\(|A|=\left|\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right|=-4-15=-19 \neq 0\)
Matrix formed by cofactor of each element in IAI
\(A_{11}=-2 \quad A_{12}=-5 \quad A_{21}=-3 \quad A_{22}=2\)
\(\therefore \operatorname{adj} A =\left[\begin{array}{cc} -2 & -5 \\ -3 & 2 \end{array}\right]^{\prime}=\left[\begin{array}{lr} -2 & -3 \\ -5 & 2 \end{array}\right] \)
\(\therefore A^{-1} =\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-19}\left[\begin{array}{rr} -2 & -3 \\ -5 & 2 \end{array}\right] \)
\(=\frac{-1}{-19}\left[\begin{array}{rr} 2 & 3 \\ 5 & -2 \end{array}\right]=\frac{1}{19} A . \)
\(\therefore A^{-1} =k A \)
\(\text { So, } k =\frac{1}{19}\)

Let \(A=\left[\begin{array}{rrr} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right]\) 
Now, \(|A|=\left|\begin{array}{rrr} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{array}\right|\) 
= 2(-1-0) -1 (4 -0) + 3(8-7)
= -2 - 4 + 3 = -6 + 3
= -3 ≠ 0
Hence, A is a non-singular matrix.

土15

x=2

We have:
\(A=\left| \begin{matrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{matrix} \right| \)
\(\therefore\ B=A'=\begin{bmatrix} 1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}\)
Now \(|B|=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1 \end{bmatrix}\)
= (1)(-1-8)-0-2(-8+3)
= -9 + 10 = 1 ≠ 0
= B^-1 Exsits
Now \(B_{11}=\begin{vmatrix}-1&2\\4&1 \end{vmatrix}=-1-8=-9;\)
\(B_{12}=-\begin{vmatrix}-2&2\\3&1 \end{vmatrix}=-(-2-6)=8;\)
\(B_{13}=\begin{vmatrix}-2&-1\\3&4 \end{vmatrix}=-8+3=-5\)
\(B_{21}=-\begin{vmatrix} 0&-2\\4&1\end{vmatrix}=-(0+8)=-8;\)
\(B_{22}=\begin{vmatrix} 1&-2\\3&1\end{vmatrix}=1+6=7;\)
\(B_{23}=-\begin{vmatrix} 1&0\\3&4\end{vmatrix}=-(4-0)=-4\)
\(B_{31}=\begin{vmatrix}0&-2\\-1&2 \end{vmatrix}=0-2=-2;\)
\(B_{32}=-\begin{vmatrix} 1&-2\\-2&2\end{vmatrix}=-(2-4)=2;\)
\(B_{33}=\begin{vmatrix} 1&0\\-2&-1\end{vmatrix}=-1+0=-1\)
\(\therefore\ adj.B=\begin{bmatrix}B_{11 }&B_{12}&B_{13}\\B_{21}&B_{22}&B_{23}\\B_{31}&B_{32}&B_{33}\end{bmatrix}=\begin{bmatrix} -9&8&-5\\-8&7&-4\\-2&2/&-1\end{bmatrix}\)
\(=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
\(\therefore\ B^{-1}={adj.B\over |B|}={1\over1}\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
\(=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
Hence, \((A')^{-1}=B^{-1}=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)

The given system of equations is:
5x + 2y = 3
3x + 2y = 5
These can be written as AX = B
\(\Rightarrow X=A^{-1}B\)
where \(A=\begin{bmatrix}5&2\\3&2 \end{bmatrix}, x=\begin{bmatrix}X\\Y \end{bmatrix}and\ B=\begin{bmatrix}3\\5 \end{bmatrix}\)
\(\therefore|A|=\begin{bmatrix} 5&2\\3&2\end{bmatrix}=10-6=4\neq0\Rightarrow A^{-1}\) exists.
\(\therefore A^{-1}={1\over|A|}(adj\ A)={1\over4}\begin{bmatrix} 2&-2\\-3&5\end{bmatrix}\)
From (1), \(X={1\over4} \begin{bmatrix} 2&-2\\-3&5\end{bmatrix}\begin{bmatrix} 3\\5\end{bmatrix}\)
\(={1\over4}\begin{bmatrix}6-10\\-9+25 \end{bmatrix}={1\over4}\begin{bmatrix}-4\\16 \end{bmatrix}\)
\(\Rightarrow\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix} -1\\4\end{bmatrix}\)
Hence, x = 1, y = 4.

