Miscellaneous Exercise (Revised) - Chapter 4 - Determinants - Ncert Solutions class 12 - Maths

---

You can Download the Miscellaneous Exercise (Revised) - Chapter 4 - Determinants - Ncert Solutions class 12 - Maths with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Chapter 4 - Determinants NCERT Solutions Class 12 Maths | Step-by-Step Solutions & Explanations

Miscellaneous Exercise Question 1:
$
\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|_{\text {is independent of } \theta .}
$

Answer
$
\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right| \\
& =x\left(x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta) \\
& =x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta \\
& =x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right) \\
& =x^3-x+x \\
& =x^3 \text { (Independent of } \theta \text { ) }
\end{aligned}
$

Hence, $\Delta$ is independent of $\theta$.

Miscellaneous Exercise Question 2:
Evaluate $\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right|$

Answer
$
\Delta=\left|\begin{array}{ccc}
\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\
-\sin \beta & \cos \beta & 0 \\
\sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha
\end{array}\right|
$

Expanding along $C_3$, we have:

$\begin{aligned}
\Delta & =-\sin \alpha\left(-\sin \alpha \sin ^2 \beta-\cos ^2 \beta \sin \alpha\right)+\cos \alpha\left(\cos \alpha \cos ^2 \beta+\cos \alpha \sin ^2 \beta\right) \\
& =\sin ^2 \alpha\left(\sin ^2 \beta+\cos ^2 \beta\right)+\cos ^2 \alpha\left(\cos ^2 \beta+\sin ^2 \beta\right) \\
& =\sin ^2 \alpha(1)+\cos ^2 \alpha(1) \\
& =1
\end{aligned}$

Miscellaneous Exercise Question 3:

If $A^{-1}=\left[\begin{array}{ccc}3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right]$, Find $(A B)^{-1}$

Answer

We know that
$
(A B)^{-1}=B^{-1} A^{-1}
$

We are given $A^{-1}$, so calculating $B^{-1}$

Calculating $\mathrm{B}^{-1}$
We know that
$
\begin{aligned}
B^{-1}= & \frac{1}{|B|} \operatorname{adj}(B) \\
& \text { exists if }|B| \neq 0
\end{aligned}
$

$
\begin{aligned}
|\mathbf{B}| & =\left|\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right| \\
& =1(3-0)-2(-1-0)-2(2-0)=1(3)-2(-1)-2(2) \\
& =3+2-4=1
\end{aligned}
$

Since $|\mathbf{B}| \neq \mathbf{0}$
Thus, $B^{-1}$ exists

Calculating adj B

Now, adj $B=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]$

$\begin{aligned}
& B=\left[\begin{array}{ccc}
1 & 2 & -2 \\
-1 & 3 & 0 \\
0 & -2 & 1
\end{array}\right] \\
& M_{11}=\left|\begin{array}{cc}
3 & 0 \\
-2 & 1
\end{array}\right|=3(1)-(-2) 0=3 \\
& M_{12}=\left|\begin{array}{cc}
-1 & 0 \\
0 & 1
\end{array}\right|=-1(1)-0(0)=-1 \\
& M_{13}=\left|\begin{array}{cc}
-1 & 3 \\
0 & -2
\end{array}\right|=(-1)(-2)-0(3)=2 \\
& M_{21}=\left|\begin{array}{cc}
2 & -2 \\
-2 & 1
\end{array}\right|=2(1)-(-2)(-2)=-2 \\
& M_{22}=\left|\begin{array}{cc}
1 & -2 \\
0 & 1
\end{array}\right|=1(1)-0(-2)=1
\end{aligned}$

$\begin{aligned}
& M_{23}=\left|\begin{array}{cc}
1 & 2 \\
0 & -2
\end{array}\right|=1(-2)-0(2)=-2 \\
& M_{31}=\left|\begin{array}{cc}
2 & -2 \\
3 & 0
\end{array}\right|=2(0)-3(-2)=6 \\
& M_{32}=\left|\begin{array}{cc}
1 & -2 \\
-1 & 0
\end{array}\right|=1(0)-(-1)(-2)=-2 \\
& M_{33}=\left|\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right|=1(3)-(-1) 2=5
\end{aligned}$

