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Provide solution for rd sharma class 12 chapter 6 Adjoint and Inverse of Matrix excercise 6.2 question 2

Answers (1)

Answer: $A^{-1}=\begin{bmatrix} 1 &-2 \\ -2& 5 \end{bmatrix}$

Hint: Here, we use the basic of matrix transpose

Given: $\begin{bmatrix} 5 &2 \\ 2& 1 \end{bmatrix}$

Solution: Let $A = IA$
$A = \begin{bmatrix} 5 &2 \\ 2& 1 \end{bmatrix}, I = \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 5 &2 \\ 2& 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0& 1 \end{bmatrix}A$

Applying $R_1 = \frac{1}{5}R_1$

$\Rightarrow \begin{bmatrix} 1 &\frac{2}{5} \\ 0&\frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & 0\\ 0& 1 \end{bmatrix}A$

Applying $R_2 = R_2 - 2R_1$

$\Rightarrow \begin{bmatrix} 1 &\frac{2}{5} \\ 0&\frac{1}{5} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} & 0\\ \frac{-2}{5}& 1 \end{bmatrix}A$

$\Rightarrow \begin{bmatrix} 1 & \frac{2}{5}\\ 0&1 \end{bmatrix}=\begin{bmatrix} \frac{1}{5}&0 \\ -2& 5 \end{bmatrix}A$

Applying $R_2\rightarrow 5R_2$

$\Rightarrow \begin{bmatrix} 1 &\frac{2}{5} \\ 0 &1 \end{bmatrix}=\begin{bmatrix} \frac{1}{5} &0 \\ -2& 5 \end{bmatrix}A$

Applying $R_1 \rightarrow R_1-\frac{2}{5}R_2$

$\Rightarrow \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}=\begin{bmatrix} 1 &-2 \\ -2& 5 \end{bmatrix}A$

So, Here $A^{-1}=\begin{bmatrix} 1 &-2 \\ -2& 5 \end{bmatrix}$

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