Question

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

Solution

## As, $$A=IA$$, $$\therefore \quad \begin{bmatrix} 2 & 3 & -3 \\ -1 & -2 & 2 \\ 1 & 1 & -1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}A$$And we know that $$I=A^{-1}A$$.So, by row transformation we will try to convert $$LHS$$ into $$I$$$$\begin{bmatrix} \because R_{ 2 }\rightarrow & R_{ 2 }+ & R_{ 3 } \\ and\quad R_{ 1 }\rightarrow & R_{ 1 }- & 2R_{ 3 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} 0 & 1 & -1 \\ 0 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}A$$$$\begin{bmatrix} \because R_{ 2 }\rightarrow & R_{ 2 }+ & R_{ 1 } \end{bmatrix}$$$$\Rightarrow \begin{bmatrix} 0 & 1 & -1 \\ 0 & 0 & 0 \\ 1 & 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 & -2 \\ 2 & 1 & -2 \\ 0 & 0 & 1 \end{bmatrix}A$$ Since, second row of the matrix $$A$$ on LHS is containing all zeroes, so we can say that every inverse of matrix $$A$$ does not exist.Mathematics

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