Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

    Q8     \begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}

Answers (1)
S seema garhwal

A=\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

             R_1\rightarrow R_1-R_2

\Rightarrow          \begin{bmatrix} 1 &1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A

                  

\Rightarrow          R_2\rightarrow R_2-3R_1

\Rightarrow          \begin{bmatrix}1&1\\0&1 \end{bmatrix} = \begin{bmatrix}1&-1\\-3&4 \end{bmatrix}A

                 R_1\rightarrow R_1-R_2

 \Rightarrow          \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}A

Thus using elementary transformation inverse of A is obtained as

\therefore A^{-1}=\begin{bmatrix}4&-5\\-3&4 \end{bmatrix}.

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