Now,
$
\begin{aligned}
& A_{11}=(-1)^{1+1} M_{11}=(-1)^2 \cdot 3=3 \\
& A_{12}=(-1)^{1+2} M_{12}=(-1)^3(-1)=1 \\
& A_{13}=(-1)^{1+3} M_{13}=(-1)^4 2=2 \\
& A_{21}=(-1)^{2+1} M_{21}=(-1)^3(-2)=2 \\
& A_{22}=(-1)^{2+2} M_{22}=(-1)^4 \cdot 1=1 \\
& A_{23}=(-1)^{2+3} M_{23}=(-1)^5(-2)=2 \\
& A_{31}=(-1)^{3+1} M_{31}=(-1)^4 \cdot 6=6 \\
& A_{32}=(-1)^{3+2} M_{32}=(-1)^5(-2)=2 \\
& A_{33}=(-1)^{3+3} M_{33}=(-1)^6 \cdot 5=5
\end{aligned}
$
$
\text { Thus, } \operatorname{adj}(B)=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]
$

Now,
$
\mathbf{B}^{-1}=\frac{1}{|\mathbf{B}|} \operatorname{adj}(\mathbf{B})
$

Putting values
$
=\frac{1}{1}\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]
$

Also,
$
\begin{aligned}
& (A B)^{-1} \\
& =B^{-1} A^{-1} \\
& =\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right]\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right] \\
& =\left[\begin{array}{lll}
3(3)+2(-15)+6(-5) & 3(-1)+2(6)+6(-2) & 3(1)+2(-5)+6(2) \\
1(3)+1(-15)+2(-5) & 1(-1)+1(6)+2(-2) & 1(1)+1(-5)+2(2) \\
2(3)+2(-15)+5(-5) & 2(-1)+2(6)+5(-2) & 2(1)+2(-5)+5(2)
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
& =\left[\begin{array}{ccc}
9-30+30 & -3+12-12 & 3-10+12 \\
3-15+10 & -1+6-4 & 1-5+4 \\
6-30+25 & -2+12-10 & 2-10+10
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\mathbf{9} & -\mathbf{3} & \mathbf{5} \\
\mathbf{- 2} & \mathbf{1} & \mathbf{0} \\
\mathbf{1} & \mathbf{0} & \mathbf{2}
\end{array}\right]
\end{aligned}$

Miscellaneous Exercise Question 4: (i)

Let $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$ verify that
(i) $[\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right)$

Answer:

First we will calculate adj $(A) \& A^{-1}$
$\operatorname{adj} A=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right]^{\prime}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]$
$
A=\left[\begin{array}{ccc}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right]
$

$\begin{aligned}
& M_{11}=\left|\begin{array}{ll}
3 & 1 \\
1 & 5
\end{array}\right|=15-1=14 \\
& M_{12}=\left|\begin{array}{cc}
-2 & 1 \\
1 & 5
\end{array}\right|=-10-1=-11 \\
& M_{13}=\left|\begin{array}{cc}
-2 & 3 \\
1 & 1
\end{array}\right|=-2-3=-5 \\
& M_{21}=\left|\begin{array}{cc}
-2 & 1 \\
1 & 5
\end{array}\right|=-10-4=-11 \\
& M_{22}=\left|\begin{array}{cc}
1 & 1 \\
1 & 5
\end{array}\right|=5-1=4 \\
& M_{23}=\left|\begin{array}{cc}
1 & -2 \\
1 & 1
\end{array}\right|=1+2=3 \\
& M_{31}=\left|\begin{array}{cc}
-2 & 1 \\
3 & 1
\end{array}\right|=-2-3=-5 \\
& M_{32}=\left|\begin{array}{cc}
1 & 1 \\
-2 & 1
\end{array}\right|=1+2=3 \\
& M_{33}=\left|\begin{array}{cc}
1 & -2 \\
-2 & 3
\end{array}\right|=3-4=-5
\end{aligned}$