Let \(A=\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1 \end{bmatrix}\)
\(\therefore |A|=\begin{bmatrix}2&1&3\\4&-1&0\\-7&2&1 \end{bmatrix}\)
\(=2(-1-0)-(1)(4+0)+3(8-7)\)
\(=-2-4+3=-3\neq0\)
A is non-singular = A^-1 exists.
Now \(A_{11}=(-1)^{1+1}\begin{vmatrix} -1&0\\2&1\end{vmatrix}=(+1)(-1-0)=1;\)
\(A_{12}=(-1)^{1+2}\begin{vmatrix}4&0\\-7&1 \end{vmatrix}=(-1)(4)=-4;\)
\(A_{13}=(-1)^{1+3}\begin{vmatrix} 4&-1\\-7&2\end{vmatrix}=(+1)(1)=1;\)
\(A_{21}=(-1)^{2+1}\begin{vmatrix} 1&3\\2&1\end{vmatrix}=(-1)(-5)=5;\)
\(A_{22}=(-1)^{2+2}\begin{vmatrix} 2&3\\-7&1\end{vmatrix}=(1)(23)=23;\)
\(A_{23}=(-1)^{2+3}\begin{vmatrix}2&1\\-7&2 \end{vmatrix}=(-1)(4+7)=-11;\)
\(A_{31}=(-1)^{3+1}\begin{vmatrix} 1&3\\-1&0\end{vmatrix}=(+1)(3)=3;\)
\(A_{32}=(-1)^{3+2}\begin{vmatrix} 2&3\\4&0\end{vmatrix}=(-1)(-12)=12;\)
\(A_{33}=(-1)^{3+3}\begin{vmatrix} 2&1\\4&-1\end{vmatrix}=(+1)(-2-4)=-6\)
\(\therefore\ adj\ A\begin{bmatrix}-1&5&3\\-4&23&12\\3&12&-6 \end{bmatrix}=\begin{bmatrix} -1&5&3\\-4&23&12\\1&-11&-6\end{bmatrix}\)
\(\therefore\ adj\ A^{-1}={1\over|A|}(adj\ A)\)
\(={1\over-3}\begin{bmatrix}-1&5&3\\-4&23&12\\1&-11&-6 \end{bmatrix}\)
\(=\left[ \begin{matrix} \frac { 1 }{ 3 } & \frac { -5 }{ 3 } & -1 \\ \frac { 4 }{ 3 } & \frac { -23 }{ 3 } & -4 \\ -\frac { 1 }{ 3 } & \frac { 11 }{ 3 } & 2 \end{matrix} \right] \)

Let \(A=\left[\begin{matrix}2&3\\-4&-6\end{matrix}\right]\)
Then \(A_{11}=(-1)^{1+1}M_{11}=(+1)(-6)=-6\)
\(A(12)=(-1){1+2}M_{12}=(-1)(-4)=4;\)
\(A_{12}=(-1)^{2+1}M_{21}=(-1)(3)=-3\)
\(A_{22}=(-1)^{2+2}M_{22}=(+1)(2)=2\)
\(adj\ A=\begin{vmatrix} -6&4\\-3&2\end{vmatrix}'\)
\(=\begin{vmatrix}-6&3\\4&2 \end{vmatrix}\)
Also   \(|A|=\begin{vmatrix}2&3\\-4&-6 \end{vmatrix}=-12+12\)
= 0 ...(1)
Now \(A(adj\ A)=\begin{vmatrix} 2&3\\-4&-6\end{vmatrix}\begin{vmatrix}-6&-3\\4&2 \end{vmatrix} \)
\(=\begin{vmatrix} -12+12&-6+6\\24-24&12-12\end{vmatrix}\)
\(=\begin{vmatrix} 0&0\\0&0\end{vmatrix} \)
= |A|I
And \((adj\ A)A=\begin{vmatrix}-6&-3\\4&2 \end{vmatrix}\begin{vmatrix} 2&3\\-4&-6\end{vmatrix} \)
\(=\begin{vmatrix}-12+12&-18+18\\8-8&12-12 \end{vmatrix}\)
\(=\begin{vmatrix} 0&0\\0&0\end{vmatrix}\)
= |A|I
Hence A (adj A)  = (adj A) = |A|I

