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

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

    Q15.    \begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}

Answers (1)

best_answer

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

                A=IA

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

             R_1\rightarrow R_1-R_3

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

            R_1\rightarrow -R_1

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

\Rightarrow          R_2\rightarrow R_2-2R_1

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

                    R_3\rightarrow R_3-3R_1

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

                               R_2\leftrightarrow R_3             

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

                      R_2\rightarrow \frac{-R_2}{5}

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

                 R_1\rightarrow R_1-R_2

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

                 R_3\rightarrow \frac{R_3}{5}

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

                    R_2\rightarrow R_2+R_3

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

Thos the Inverse of A is

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

Posted by

seema garhwal

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