Please Solve RD Sharma Class 12 Chapter 6 Adjoint and Inverese of Matrices Exercise 6.1 Question 23 Maths Textbook Solution.

$A^{-1}=\left[\begin{array}{cc} 6 & -5 \\ -7 & 6 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]$

Solution:

\begin{aligned} &A^{2}-12 A+I=0 \\\\ &A^{2}=\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]=\left[\begin{array}{cc} 36+35 & 30+30 \\ 42+42 & 35+36 \end{array}\right]=\left[\begin{array}{cc} 71 & 60 \\ 84 & 71 \end{array}\right] \end{aligned}

Now,

\begin{aligned} &A^{2}-12 A+1=0 \\\\ &{\left[\begin{array}{ll} 71 & 60 \\ 84 & 71 \end{array}\right]-12\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\\\ &=\left[\begin{array}{cc} 71-72+1 & 60-60+0 \\ 84-89+0 & 1-72+1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \end{aligned}

Also,

\begin{aligned} &A^{2}-12 A+1=0 \\\\ &A-12 I+A^{-1}=0 \\\\ &A^{-1}-12 I=A \end{aligned}

\begin{aligned} &A^{-1}=\left[\begin{array}{cc} 12 & 0 \\ 0 & 12 \end{array}\right]-\left[\begin{array}{ll} 6 & 5 \\ 7 & 6 \end{array}\right]=\left[\begin{array}{cc} 12-6 & -5 \\ -7 & 12-6 \end{array}\right] \\\\ &A^{-1}=\left[\begin{array}{cc} 6 & -5 \\ -7 & 6 \end{array}\right] \end{aligned}