$\begin{aligned}
& A_{11}=(-1)^{1+1} M_{11}=(-1)^2-14=14 \\
& A_{12}=(-1)^{1+2} M_{12}=(-1)^3(-11)=11 \\
& A_{13}=(-1)^{1+3} M_{13}=(-1)^4(-5)=-5 \\
& A_{21}=(-1)^{2+1} M_{21}=(-1)^3 \cdot(-11)=11 \\
& A_{22}=(-1)^{2+2} M_{22}=(-1)^4 \cdot 4=4 \\
& A_{23}=(-1)^{2+3} M_{23}=(-1)^5(3)=-3 \\
& A_{31}=(-1)^{3+1} M_{31}=(-1)^4 \cdot(-5)=-5 \\
& A_{32}=(-1)^{3+2} M_{32}=(-1)^5 \cdot(3)=-3 \\
& A_{33}=(-1)^{3+3} M_{33}=(-1)^6 \cdot(-5)=-1 \\
& \text { Thus, adj (A) }=\left[\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right]
\end{aligned}$

$=\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]$

Now,
$
A^{-1}=\frac{1}{|A|} \operatorname{adj}(B)
$

Finding $|A|$
$
\begin{aligned}
|A| & =\left|\begin{array}{ccc}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right| \\
& =1(15-1)+2(-10-1)+1(-2-3)=14-22-5=-13
\end{aligned}
$

Therefore
$
A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=\frac{1}{13}\left[\begin{array}{ccc}
\mathbf{1 4} & \mathbf{1 1} & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]
$

We need to verify
$
[\operatorname{adj} A]^{-1}=\operatorname{adj}\left(A^{-1}\right)
$

Solving L.H.S
$
(\operatorname{adj} A)^{-1}
$

Let $B=\operatorname{adj}(A)$
$
B=\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]
$

Now,
$
B^{-1}=\frac{1}{|B|} \operatorname{adj}(B)
$
$
\text { exists if }|B| \neq 0
$

$
\begin{aligned}
|B| & =\left|\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right| \\
& =14(-4-9)+1(-11-15)-5(-33+20) \\
& =14(-13)-11(-26)-5(-13)=-182+286+65=169
\end{aligned}
$
Thus $|\mathrm{B}|=169 \neq 0$
$\therefore \mathrm{B}^{-1}$ exist

Now, calculating adj B
$
\operatorname{adj} B=\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]^{\prime}=\left[\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right]
$

Here $A_{i j}$ are the cofactors of matrix $B$

$\begin{aligned}
& B=\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right] \\
& M_{11}=\left|\begin{array}{cc}
4 & -3 \\
-3 & -1
\end{array}\right|=-4-9=-13 \\
& M_{12}=\left|\begin{array}{cc}
11 & -3 \\
-5 & -1
\end{array}\right|=-11-15=-26 \\
& M_{13}=\left|\begin{array}{cc}
11 & 4 \\
-5 & -3
\end{array}\right|=-33+20=-13 \\
& M_{21}=\left|\begin{array}{cc}
11 & -5 \\
-3 & -1
\end{array}\right|=-11-15=-26 \\
& M_{22}=\left|\begin{array}{cc}
14 & -5 \\
-5 & -1
\end{array}\right|=-14-25=-39 \\
& M_{23}=\left|\begin{array}{cc}
14 & 11 \\
-5 & -3
\end{array}\right|=(-42+55)=+13
\end{aligned}$

$\begin{aligned}
&\begin{aligned}
& M_{31}=\left|\begin{array}{cc}
11 & -5 \\
4 & -3
\end{array}\right|=-33-20=-13 \\
& M_{32}=\left|\begin{array}{cc}
14 & 5 \\
-11 & 3
\end{array}\right|=-42+55=13 \\
& M_{33}=\left|\begin{array}{cc}
14 & 11 \\
-11 & 4
\end{array}\right|=56-121=-65
\end{aligned}\\
&\begin{aligned}
& A_{11}=(-1)^{1+1} M_{11}=(-1)^2(-13)=-13 \\
& A_{12}=(-1)^{1+2} M_{12}=(-1)^3(-26)=26 \\
& A_{13}=(-1)^{1+3} M_{13}=(-1)^4 \cdot(-13)=-13 \\
& A_{21}=(-1)^{2+1} M_{21}=(-1)^3 \cdot(-26)=26 \\
& A_{22}=(-1)^{2+2} M_{22}=(-1)^4 \cdot(-39)=-39 \\
& A_{23}=(-1)^{2+3} M_{23}=(-1)^5 \cdot(-13)=-13 \\
& A_{31}=(-1)^{3+1} M_{31}=(-1)^4 \cdot(-13)=-13
\end{aligned}\\
&\begin{aligned}
& A_{32}=(-1)^{3+2} M_{32}=(-1)^5 \cdot(13)=-13 \\
& A_{33}=(-1)^{3+3} M_{33}=(-1)^6 \cdot(-65)=-65
\end{aligned}
\end{aligned}$

