 & 1 & 1 \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0 \\
& \left(\because \mathrm{R}_1=\mathrm{R}_3\right) \\
&
\end{aligned}
$

Question 16.
If $A$ is a square matrix and $|A|=2$, find the value of $\left|A^T\right|$.
Solution:
$
\begin{aligned}
& |\mathrm{A}|=2\left(\text { Given) }\left|\mathrm{A}^{\mathrm{T}}\right|=2\right. \\
& \text { Now }\left|\mathrm{AA}^{\mathrm{T}}\right|=|\mathrm{A}|\left|\mathrm{A}^{\mathrm{T}}\right|=2 \times 2=4
\end{aligned}
$
Question 17.
If $A$ and $B$ are square matrices of order 3 such that $|A|=-1$ and $|B|=3$, find the value of $|3 A B|$.
Solution:
Given $|\mathrm{A}|=-1:|\mathrm{B}|=3$
Given $A$ and $B$ are square matrices of order 3 .
$
\begin{aligned}
& \therefore|\mathrm{kAB}|=\mathrm{k}^3|\mathrm{AB}| \\
& \text { Here } \mathrm{k}=3 \therefore|3 \mathrm{AB}|=3^3|\mathrm{AB}| \\
& =27|\mathrm{AB}| \\
& =27(-1)(3) \\
& =-81
\end{aligned}
$

Question 18.
$
\left|\begin{array}{ccc}
0 & 2 \lambda & 1 \\
\lambda^2 & 0 & 3 \lambda^2+1 \\
-1 & 6 \lambda-1 & 0
\end{array}\right|
$
If $\lambda=-2$, determine the value of

Solution:
Given $\lambda=-2$
$
\therefore 2 \lambda=-4 ; \lambda^2=(-2)^2 ; 3 \lambda^2+1=3(4)+1=13
$
$6 \lambda-1=6(-2)-1=-13$
So
$
\left|\begin{array}{ccc}
0 & 2 \lambda & 1 \\
\lambda^2 & 0 & 3 \lambda^2+1 \\
-1 & 6 \lambda-1 & 0
\end{array}\right|=\left|\begin{array}{ccc}
0 & -4 & 1 \\
4 & 0 & 13 \\
-1 & -13 & 0
\end{array}\right|
$
expanding along $R_1$
$
0(0)+4(0+13)+1(-52+0)=52-52=0
$
Aliter: The determinant value of a skew-symmetric matrix is zero

Question 19.

Determin
$
\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & 5 \\
1 & 2 x & 5 x^2
\end{array}\right|=0
$

Solution:
Now
$
\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & 5 \\
1 & 2 x & 5 x^2
\end{array}\right|=(5)\left|\begin{array}{ccc}
1 & 4 & 4 \\
1 & -2 & 1 \\
1 & 2 x & x^2
\end{array}\right|
$
(Taking 5 as a common factor from $\mathrm{C}_3$ )
$
\quad=(5)(2)\left|\begin{array}{ccc}
1 & 2 & 4 \\
1 & -1 & 1 \\
1 & x & x^2
\end{array}\right|
$
(Taking 2 as a common factor from $\mathrm{C}_2$ )
$
=10\left|\begin{array}{ccc}
1 & 2 & 4 \\
1 & -1 & 1 \\
1 & x & x^2
\end{array}\right|
$
expanding along $\mathrm{C}_1$
$
=10\left|\begin{array}{ccc}
0 & 3 & 3 \\
0 & -1-x & 1-x^2 \\
1 & x & x^2
\end{array}\right| \begin{aligned}
& \mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2 \\
& \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3
\end{aligned}
$
$10\left\{1\left[3\left(1-x^2\right)-(-1-x)(3)\right]\right\}$
$
\begin{aligned}
& =10\left[3\left(1-x^2\right)+3(1+x)\right] \\
& =10[3(1+x)(1-x)+3(1+x)] \\
& =10[3(1+x)(1-x+1)] \\
& =30(1+x)(2-x)
\end{aligned}
$