Note that in the third column, two entries are zero. So expanding along third column (C₃), we get
\(\Delta =\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| -0\left| \begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix} \right| +0\left| \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right| \)
= 4 (–1 – 12) – 0 + 0 = – 52

We have
\(\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}=(x)(x)-(x-1)(x-1)\)
\(=x^{2}-\left(x^{2}-1\right)=x^{2}-x^{2}+1=1\)

The system of equations can be written in the form AX = B, where
\(\mathrm{A}=\left[\begin{array}{ccc} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{array}\right], \mathrm{X}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right]\)
We see that
|A| = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 \(\ne\) 0
Hence, A is nonsingular and so its inverse exists. Now
A₁₁ = –1, A₁₂ = – 8, A₁₃ = –10
A₂₁ = –5, A₂₂ = – 6, A₂₃ = 1
A₃₁ = –1, A₃₂ = 9, A₃₃ = 7
Therefore \(\mathrm{A}^{-1}=-\frac{1}{17}\left[\begin{array}{ccc} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]\)
so \(X=A^{-1} B=-\frac{1}{17}\left[\begin{array}{ccc} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{array}\right]\left[\begin{array}{l} 8 \\ 1 \\ 4 \end{array}\right] \)
\({\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=-\frac{1}{17}\left[\begin{array}{l} -17 \\ -34 \\ -51 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]}\)
Hence x = 1, y = 2 and z = 3.

\(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] =8I\)
where I is the identity matrix
Let \(A=\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(\Rightarrow AB=8I\)
Post Multiplyingbith sides by B ^-1, we get 
\(AB{ B }^{ -1 }=8I{ B }^{ -1 }\)
\(\Rightarrow A=8{ B }^{ -1 }\)
\(\Rightarrow { B }^{ -1 }=\frac { A }{ 8 } \)
Given Equations are:
x - y + z = 4
x - 2y - 2z = 9
and 2x + y + 3z = 1
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow BX=C\)
where \(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \ and\ C=\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow X={ B }^{ -1 }C\)
Using \({ B }^{ -1 }=\frac { A }{ 8 } \)
\(\Rightarrow X=\frac { A }{ 8 } C\)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
(x, y, z)  = (3, -2, -1)

Given \(A=\left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \) 
|A| = 1(1+3)-2(-1-1)+1(3-1)
= 4 + 4 + 2 = 10
Co-factor Matrix is :
\(=\left[ \begin{matrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{matrix} \right] \)
\(\therefore\) adj A = transpose of above matrix
\(=\left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\therefore { A }^{ -1 }=\frac { adj\quad A }{ |A| } \)
\(=\frac { 1 }{ 10 } \left[ \begin{matrix} 4 & -5 & 2 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{matrix} \right] \quad \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \)
Given set of equations are:
x + 2y + z = 4
-x + y + z = 0
x - 3y + z = 4
\(\Rightarrow \quad \left[ \begin{matrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \quad A.X=B\)
Multiplying both sides by A^-1 , we get
\({ A }^{ -1 }AX={ A }^{ -1 }B\)
\(\Rightarrow X={ A }^{ -1 }B\)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2/5 & -1/2 & 1/10 \\ 1/5 & 0 & -1/5 \\ 1/5 & 1/2 & 3/10 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 0 \\ 4 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} \frac { 8 }{ 5 } +\frac { 2 }{ 5 } \\ \frac { 4 }{ 5 } -\frac { 4 }{ 5 } \\ \frac { 4 }{ 5 } +\frac { 6 }{ 5 } \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 0 \\ 2 \end{matrix} \right] \)
\(\therefore \ x=2,y=0,z=2\)

\(A=\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] \)
then \(adj\quad A={ \left[ \begin{matrix} 4 & -3 \\ -1 & -2 \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
Taking \(A(adj\quad A)=\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] { \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
\(=\left[ \begin{matrix} -11 & 0 \\ 0 & -11 \end{matrix} \right] =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Taking \((adj\ A)A={ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }{ \left[ \begin{matrix} 4 & -1 \\ -3 & -2 \end{matrix} \right] }\)
\(=\left[ \begin{matrix} -11 & 0 \\ 0 & -11 \end{matrix} \right] =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Getting \(\left| A \right| =\left[ \begin{matrix} -2 & 1 \\ 3 & 4 \end{matrix} \right] =-11\)
\(\therefore \left| A \right| =-11\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
A.(adj A) = (adj A).A = |A| Hence proved.