Thus, adj $B=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right]=\left[\begin{array}{ccc}-13 & 26 & 13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]$

Now,
$
\mathrm{B}^{-1}=\frac{1}{|\mathrm{~B}|}(\operatorname{adj} \mathrm{B})=\frac{1}{169}\left[\begin{array}{ccc}
-13 & 26 & 13 \\
26 & -39 & -13 \\
-13 & -13 & -65
\end{array}\right]
$

Taking 13 common from all elements of the matrix
$
\begin{aligned}
& =\frac{\mathbf{1 3}}{169}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right] \\
& =\frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]
\end{aligned}
$

Since 13 is common in all elements of the matrix, 13 gets multiplied to the matrix

Thus, $[\operatorname{adj} A]^{-1}=B^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]$
Solving R.H.S
$\operatorname{adj}\left(\mathrm{A}^{-1}\right)$
$
A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}
-14 & -11 & 5 \\
-11 & -4 & 3 \\
5 & 3 & 1
\end{array}\right]
$

Let $C=A^{-1}$
$
\begin{aligned}
C & =\frac{1}{13}\left[\begin{array}{ccc}
-14 & -11 & 5 \\
-11 & -4 & 3 \\
5 & 3 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\frac{-14}{13} & \frac{-11}{13} & \frac{5}{13} \\
\frac{-11}{13} & \frac{-4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right]
\end{aligned}
$

Since 13 is divided to the matrix, 13 gets divided to all elements of the matrix)

Now, $\operatorname{adj} C=\operatorname{adj}\left(A^{-1}\right)$
$
\operatorname{adj} C=\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]^{\prime}=\left[\begin{array}{lll}
A_{11} & A_{21} & A_{31} \\
A_{12} & A_{22} & A_{32} \\
A_{13} & A_{23} & A_{33}
\end{array}\right]
$

Here $\mathrm{A}_{\mathrm{ij}}$ are the cofactors of matrix $\mathrm{C}$
$
\begin{gathered}
C=\left[\begin{array}{ccc}
\frac{-14}{13} & \frac{-11}{13} & \frac{5}{13} \\
\frac{-11}{13} & \frac{-4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right] \\
M_{11}=\left|\begin{array}{cc}
\frac{-4}{13} & \frac{-3}{13} \\
\frac{3}{13} & \frac{1}{13}
\end{array}\right|=\frac{-4}{169}-\frac{9}{169}=\frac{-13}{169}=\frac{-1}{13} \\
M_{12}=\left|\begin{array}{cc}
\frac{-11}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{1}{13}
\end{array}\right|=\frac{-11}{169}-\frac{15}{169}=\frac{-26}{169}=\frac{-2}{13}
\end{gathered}
$

$\begin{aligned}
& M_{13}=\left|\begin{array}{cc}
\frac{-11}{13} & \frac{-4}{13} \\
\frac{5}{13} & \frac{3}{13}
\end{array}\right|=\frac{-33}{169}+\frac{20}{169}=\frac{-13}{169}=\frac{-1}{13} \\
& M_{21}=\left|\begin{array}{ll}
\frac{-11}{13} & \frac{5}{13} \\
\frac{3}{13} & \frac{1}{13}
\end{array}\right|=\frac{-11}{169}-\frac{55}{169}=\frac{-26}{169}=\frac{-2}{13} \\
& M_{22}=\left|\begin{array}{ll}
\frac{-14}{13} & \frac{5}{13} \\
\frac{5}{13} & \frac{1}{13}
\end{array}\right|=\frac{-14}{169}-\frac{25}{169}=\frac{-39}{169}=\frac{-3}{13}
\end{aligned}$

