Q.Using elementary transformations, find the inverse of the matrix $A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$.

Write $[A\,|\,I]$ and reduce the left block to $I$ using elementary row operations; the right block becomes $A^{-1}$.

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

Write $[A\,|\,I_3]$:

$\left[\begin{array}{ccc|ccc}1&2&-2&1&0&0\\-1&3&0&0&1&0\\0&-2&1&0&0&1\end{array}\right]$

$R_2\to R_2+R_1$:

$\left[\begin{array}{ccc|ccc}1&2&-2&1&0&0\\0&5&-2&1&1&0\\0&-2&1&0&0&1\end{array}\right]$

$R_3\to 5R_3+2R_2$ (scaling by $5$ to clear the fraction, an allowed combination of elementary operations):

$5R_3+2R_2:\ \big(0,\,-10+10,\,5-4\,\big|\,0+2,\,0+2,\,5+0\big)=(0,0,1\,|\,2,2,5)$

$\left[\begin{array}{ccc|ccc}1&2&-2&1&0&0\\0&5&-2&1&1&0\\0&0&1&2&2&5\end{array}\right]$

$R_1\to R_1+2R_3$: $(1,2,-2+2\,|\,1+4,0+4,0+10)=(1,2,0\,|\,5,4,10)$

$R_2\to R_2+2R_3$: $(0,5,-2+2\,|\,1+4,1+4,0+10)=(0,5,0\,|\,5,5,10)$

$\left[\begin{array}{ccc|ccc}1&2&0&5&4&10\\0&5&0&5&5&10\\0&0&1&2&2&5\end{array}\right]$

$R_2\to \dfrac15R_2$: $(0,1,0\,|\,1,1,2)$

$R_1\to R_1-2R_2$: $(1,2-2,0\,|\,5-2,4-2,10-4)=(1,0,0\,|\,3,2,6)$

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