Here, \(\left| A \right| =\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
= 1(0+0) + 1(9+2) + 2(0-0)
= 11
\(\Rightarrow \left| A \right| =\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ...(i)\)
\(adj\quad A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \quad \)
\(A(adj\quad A)=\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| \)
\((adj\quad A)A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ....(i)\)
Thus, it is verified that A(adj A) = (adj A)A = |A|I

Given, \(A=\left[\begin{array}{ll}p & 2 \\ 2 & p\end{array}\right]\)
\(\therefore |A|=\left|\begin{array}{cc} p & 2 \\ 2 & p \end{array}\right|=p^2-4\)
and \(\left|A^3\right|=125\)   \(\left[\because|A|^3=\left|A^3\right|\right]\)
\(\Rightarrow |A|^3=125\)
\(\Rightarrow \left(p^2-4\right)^3=125 \Rightarrow p^2-4=5\)
\(\Rightarrow p^2=9 \Rightarrow p= \pm 3\)

We have, A²=2 A
\(\Rightarrow A \cdot A=2 A\)
\(\Rightarrow  |A \cdot A|=|2 A| \)
\(\Rightarrow  |A| \cdot|A|=2^3|A|\)
\({\left[\because|A B|=|A||B| \text { and }|k A|=k^n|A|\right]_{(1)}}\)
\(\Rightarrow |A|^2=8|A| \)
\(\Rightarrow  |A|^2-8|A|=0 \)
\(\Rightarrow |A|(|A|-8)=0 \)
\(\Rightarrow |A|=0 \text { or } 8\)

We know that a matrix A is said to be singular, if |A| = 0.
\(\left|\begin{array}{cc} 2 x+2 & 2 x \\ x & x-2 \end{array}\right|  =0\)
\(\Rightarrow  (2 x+2)(x-2)-2 x^2  =0\)
\(\Rightarrow  2 x^2-2 x-4-2 x^2  =0 \)
\(\Rightarrow  -2 x  =4\)
x =-2

Given, \(A(\operatorname{adj} A)=\left[\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right] \Rightarrow|A(\operatorname{adj} A)|=\left|\begin{array}{ll}8 & 0 \\ 0 & 8\end{array}\right|\)
\( \Rightarrow  |A||\operatorname{adj}(A)| =64-0\)
\(\Rightarrow |A||A|^{2-1}  =64\)  \(\left[\because|\operatorname{adj} A|=|A|^{n-1}\right]\)
\(\Rightarrow  |A|^2  =64\)
\(\Rightarrow  |A|  = \pm 8\)

We have, \(A=\left[\begin{array}{cc}2 & -1 \\ 4 & 3\end{array}\right]\)
Cofactors of |A| are
\(A_{11}=3, A_{12}=-4, A_{21}=1 \text { and } A_{22}=2\)
\(\therefore  \operatorname{adj}(A)=\left[\begin{array}{ll} A_{11} & A_{21} \\ A_{12} & A_{22} \end{array}\right]=\left[\begin{array}{cc} 3 & 1 \\ -4 & 2 \end{array}\right]\)

Given IAI = 12.
Determinant formed by cofactors is (adj A)'
Also I(adj A)'I = I(adj A)I = IAI2 [ IAI is of order 3]
= (12)² = 144.

Compute A^-I and A² and prove the result.

(A³)^-1=(A-1)³

Let \(A=\left[\begin{array}{rr} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]\) then
\(|A|=\left|\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right|=\cos ^{2} \theta+\sin ^{2} \theta=1 \neq 0 \)
\(\therefore A^{-1} \text {exists. } \)
\(\text { Also, } \operatorname{adj}(A)=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \) 
\(\left[\begin{array}{ll} \because \text { adj } \end{array}\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]\right]\) 
Now,\(A^{-1}=\frac{1}{|A|} \operatorname{adj}(A)=\left[\begin{array}{cr} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\)

we have,\(A=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]\) 
Here, \(|A|=\left|\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right|=14-12=2 \neq 0\) 
\(\therefore\) A^-1 exists.
Clearly,\(\operatorname{adj}(A)=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]\) 
\(\left[\text { if } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \text { , then adj }(A)=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]\right]\) 
\( \therefore A^{-1} =\frac{1}{|A|} \operatorname{adj}(A) \)
\(=\frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right] \) 
Now, consider RHS = 9 I - A
\(=9\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right] \)
\(=\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]\) 
= 2A^-I [using Eq. (i)]
=LHS Hence proved