$\begin{aligned}
& M_{23}=\left|\begin{array}{cc}
\frac{-14}{13} & \frac{-11}{13} \\
\frac{5}{13} & \frac{3}{13}
\end{array}\right|=\left(\frac{-42}{169}+\frac{55}{169}\right)=\frac{13}{169}=\frac{1}{13} \\
& M_{31}=\left|\begin{array}{cc}
\frac{-11}{13} & \frac{5}{13} \\
\frac{-4}{13} & \frac{3}{13}
\end{array}\right|=\frac{-33}{169}+\frac{20}{167}=\frac{-13}{169}=\frac{-1}{169} \\
& M_{32}=\left|\begin{array}{cc}
\frac{-14}{13} & \frac{5}{13} \\
\frac{-11}{13} & \frac{3}{13}
\end{array}\right|=\frac{-42}{169}+\frac{55}{169}=\frac{13}{169}=\frac{1}{13} \\
& M_{33}=\left|\begin{array}{cc}
\frac{-14}{13} & \frac{-11}{13} \\
\frac{-11}{13} & \frac{-4}{13}
\end{array}\right|=\frac{56}{169}-\frac{-121}{169}=\frac{-65}{169}=\frac{-5}{13}
\end{aligned}$

$\begin{aligned}
& A_{11}=(-1)^{1+1} M_{11}=(-1)^2\left(\frac{-1}{13}\right)=\frac{-1}{13} \\
& A_{12}=(-1)^{1+2} M_{12}=(-1)^3\left(\frac{-2}{13}\right)=\frac{2}{13} \\
& A_{13}=(-1)^{1+3} M_{13}=(-1)^4 \cdot \frac{-1}{13}=\frac{-1}{13} \\
& A_{21}=(-1)^{2+1} M_{21}=(-1)^3 \cdot\left(\frac{-2}{13}\right)=\frac{2}{13} \\
& A_{22}=(-1)^{2+2} M_{22}=(-1)^4 \cdot\left(\frac{-3}{13}\right)=\frac{-3}{13} \\
& A_{23}=(-1)^{2+3} M_{23}=(-1)^5 \cdot\left(\frac{1}{13}\right)=\frac{-1}{13} \\
& A_{31}=(-1)^{3+1} M_{31}=(-1)^4 \cdot\left(\frac{-1}{169}\right)=\frac{-1}{13} \\
& A_{32}=(-1)^{3+2} M_{32}=(-1)^5 \cdot\left(\frac{1}{13}\right)=\frac{-1}{13} \\
& A_{33}=(-1)^{3+3} M_{33}=(-1)^6 \cdot\left(\frac{-5}{13}\right)=\frac{-5}{13} \\
& \text { Thus, adj } C=\left[\begin{array}{ccc}
\frac{-1}{13} & \frac{2}{13} & \frac{-1}{13} \\
\frac{2}{13} & \frac{-3}{13} & \frac{-1}{13} \\
\frac{-1}{13} & \frac{-1}{13} & \frac{-5}{13}
\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]
\end{aligned}$

$
\therefore \operatorname{adj}\left(A^{-1}\right)=\operatorname{adj} C=\frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]=\text { R.H.S }
$

Hence L.H.S = R.H.S
$
\therefore(\operatorname{adj} A)^{-1}=\operatorname{adj}\left(A^{-1}\right)
$

Miscellaneous Exercise Question 4 (ii)

Let $A=\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]$ verify that
(ii) $\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}$

We have to find $\left(A^{-1}\right)^{-1}$
So, $\left(A^{-1}\right)^{-1}=\frac{1}{\left|A^{-1}\right|} \operatorname{adj}\left(A^{-1}\right)$

From First part,
$
A^{-1}=\frac{1}{13}\left[\begin{array}{ccc}
-14 & -11 & 5 \\
-11 & -4 & 3 \\
5 & 3 & 1
\end{array}\right]
$
Calculating $\left|A^{-1}\right|$
$
\left|A^{-1}\right|=\left|\frac{1}{13}\left[\begin{array}{ccc}
-14 & -11 & 5 \\
-11 & -4 & 3 \\
5 & 3 & 1
\end{array}\right]\right|
$

Using $|\boldsymbol{k A}|=\boldsymbol{k n}|\boldsymbol{A}|$ Where $n$ is order of $A$
$
\begin{aligned}
& =\left(\frac{1}{13}\right)^3\left(-14\left|\begin{array}{cc}
-4 & 3 \\
3 & 1
\end{array}\right|-(-11)\left|\begin{array}{cc}
-11 & 3 \\
5 & 1
\end{array}\right|+5\left|\begin{array}{cc}
-11 & -4 \\
5 & 3
\end{array}\right|\right) \\
& =\left(\frac{1}{13}\right)^3(-14(-4-9)+11(-11-15)+5(-33+20)) \\
& =\left(\frac{1}{13}\right)^3(-14(-13)+11(-26)+5(-13)) \\
& =\left(\frac{1}{13}\right)^3(182-286-65)=\left(\frac{1}{13}\right)^3(-169)=\frac{-169}{13 \times 13 \times 13} \\
& =\frac{1}{-13}
\end{aligned}
$

