#### Need solution for RD Sharma Maths Class 12 Chapter 6 Adjoint and Inverse of Matrices Excercise 6.1 Question 7 Subquestion (iv)

$A^{-1}=\frac{1}{17}\left[\begin{array}{cc} 1 & -5 \\ 3 & 2 \end{array}\right]$

Hint:

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

Given:

$A=\left[\begin{array}{cc} 2 & 5 \\ -3 & 1 \end{array}\right]$

Solution:

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

So let’s find  $\left | A \right |$

$|A|=\left|\begin{array}{ll} 2 & 5 \\ -3 & 1 \end{array}\right|=2-[(5) \times(-3)]=2+15=17$

Let’s find  $Adj\left ( A \right )$

\begin{aligned} &C_{11}=1, C_{12}=3 \\ &C_{21}=-5, C_{22}=2 \end{aligned}

So,

\begin{aligned} &C_{i j}=\left[\begin{array}{cc} 1 & 3 \\ -5 & 2 \end{array}\right] \\\\ &\operatorname{Adj}(A)=C_{i j}^{T} \\\\ &\operatorname{Adj}(A)=\left[\begin{array}{cc} 1 & -5 \\ 3 & 2 \end{array}\right] \end{aligned}

Let’s put the values in formula

$A^{-1}=\frac{1}{17}\left[\begin{array}{cc} 1 & -5 \\ 3 & 2 \end{array}\right]$