Ex 4.5 Class 12 Maths Ncert Solutions

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix [Aij]n \( \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( \mathrm{A}_{11}=4, \mathrm{~A}_{12}=-3, \mathrm{~A}_{21}=-2, \mathrm{~A}_{22}=1 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left|\begin{array}{ll}A_{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right| \)
\(=\left|\begin{array}{cc}
4 & -2 \\
-3 & 1
\end{array}\right|\)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[\mathrm{A}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( \mathrm{A}_{11}=1\{(3 \times 1-0 \times 5)\}=3 \)
Similarly,
\( A_{12}=-12, A_{13}=6, A_{21}=1, A_{22}=5, A_{23}=2, A_{31}=-11, \)\(A_{32}=-1, A_{33}= 5 \).
\( \therefore \) Adj \( \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right] \)
\( =\left|\begin{array}{ccc}3 & 1 & -11 \\ -12 & 5 & -1 \\ 6 & 2 & 5\end{array}\right| \)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[\mathrm{A}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=-6, A_{12}=4, A_{21}=-3, A_{22}=2 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left|\begin{array}{ll}A_{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right| \)
\( =\left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right] \)
So LHS \( =A(\operatorname{Adj} A)=\left[\begin{array}{cc}2 & 3 \\ -4 & 6\end{array}\right]\left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \)
Also Adj \( A(A)=\left[\begin{array}{cc}-6 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -4 & -6\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \)
Determinant of \( A=|A|=2(-6)-(3)(-4)=0 \)
So RHS \( =|A|\mathrm{I}=0 \)
Hence \( A(\operatorname{Adj} A)=\operatorname{Adj} A(A)=|A|\mathrm{I}=0 \) {hence proved}

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[\mathrm{A}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=0, A_{12}=-11, A_{13}=0, A_{21}=3, A_{22}=1, A_{23}=-1, A_{31}=2 \), \( \mathrm{A}_{32}=8, \mathrm{~A}_{33}=3 \)
\( \therefore \) Adj \( \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33}\end{array}\right] \)
\( =\left[\begin{array}{ccc}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right] \)
So, LHS \( = \) A \( (\operatorname{Adj} \mathrm{A}) \)
\( =\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]\left[\begin{array}{ccc}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \)
Also Adj
\( A(A)=\left[\begin{array}{ccc}0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \)
Determinant of \( A=|A|=11 \)
So RHS \( =|\mathrm{A}| \mathrm{I}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \)
Hence
\( A(\operatorname{Adj} A)=\operatorname{Adj} A(A)=|A| \mathrm{I}=\left[\begin{array}{ccc}11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11\end{array}\right] \)
{hence proved}

We know that \( \mathrm{A}^{-1}=\frac{1}{|A|} \) Adj A
Adjoint of the matrix \( \mathrm{A}=[\mathrm{aij}] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix [Aij]n \( \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( \mathrm{A}_{11}=3, \mathrm{~A}_{12}=-4, \mathrm{~A}_{21}=2, \mathrm{~A}_{22}=2 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left|\begin{array}{ll}A_{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right| \)
\( =\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right] \)
And \( |\mathrm{A}|=2(3)-(-2)(4)=14 \)
So \( \mathrm{A}^{-1}=\frac{1}{|A|} \) Adj
\( \mathrm{A}=\frac{1}{14}\left[\begin{array}{cc}3 & 2 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}\frac{3}{14} & \frac{2}{14} \\ -\frac{4}{14} & \frac{2}{14}\end{array}\right]=\left[\begin{array}{cc}\frac{3}{7} & \frac{1}{7} \\ -\frac{2}{7} & \frac{1}{7}\end{array}\right] \).

