Question

# If possible, using elementary row transformations, find the inverse of the following matrix.$$\begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix}$$

Solution

## For getting the inverse of the given matrix $$A$$ by row elementary operations we may write the given matrix as $$A=IA$$ And, we know $$AA^{-1}=I$$Now $$|A|=2\times 7+1\times (-12)+3\times (-1)=14-12-3=-1\neq0$$So, $$A^{-1}$$ is possible.$$\because \begin{bmatrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}A$$$$[R_2\rightarrow R_2+R_1]$$$$\Rightarrow \begin{bmatrix} 2 & -1 & 3 \\ -3 & 2 & 4 \\ -3 & 2 & 3 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}A$$$$\begin{bmatrix} \because R_{ 3 }\rightarrow & R_{ 3 }-& R_{ 2 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} 2 & -1 & 3 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix}A$$$$\begin{bmatrix} R_{ 1 }\rightarrow & R_{ 1 }+ & R_{ 2 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} -1 & 1 & 7 \\ -3 & 2 & 4 \\ 0 & 0 & -1 \end{bmatrix}=\begin{bmatrix} 2 & 1 & 0 \\ 1 & 1 & 0 \\ -1 & -1 & 1 \end{bmatrix}A$$$$\begin{bmatrix} \because R_{ 2 }\rightarrow & R_{ 2 }- & 3R_{ 1 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} -1 & 1 & 7 \\ 0 & -1 & -17 \\ 0 & 0 & -1 \end{bmatrix}=\begin{bmatrix} 2 & 1 & 0 \\ -5 & -2 & 0 \\ -1 & -1 & 1 \end{bmatrix}A$$$$\begin{bmatrix} \because R_{ 1 }\rightarrow & R_{ 1 }+ & R_{ 2 } \\ and\quad R_{ 3 }\rightarrow & -1 &\times R_{ 3 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} -1 & 0 & -10 \\ 0 & -1 & -17 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -3 & -1 & 0 \\ -5 & -2 & 0 \\ 1 & 1 & -1 \end{bmatrix}A$$$$\begin{bmatrix} \because R_{ 1 }\rightarrow & R_{ 1 }+ & 10R_{ 3 } \\ and\quad R_{ 2 }\rightarrow & R_{ 2 }+ & 17R_{ 3 } \end{bmatrix}$$ $$\Rightarrow \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ 1 & 1 & -1 \end{bmatrix}A$$$$\begin{bmatrix} \because R_{ 1 }\rightarrow & -1R_{ 1 } \\ and\quad R_{ 2 }\rightarrow & -1R_{ 2 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=\begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix}A$$As, $$I=A^{-1}A=AA^{-1}$$So, the inverse of $$A$$ is $$A^{-1}=\begin{bmatrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{bmatrix}$$Mathematics

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