#### Explain solution RD Sharma class 12 Chapter 6 Adjoint and Inverse of a Matrix exercise 6.1 question 9 Subquestion (i)

$A^{-1}=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]$

Hint:

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

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

Given:

$A=\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right]$

Solution:

Here, let’s find  $\left | A \right |$

\begin{aligned} &|A|=1\left|\begin{array}{ll} 4 & 3 \\ 3 & 4 \end{array}\right|-3\left|\begin{array}{ll} 1 & 3 \\ 1 & 4 \end{array}\right|+3\left|\begin{array}{ll} 1 & 4 \\ 1 & 3 \end{array}\right| \\ &=1(16-9)-3(4-3)+3(3-4) \\ &=7-3-3=1 \end{aligned}

Hence $A^{-1}$  exist

Let’s find cofactor of A

\begin{aligned} &C_{11}=7, C_{21}=-3, C_{31}=-3 \\ &C_{12}=-1, C_{22}=1, C_{32}=0 \\ &C_{13}=-1, C_{23}=0, C_{33}=1 \\ &A d j(A)=C_{{ij}}^{T} \end{aligned}

$\begin{gathered} \operatorname{Adj}(A)=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \\ \end{gathered}$

$A^{-1}=\frac{1}{|A|} \times A d j(A)$

\begin{aligned} &A^{-1}=\frac{1}{1}\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right]=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \\ &A^{-1} \times A=\left[\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right] \end{aligned}

\begin{aligned} &=\left[\begin{array}{ccc} 7-3-3 & 21-12-9 & 21-9-12 \\ -1+1+0 & -3+4+0 & -3+3+0 \\ -1+0+1 & -3+0+3 & -3+0+4 \end{array}\right] \\ &A^{-1} \times A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=I \end{aligned}