We know that \( \mathrm{A}^{-1}=\frac{1}{|A|} \) Adj A
Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[\mathrm{A}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( \mathrm{A}_{11}=2, \mathrm{~A}_{12}=3, \mathrm{~A}_{21}=-5, \mathrm{~A}_{22}=-1 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left|\begin{array}{ll}A_{11} & A_{21} \\ A_{12} & A_{22}\end{array}\right| \)
\( =\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right] \)
And \( |A|=-1(2)-(-3)(5)=13 \)
So \( A^{-1}=\frac{1}{|A|} \operatorname{Adj} A=\frac{1}{13}\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]=\left[\begin{array}{ll}\frac{2}{13} & \frac{-5}{13} \\ \frac{3}{13} & \frac{-1}{13}\end{array}\right] \)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[A_{i j}\right] n \times n \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=10, A_{12}=0, A_{13}=0, A_{21}=-10, A_{22}=5, A_{23}=0, A_{31}=2 \), \( \mathrm{A}_{32}=-4, \mathrm{~A}_{33}=2 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right] \)
\( =\left[\begin{array}{ccc}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right] \)
And \( |\mathrm{A}|=10 \).
\( A^{-1}=\frac{1}{|A|} \operatorname{Adj} A=\frac{1}{10}\left[\begin{array}{ccc}10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2\end{array}\right]=\left[\begin{array}{ccc}\frac{10}{10} & -\frac{10}{10} & \frac{2}{10} \\ 0 & \frac{5}{10} & -\frac{4}{10} \\ 0 & 0 & \frac{2}{10}\end{array}\right] \)
\( \mathrm{A}-1=\left|\begin{array}{ccc}1 & -1 & \frac{1}{5} \\ 0 & \frac{1}{2} & -\frac{2}{5} \\ 0 & 0 & \frac{1}{5}\end{array}\right| \)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[\mathrm{A}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=-3, A_{12}=3, A_{13}=-9, A_{21}=0, A_{22}=-1, A_{23}=-2, A_{31}=0 \), \( \mathrm{A}_{32}=0, \mathrm{~A}_{33}=3 \)
\( \therefore \operatorname{Adj} \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right] \)
\( =\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right] \)
And \( |A|=-3 \).
\( \mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{Adj} \mathrm{A}=\frac{1}{-3}\left[\begin{array}{ccc}-3 & 0 & 0 \\ 3 & -1 & 0 \\ -9 & -2 & 3\end{array}\right]=\left[\begin{array}{ccc}\frac{-3}{-3} & 0 & 0 \\ \frac{3}{-3} & \frac{-1}{-3} & 0 \\ \frac{-9}{-3} & \frac{-2}{-2} & \frac{3}{-3}\end{array}\right]\)\(=\left[\begin{array}{ccc}1 & 0 & 0 \\ -1 & \frac{1}{3} & 0 \\ 3 & \frac{2}{3} & -1\end{array}\right] \)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( [\mathrm{Aij}] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=-1, \mathrm{~A}_{12}=-4, \mathrm{~A}_{13}=1, \mathrm{~A}_{21}=5, \mathrm{~A}_{22}=23, \mathrm{~A}_{23}=-11, \) \(\mathrm{~A}_{31}=3, \mathrm{A}_{32}=12, \mathrm{~A}_{33}=-6 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right] \)
\(=\left[\begin{array}{ccc}
-1 & 5 & 3 \\
-4 & 23 & 12 \\
1 & -11 & -6
\end{array}\right]\)
And \( |A|=-3 \).
\( \mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{Adj} \mathrm{A}=\frac{1}{-3}\left[\begin{array}{ccc}-1 & 5 & 3 \\ -4 & 23 & 12 \\ 1 & 11 & -6\end{array}\right]=\left[\begin{array}{ccc}\frac{-1}{-3} & \frac{5}{-3} & \frac{3}{-3} \\ \frac{-4}{13} & \frac{23}{-3} & \frac{12}{-3} \\ \frac{1}{-3} & \frac{-11}{-3} & \frac{-6}{-3}\end{array}\right] \)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( [A \mathrm{ij}] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=2, A_{12}=-9, A_{13}=-6, A_{21}=0, A_{22}=-2, A_{23}=-1, \) \( A_{31}=-1 ,\mathrm{A}_{32}=3, \mathrm{~A}_{33}=2 \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right] \)
\(=\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]\)
And \( |A|=-1 \).
\(\begin{array}{l}
A^{-1}=\frac{1}{|A|} \operatorname{Adj} A=\frac{1}{-1}\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right] \\
A^{-1}=\left[\begin{array}{ccc}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]
\end{array}\)

