#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 18

$A^{-1}=\frac{1}{42}\left[\begin{array}{cc} -4 & 5 \\ 2 & 8 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]$

Solution:

\begin{aligned} &A^{2}=\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} 64+10 & -40+20 \\ -16+8 & 10+16 \end{array}\right]=\left[\begin{array}{cc} 74 & -20 \\ -8 & 26 \end{array}\right] \\\\ &4 A=4 \times\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]=\left[\begin{array}{cc} -32 & 20 \\ 8 & 16 \end{array}\right] \\\\ &42 I=42 \times\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 42 & 0 \\ 0 & 42 \end{array}\right] \end{aligned}

Now,

$A^{2}+4 A-42 I=\left[\begin{array}{cc} 74 & -20 \\ -8 & 26 \end{array}\right]+\left[\begin{array}{cc} -32 & 20 \\ 8 & 16 \end{array}\right]-\left[\begin{array}{cc} 42 & 0 \\ 0 & 42 \end{array}\right]$

$=\left[\begin{array}{cc} 74-74 & -20+20 \\ -2+8 & 42-42 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\$

$A^{2}+4 A-42 I=0$

Multiplying by $A^{-1}$ both sides

\begin{aligned} &A \times A \times A^{-1}+4 A \times A^{-1}-42 I \times A^{-1}=0 \\ &A+4 I-42 A^{-1}=0 \\ &42 A^{-1}=A+4 I \end{aligned}

So,

\begin{aligned} A^{-1} &=\frac{1}{42}[A+4 I] \\\\ A^{-1} &=\frac{1}{42}(\left[\begin{array}{cc} -8 & 5 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right] )\\\\ A^{-1} &=\frac{1}{42}\left[\begin{array}{cc} -4 & 5 \\ 2 & 8 \end{array}\right] \end{aligned}