Now,
$
\left(A^{-1}\right)^{-1}=\frac{1}{\left|A^{-1}\right|}\left(\operatorname{adj} A^{-1}\right)
$

Putting values

$=\frac{1}{\frac{-1}{13}} \times \frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]$

$=-13 \times \frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]$

$\begin{aligned}
& =-\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]=\left[\begin{array}{ccc}
-\mathbf{1} & \mathbf{2} & \mathbf{- 1} \\
\mathbf{2} & \mathbf{- 3} & \mathbf{- 1} \\
\mathbf{- 1} & \mathbf{- 1} & \mathbf{- 5}
\end{array}\right] \\
& =A
\end{aligned}$

Thus, $\left(A^{-1}\right)^{-1}=A$
Hence Proved

Miscellaneous Exercise Question 5:
Evaluate $\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right|$
Answer
$
\Delta=\left|\begin{array}{ccc}
x & y & x+y \\
y & x+y & x \\
x+y & x & y
\end{array}\right|
$

Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3$, we have:
$
\Delta=\left|\begin{array}{ccc}
2(x+y) & 2(x+y) & 2(x+y) \\
y & x+y & x \\
x+y & x & y
\end{array}\right|
$

$
=2(x+y)\left|\begin{array}{ccc}
1 & 1 & 1 \\
y & x+y & x \\
x+y & x & y
\end{array}\right|
$

Applying $\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1$ and $\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1$, we have:
$
\Delta=2(x+y)\left|\begin{array}{ccc}
1 & 0 & 0 \\
y & x & x-y \\
x+y & -y & -x
\end{array}\right|
$

Expanding along $R_1$, we have:
$
\begin{aligned}
\Delta & =2(x+y)\left[-x^2+y(x-y)\right] \\
& =-2(x+y)\left(x^2+y^2-y x\right) \\
& =-2\left(x^3+y^3\right)
\end{aligned}
$

Miscellaneous Exercise Question 6:
Evaluate $\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right|$
Answer
$
\Delta=\left|\begin{array}{ccc}
1 & x & y \\
1 & x+y & y \\
1 & x & x+y
\end{array}\right|
$

Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$, we have:
$
\Delta=\left|\begin{array}{lll}
1 & x & y \\
0 & y & 0 \\
0 & 0 & x
\end{array}\right|
$

Expanding along $C_1$, we have:
$
\Delta=1(x y-0)=x y
$

Miscellaneous Exercise Question 7:
Solve the system of the following equations
$
\begin{aligned}
& \frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4 \\
& \frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1 \\
& \frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2
\end{aligned}
$

Answer
Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$.
Then the given system of equations is as follows:
$
\begin{aligned}
& 2 p+3 q+10 r=4 \\
& 4 p-6 q+5 r=1 \\
& 6 p+9 q-20 r=2
\end{aligned}
$

This system can be written in the form of $A X=B$, where
$
A=\left[\begin{array}{rrc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right], X=\left[\begin{array}{l}
p \\
q \\
r
\end{array}\right] \text { and } B=\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right] \text {. }
$.

Now
$
\begin{aligned}
|A| & =2(120-45)-3(-80-30)+10(36+36) \\
& =150+330+720 \\
& =1200
\end{aligned}
$

Thus, $A$ is non-singular. Therefore, its inverse exists.
Now,
$
\begin{aligned}
A_{11} & =75, A_{12}=110, A_{13}=72 \\
A_{21} & =150, A_{22}=-100, A_{23}=0 \\
A_{31} & =75, A_{32}=30, A_{33}=-24 \\
\therefore A^{-1} & =\frac{1}{|A|} \operatorname{adj} A \\
& =\frac{1}{1200}\left[\begin{array}{lll}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]
\end{aligned}
$

Now,
$
\begin{aligned}
& X=A^{-1} B \\
& \Rightarrow\left[\begin{array}{l}
p \\
q \\
r
\end{array}\right]=\frac{1}{1200}\left[\begin{array}{lll}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right] \\
& \quad=\frac{1}{1200}\left[\begin{array}{c}
300+150+150 \\
440-100+60 \\
288+0-48
\end{array}\right]
\end{aligned}
$