Adjoint of the matrix \( \mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) is defined as the transpose of the matrix \( \left[\mathrm{A}_{\mathrm{ij}}\right] \mathrm{n} \times \mathrm{n} \) where Aij is the co-factor of the element aij.
Let's find the cofactors for all the positions first-
Here, \( A_{11}=-1, A_{12}=0, A_{13}=0, A_{21}=0, A_{22}=-\cos \alpha, A_{23}=-\sin \alpha, \) \(A_{31} =0, \mathrm{~A}_{32}=-\sin \alpha, \mathrm{A}_{33}=\cos \alpha \).
\( \therefore \operatorname{Adj} \mathrm{A}=\left[\begin{array}{lll}A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33}\end{array}\right] \)
\( =\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right] \)
And \( |\mathrm{A}|=1 \).
\( \mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{Adj} \mathrm{A} \)
\( =\frac{1}{-1}\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right]=\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\cos \alpha & -\sin \alpha \\ 0 & -\sin \alpha & \cos \alpha\end{array}\right] \)

We have
\( \mathrm{AB}=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right]=\left[\begin{array}{ll}67 & 87 \\ 47 & 61\end{array}\right]=(61)(67)-(47)(87)=-2 \)
Here determinant of matrix \( =|A B| \neq 0 \) hence \( (A B)^{-1} \) exists.
\( (A B)^{-1}=\frac{1}{|A B|} \operatorname{Adj}(A B)=-\frac{1}{2}\left[\begin{array}{cc}61 & -47 \\ -87 & 67\end{array}\right]=\left[\begin{array}{cc}\frac{61}{-2} & \frac{-47}{-2} \\ \frac{-87}{-2} & \frac{67}{-2}\end{array}\right] \)
\( \left\{\therefore \operatorname{Adj}(\mathrm{AB})=\left[\begin{array}{cc}61 & -47 \\ -87 & 67\end{array}\right]\right\} \)
So, \( (\mathrm{AB})^{-1}=\left[\begin{array}{cc}-\frac{61}{2} & \frac{47}{2} \\ \frac{87}{2} & -\frac{67}{2}\end{array}\right] \)
Also \( |A|=1 \neq 0 \) and \( |B|=-2 \neq 0 \).
\( \therefore \mathrm{A}^{-1} \) and \( \mathrm{B}^{-1} \) will also exist and are given by-
(A) \( -1=\frac{1}{|A|} \operatorname{Adj} \mathrm{A}=\frac{1}{1}\left[\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right] \)
(B) \( -1=\frac{1}{|B|} \) Adj \( A=-\frac{1}{2}\left[\begin{array}{cc}9 & -7 \\ -8 & 6\end{array}\right] \)
And hence,
\( (\mathrm{B})^{-1}(\mathrm{A})^{-1}=-\frac{1}{2}\left[\begin{array}{cc}9 & -7 \\ -8 & 6\end{array}\right]\left[\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right]=-\frac{1}{2}\left[\begin{array}{cc}61 & -47 \\ -87 & 67\end{array}\right]= \) \( \left[\begin{array}{ll}\frac{61}{-2} & \frac{-47}{-2} \\ \frac{-87}{-2} & \frac{67}{-2}\end{array}\right] \)
{Hence proved}

