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

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

    Q17.    \begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}

Answers (1)

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

                A=IA

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

             R_1\rightarrow \frac{R_1}{2}     

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

            R_2\rightarrow R_2-5R_1             

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

\Rightarrow          R_3\rightarrow R_3-R_2

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

                       R_3\rightarrow 2R_3

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

                               R_1\rightarrow R_1+\frac{R_3}{2}          and  R_2\rightarrow R_2-\frac{5}{2}R_3              

 \Rightarrow        \begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3&-1&1\\-15&6&-5 \\ 5&-2&2 \end{bmatrix}A

      Thus the inverse of A is obtained as                

      \therefore A^{-1}=.\begin{bmatrix} 3&-1&1\\-15&6&-5 \\ 5&-2&2 \end{bmatrix}.

Posted by

seema garhwal

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