Get Answers to all your Questions

header-bg qa

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

    Q10.    \begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}

Answers (1)

best_answer

A=\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

             R_2\rightarrow R_2+R_1

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

\Rightarrow          R_1\rightarrow R_1+2R_2

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

            R_2\rightarrow R_2+R_1 

         \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} = \begin{bmatrix}3&2\\4&3 \end{bmatrix}A 

        R_2\rightarrow \frac{R_2}{2}

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

             R_1\rightarrow R_1-R_2

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

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

Thus the inverse of A is obtained using elementary transformation.

Posted by

seema garhwal

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads