#### Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 19

$A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & -1 \\ 1 & 3 \end{array}\right]$

Hint:

Here, we use basic concept of determinant and inverse of matrix

$A^{-1}=\frac{1}{|A|} \times \operatorname{Adj}(A)$

Given:

$A=\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]$   Show that  $A^{2}-5 A+7 I=0$

Solution:

\begin{aligned} &A=\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right] \\\\ &A^{2}=\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} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right]=\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right] \end{aligned}

Now,

\begin{aligned} &A^{2}-5 A+7 I \\\\ &=\left[\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right]-5 \times\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\ &=\left[\begin{array}{ll} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\\\ &A^{2}-5 A+7 I=0 \end{aligned}

Multiply by $A^{-1}$  both sides

\begin{aligned} &A \times A \times A^{-1}-5 A \times A^{-1}+7 I \times A^{-1}=0 \\\\ &A-5 I+7 A^{-1}=0 \\\\ &A^{-1}=\frac{1}{7}[5 I-A] \end{aligned}

\begin{aligned} &A^{-1}=\frac{1}{7}\left[\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right]-\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]\right] \\\\ &A^{-1}=\frac{1}{7}\left[\begin{array}{cc} 2 & -1 \\ 1 & 3 \end{array}\right] \end{aligned}