Given the determinant value is 0
$
\begin{aligned}
& \Rightarrow 30(1+\mathrm{x})(2-\mathrm{x})=0 \\
& \Rightarrow 1+\mathrm{x}=0 \text { or } 2-\mathrm{x}=0 \\
& \Rightarrow \mathrm{x}=-1 \text { or } \mathrm{x}=2
\end{aligned}
$
So, $\mathrm{x}=-1$ or 2 .
Question 20.
Verify that $\operatorname{det}(\mathrm{AB})=(\operatorname{det} A)(\operatorname{det} \mathrm{B})$ for
$
A=\left[\begin{array}{rrr}
4 & 3 & -2 \\
1 & 0 & 7 \\
2 & 3 & -5
\end{array}\right] \text { and } B=\left[\begin{array}{rrr}
1 & 3 & 3 \\
-2 & 4 & 0 \\
9 & 7 & 5
\end{array}\right]
$
Solution:
$
A=\left(\begin{array}{rrr}
4 & 3 & -2 \\
1 & 0 & 7 \\
2 & 3 & -5
\end{array}\right) \text { and } B=\left(\begin{array}{ccc}
1 & 3 & 3 \\
-2 & 4 & 0 \\
9 & 7 & 5
\end{array}\right)
$
$
\text { Now } \mathrm{AB}=\left(\overrightarrow{\left(\begin{array}{ccc}
4 & 3 & -2 \\
1 & 0 & 7 \\
2 & 3 & -5
\end{array}\right)} \downarrow\left(\begin{array}{ccc}
1 & 3 & 3 \\
-2 & 4 & 0 \\
9 & 7 & 5
\end{array}\right)\right.
$
$
=\left[\begin{array}{ccc}
-20 & 10 & 2 \\
64 & 52 & 38 \\
-49 & -17 & -19
\end{array}\right]
$
Now
(expanding along $\mathrm{R}_2$ )
$
\begin{aligned}
|A| & =\left|\begin{array}{ccc}
4 & 3 & -2 \\
1 & 0 & 7 \\
2 & 3 & -5
\end{array}\right| \\
& =(-1)[-15+6]-7[12-6] \\
& =(-1)(-9)-7(6) \\
& =9-42=-33=|A|=-33
\end{aligned}
$
$
|B|=\left|\begin{array}{ccc}
1 & 3 & 3 \\
-2 & 4 & 0 \\
9 & 7 & 5
\end{array}\right|
$

(expanding along $\mathrm{C}_3$ )
$
\begin{aligned}
= & 3(-14-36)-0(-14-36)+5(4+6) \\
= & 3(-50)+5(10)=-150+50=-100 \\
& |B|=-100
\end{aligned}
$
$
\begin{aligned}
& \qquad|\mathrm{AB}|=\left|\begin{array}{ccc}
-20 & 10 & 2 \\
64 & 52 & 38 \\
-49 & -17 & -19
\end{array}\right| \\
& \{(-20)(52)(-19)+(10)(38)(-49)+(2)(64)(-17)\}-\{(-49)(52)(2)+(-17)(38)(-20)+(-19)(64)(10)\} \\
& =(19760-18620-2176)-(-5096+12920-12160) \\
& =(19760+5096+12160)-(18620+2176+12920) \\
& =37016-33716=3300 \ldots(3) \\
& \text { Now }(1) \times(2)=(3) \\
& (\text { i.e., }(-33)(-100)=3300 \\
& \Rightarrow \operatorname{det}(\mathrm{AB})=(\operatorname{det} \mathrm{A}),(\operatorname{det} \mathrm{B})
\end{aligned}
$

Question 21.
Using cofactors of elements of second row, evaluate $|A|$, where
$
A=\left[\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right] .
$
Solution:
$
\begin{aligned}
A & =\left[\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right] \\
|A| & =\left[\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right]
\end{aligned}
$
expanding along $R_2$
Cofactor of $2=-\left[\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right]=(-9+16)=7$
Cofactor of $0=+\left[\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right]=15-8=7$
Cofactor of $1=-\left[\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right]=-(10-3)=-7$
$
\begin{array}{rlrl}
\therefore & |A| & =2(7)+0(7)+(1)(-7) \\
& =14-7=7
\end{array}
$