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#### Provide Solution for RD Sharma Class 12 Chapter 6 Adjoint and Inverse Matrix Exercise 6.1 Question 26

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

Hint:

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

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

Given:

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

Solution:

$A^{3}=A^{2} \times A$

$=\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]=\left[\begin{array}{ccc} 12+5+5 & -6+10-5 & 6+5+10 \\ -10-6-5 & 5+12+5 & -5-6-10 \\ 10+5+6 & -5-10-6 & 5+5+12 \end{array}\right]$

$=\left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]$

Now,

$A^{3}-6 A^{2}+9 A-4 I$

$\left[\begin{array}{ccc} 22 & -21 & 21 \\ -21 & 22 & -21 \\ 21 & -21 & 22 \end{array}\right]-6\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]+9\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]-4\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

$=\left[\begin{array}{ccc} 22-36+18-4 & -21+30-9 & 21-30+9 \\ -21+30-9 & 22-36+18-4 & -21+30-9 \\ 21-30+9-0 & -21+30-9 & 22-36+18-4 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]$

Thus,

\begin{aligned} &A^{3}-6 A^{2}+9 A-4 I \\ &A^{2}-6 A+9 I=4 A^{-1} \\ &A^{-1}=\frac{1}{4}\left[A^{2}-6 A+9 I\right] \end{aligned}

$A^{2}-6 A+9 I=\left[\begin{array}{ccc} 6 & -5 & 5 \\ -5 & 6 & -5 \\ 5 & -5 & 6 \end{array}\right]-6\left[\begin{array}{ccc} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{array}\right]+9\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$

$=\left[\begin{array}{ccc} 6-12+9 & -5+6+0 & 5-6+0 \\ -5+6+0 & 6-12+9 & -5+6+0 \\ 5-6+0 & -5-6+0 & 6-12+3 \end{array}\right]=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{array}\right]$

Hence,

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