We have \( A^{2}=A \cdot{ }^{A}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \).
So \( \mathrm{A}^{2}-5 \mathrm{A}+7 \mathrm{I}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=0 \)
Hence \( A^{2}-5 A+7 I=0 \)
\( \therefore A.A -5 \mathrm{~A}=-7 \mathrm{I} \)
Now post multiply with \( A^{-1} \)
So \(A.A.A^{-1}-5 A. A^{-1}=-7 \mathrm{I}. A^{-1} \)
\(\begin{array}{l}
\rightarrow A . I-5 I=-7 I.A^{-1}\left\{\text {since }A.A^{-1}=I\right\} \\
A-5 I=-7 A^{-1}\{\text {since }X.I=X\} \\
\rightarrow A^{-1}=\frac{5 I-A}{7}=\frac{1}{7}\left(5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\right)=-\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right] \\
\rightarrow A^{-1}=\left[\begin{array}{cc}
\frac{2}{7} & -\frac{1}{7} \\
\frac{1}{7} & \frac{3}{7}
\end{array}\right]
\end{array}\)

We have \( A^{2}=A \cdot A=\left[\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right]=\left[\begin{array}{cc}10 & 5 \\ 5 & 5\end{array}\right] \)
Since \( A^{2}+a A+b I=\left[\begin{array}{cc}10 & 5 \\ 5 & 5\end{array}\right]+a\left[\begin{array}{ll}3 & 1 \\ 1 & 2\end{array}\right]+b\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}=0\right] \)
So \( \mathrm{A}^{2}+\mathrm{aA}+\mathrm{bI}=\left[\begin{array}{cc}10+3 a+b & 5+a \\ 5+a & 5+2 a+b\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \)
Hence \( 10+3 a+b=0 \ldots \) (i)
\( 5+a=0 \)
\( 5+2 \mathrm{a}+\mathrm{b}=0 \)
From (ii) \( a=-5 \)
Putting a in (iii) we get \( b=5 \)
So \( a=-5 \) and \( b=5 \) satisfy the equation.

Here \( A^{2}=A \cdot A=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]\left[\begin{array}{ccc}1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3\end{array}\right]=\left[\begin{array}{ccc}4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14\end{array}\right] \)
And hence \( \mathrm{A}^{3}=\mathrm{A} . \mathrm{A}^{2}= \)
\(\begin{array}{l}
{\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]=\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]} \\
\therefore \mathrm{A}^{3}-6 \mathrm{A}^{2}+5 \mathrm{~A}+11 \mathrm{I}= \\
{\left[\begin{array}{ccc}
8 & 7 & 1 \\
-23 & 27 & -69 \\
32 & -13 & 58
\end{array}\right]-6\left[\begin{array}{ccc}
4 & 2 & 1 \\
-3 & 8 & -14 \\
7 & -3 & 14
\end{array}\right]+5\left[\begin{array}{ccc}
1 & 1 & 1 \\
1 & 2 & -3 \\
2 & -1 & 3
\end{array}\right]+} \\
11\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
=\left[\begin{array}{ccc}
24 & 12 & 6 \\
-18 & 48 & -84 \\
42 & -18 & 84
\end{array}\right]-\left[\begin{array}{ccc}
24 & 12 & 6 \\
-18 & 48 & -84 \\
42 & -18 & 84
\end{array}\right]=0
\end{array}\)
Thus, \( A^{3}-6 A^{2}+5 A+11 I=0 \)
Now, \( \mathrm{A}^{3}-6 \mathrm{A}^{2}+5 \mathrm{A}+11 \mathrm{I}=0 \),
\(\rightarrow (A.A.A) -6(A.A )+5 \mathrm{A}=-11 \mathrm{I}\)
Post-multiply with A-1 on both sides-
\(\rightarrow\left(A.A.A.A ^{-1}\right)-6\left(A.A.A^{-1}\right)+5 A.A ^{-1}=-11 I\cdot \mathrm{A}^{-1}\)
\(
\rightarrow(A.A.I)-6(A.I)+5 \mathrm{I}=-11 \mathrm{I} \cdot \mathrm{A}-1\left\{\text {since } A \cdot \mathrm{A}^{-1}=\mathrm{I}\right\}\)
\(
\rightarrow(A.A)-6 \mathrm{A}+5 \mathrm{I}=-11 \mathrm{A}^{-1}\{\text {since } X.I=\mathrm{X}\}
\)

