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

    Q1.    \begin{bmatrix}1&-1\\2&3 \end{bmatrix}

Answers (1)
S seema garhwal

Use the elementary transformation we can find the inverse as follows

A=\begin{bmatrix}1&-1\\2&3 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix}1&-1\\2&3 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A     

       R_2\rightarrow R_2-2R_1

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

  

            R_2\rightarrow \frac{R_2}{5}

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

                R_1\rightarrow R_1+R_2

\Rightarrow          \begin{bmatrix}1&0\\0&1 \end{bmatrix} = \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A

\therefore A^{-1}= \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}

 

 

 

 

 

 

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