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  • Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 17. [1 3 -2 -3 0 -5]

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

    Q16.    \begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}

Answers (1)

best_answer

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

                A=IA

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

             R_2\rightarrow R_2+3R_1      and      R_3\rightarrow R_3-2R_1

\Rightarrow          \begin{bmatrix} 1 &3 & -2\\ 0& 9 &-11 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix}1&0&0\\3&1&0 \\-2&0&1 \end{bmatrix}A

            R_1\rightarrow R_1+3R_3         and    R_2\rightarrow R_2+8R_3

\Rightarrow          \begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & -1 & 4 \end{bmatrix} = \begin{bmatrix}-5&0&3\\-13&1&8 \\-2&0&1 \end{bmatrix}A

\Rightarrow          R_3\rightarrow R_3+R_2

\Rightarrow             \begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 25 \end{bmatrix} = \begin{bmatrix}-5&0&3\\-13&1&8 \\-15&1&9 \end{bmatrix}A

                    R_3\rightarrow \frac{R_3}{25}

 \Rightarrow                 \begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}-5&0&3\\-13&1&8 \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A

                               R_1\rightarrow R_1-10R_3          and  R_2\rightarrow R_2-21R_3              

 \Rightarrow        \begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A

                 Thus the inverse of three by three matrix A is     

      \therefore A^{-1}=.\begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}.

Posted by

seema garhwal

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