$
\begin{gathered}
=\frac{1}{1200}\left[\begin{array}{l}
600 \\
400 \\
240
\end{array}\right]=\left[\begin{array}{c}
\frac{1}{2} \\
\frac{1}{3} \\
\frac{1}{5}
\end{array}\right] \\
\therefore p=\frac{1}{2}, q=\frac{1}{3} \text {, and } r=\frac{1}{5}
\end{gathered}
$

Hence, $x=2, v=3$, and $z=5$.

Miscellaneous Exercise Question 8:
Choose the correct answer.

$\text { If } x, y, z \text { are nonzero real numbers, then the inverse of matrix } A=\left[\begin{array}{ccc}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]_{\text {is }}$

$\text { A. }\left[\begin{array}{lll}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]$

B.

$x y z\left[\begin{array}{lll}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]$

C. $\frac{1}{x y z}\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]_{\text {D. }} \frac{1}{x y z}\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

Answer

Answer: A
$
\begin{aligned}
& A=\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right] \\
& \therefore|A|=x(y z-0)=x y z \neq 0
\end{aligned}
$

Now,
$
\begin{array}{r}
\text { Now, } A_{11}=y z, A_{12}=0, A_{13}=0 \\
A_{21}=0, A_{22}=x z, A_{23}=0 \\
A_{31}=0, A_{32}=0, A_{33}=x y \\
\therefore \text { adj } A=\left[\begin{array}{lll}
y z & 0 & 0 \\
0 & x z & 0 \\
0 & 0 & x y
\end{array}\right] \\
\therefore A^{-1}=\frac{1}{|A|} \text { adj } A
\end{array}
$

$
\begin{aligned}
= & \frac{1}{x y z}\left[\begin{array}{lll}
y z & 0 & 0 \\
0 & x z & 0 \\
0 & 0 & x y
\end{array}\right] \\
& =\left[\begin{array}{lll}
\frac{y z}{x y z} & 0 & 0 \\
0 & \frac{x z}{x y z} & 0 \\
0 & 0 & \frac{x y}{x y z}
\end{array}\right] \\
& =\left[\begin{array}{lll}
\frac{1}{x} & 0 & 0 \\
0 & \frac{1}{y} & 0 \\
0 & 0 & \frac{1}{z}
\end{array}\right]=\left[\begin{array}{lll}
x^{-1} & 0 & 0 \\
0 & y^{-1} & 0 \\
0 & 0 & z^{-1}
\end{array}\right]
\end{aligned}
$

The correct answer is A.

Miscellaneous Exercise Question 9:
Choose the correct answer.
$
A=\left[\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right] \text {, where } 0 \leq \theta \leq 2 \pi \text {, then }
$
A. $\operatorname{Det}(A)=0$
B. $\operatorname{Det}(A) \in(2, \infty)$
C. $\operatorname{Det}(A) \in(2,4)$
D. $\operatorname{Det}(A) \in[2,4]$

Answer

$
\begin{aligned}
& A=\left[\begin{array}{ccc}
1 & \sin \theta & 1 \\
-\sin \theta & 1 & \sin \theta \\
-1 & -\sin \theta & 1
\end{array}\right] \\
& \begin{aligned}
\therefore|A| & =1\left(1+\sin ^2 \theta\right)-\sin \theta(-\sin \theta+\sin \theta)+1\left(\sin ^2 \theta+1\right) \\
& =1+\sin ^2 \theta+\sin ^2 \theta+1 \\
& =2+2 \sin ^2 \theta \\
& =2\left(1+\sin ^2 \theta\right)
\end{aligned}
\end{aligned}
$

Now, $0 \leq \theta \leq 2 \pi$
$
\begin{aligned}
& \Rightarrow 0 \leq \sin \theta \leq 1 \\
& \Rightarrow 0 \leq \sin ^2 \theta \leq 1 \\
& \Rightarrow 1 \leq 1+\sin ^2 \theta \leq 2 \\
& \Rightarrow 2 \leq 2\left(1+\sin ^2 \theta\right) \leq 4 \\
& \therefore \operatorname{Det}(A) \in[2,4]
\end{aligned}
$

The correct answer is D.