Here \( A^{2}= A.A =\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]=\left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right] \)
And hence
\( A^{3}=A \cdot A^{2}=\left[\begin{array}{ccc}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]\left[\begin{array}{ccc}6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6\end{array}\right]= \)
\(\begin{array}{l}
{\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & -22
\end{array}\right]} \\
\therefore \mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{~A}-4 \mathrm{I} \\
=\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]-6\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]+9\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]- \\
4\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
=\left[\begin{array}{ccc}
40 & -30 & 30 \\
-30 & 40 & -30 \\
30 & -30 & 40
\end{array}\right]-\left[\begin{array}{ccc}
40 & -30 & 30 \\
-30 & 40 & -30 \\
30 & -30 & 40
\end{array}\right]=0
\end{array}\)
Thus, \( \mathrm{A}^{3}-6 \mathrm{~A}^{2}+9 \mathrm{A}-4 \mathrm{I}=0 \)
Now, \( A^{3}-6 A^{2}+9 A-4 I=0 \),
\(\rightarrow(A.A.A )-6(\mathrm{A} . \mathrm{A})+9 \mathrm{~A}=4 \mathrm{I}\)
Post-multiply with \( \mathrm{A}^{-1} \) on both sides-
\( \rightarrow\left(A.A.A.A^{-1}\right)-6\left(\mathrm{A} \cdot \mathrm{A} \cdot \mathrm{A}^{-1}\right)+9 \mathrm{A} \cdot \mathrm{A}^{-1}=4 \mathrm{I} \cdot \mathrm{A}^{-1}\)
\(\rightarrow(\mathrm{A} \cdot \mathrm{A} \cdot \mathrm{I})-6(\mathrm{A} \cdot \mathrm{I})+9 \mathrm{I}=4 \mathrm{I} \cdot \mathrm{A}^{-1}\left\{\text {since } \mathrm{A}\cdot \mathrm{A}^{-1}=\mathrm{I}\right\}\)
\(\rightarrow(\mathrm{A} \cdot \mathrm{A})-6 \mathrm{~A}+9 \mathrm{I}=4 \mathrm{~A}^{-1}\{\text {since }X.I =X\}\)
\(\rightarrow \mathrm{A}^{-1}=\frac{1}{4}\left(\mathrm{A}^{2}-6 \mathrm{~A}+9 \mathrm{I}\right) \)
\(\begin{array}{l}=\frac{1}{4}\left(\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]-6\left[\begin{array}{ccc}
2 & -1 & -1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]+9\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right)
\end{array}\)
\( =\frac{1}{4}\left[\begin{array}{ccc}3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3\end{array}\right]=\left[\begin{array}{ccc}\frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4}\end{array}\right] \)
Hence \( A^{-1}=\left[\begin{array}{ccc}\frac{3}{4} & \frac{1}{4} & -\frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{1}{4} & \frac{3}{4}\end{array}\right] \)

For a square matrix of order \( n \times n \),
We know that \( |\operatorname{Adj} A|=|A|^{\mathrm{n}-1} \)
So, \( \left|\operatorname{Adj} A|=| A\left|\left(3^{-1}\right)=\right| A\right|^{2} \)

\( (\mathrm{A})^{-1}=\frac{1}{|A|} \operatorname{Adj} \mathrm{A} \)
SO, \( \left|(\mathrm{A})^{-1}\right|=\left|\frac{1}{|A|} \operatorname{Adj}(A)\right|=\left.\frac{1}{|A|^{n}}\left|\operatorname{Adj}(\mathrm{A})=\frac{1}{|A|^{n}}\right| \mathrm{A}\right|^{\mathrm{n}-1}=\frac{1}{|A|^{1}} \)
\( \left\{\right. \)since \( \operatorname{adj}(A) \) is of order \( n \) and \( \left.|\operatorname{Adj}(A)|=|A|^{n-1}\right\} \)
Alternative-
We know that \( \mathrm{AA}^{-1}=\mathrm{I} \)
So \( |A|\left|A^{-1}\right|=|\mathrm{I}|=1 \)
Hence \( \left|(A)^{-1}\right|=\frac{1}{|A|} \)