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  • Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. [6 -3 -2 1]

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

    Q12.    \begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}

Answers (1)

best_answer

A=\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}

                A=IA

\Rightarrow          \begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix} = \begin{bmatrix}1&0\\0&1 \end{bmatrix}A

             R_1\rightarrow \frac{R_1}{6}

\Rightarrow               \begin{bmatrix} 1& -\frac{1}{2}\\ -2 & 1 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix}A

\Rightarrow          R_2\rightarrow R_2+2R_1

\Rightarrow         \begin{bmatrix} 1& -\frac{1}{2}\\ 0 & 0 \end{bmatrix} = \begin{bmatrix}\frac{1}{6}&0\\\frac{1}{3}&1 \end{bmatrix}A

               

Hence, we can see all the zeros in the second row of the matrix in L.H.S so A^{-1} does not exist.

Posted by

seema garhwal

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