91Ó°ÊÓ

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

We have to find the inverse of the given matrix.

$\left[Aâˆ\pounds I_3\right]=\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ -2& -3& 0& 0& 0& 1\end{array}\right]$

Now transform it into the reduced row echelon form .

Now perform the row operations $R_3=r_3-2r_1$.

$\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ -2& -3& 0& 0& 0& 1\end{array}\right]→\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]$

Perform the row operation $R_2→\frac{r_2}{-2}$.

$\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& -2& 1& 0& 1& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]→\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]$

$\left[\begin{array}{cccccc}1& -1& 1& 1& 0& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]→\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]$

Perform the row operation $R_3→r_3+5r_2$.

$\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& -5& 2& 2& 0& 1\end{array}\right]→\left[\begin{array}{ccccrr}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& -\frac{1}{2}& 2& -\frac{5}{2}& 1\end{array}\right]$

Perform the row operation $R_3→\frac{r_3}{{\displaystyle \frac{-1}{2}}}$.

$\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& -\frac{1}{2}& 2& -\frac{5}{2}& 1\end{array}\right]→\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right]$

$\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& -\frac{1}{2}& 0& -\frac{1}{2}& 0\\ 0& 0& 1& -4& 5& -2\end{array}\right]→\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& 0& -2& 2& -1\\ 0& 0& 1& -4& 5& -2\end{array}\right]$

Perform the row operation $R_1→r_1-\frac{r_3}{2}$.

$\left[\begin{array}{cccccc}1& 0& \frac{1}{2}& 1& -\frac{1}{2}& 0\\ 0& 1& 0& -2& 2& -1\\ 0& 0& 1& -4& 5& -2\end{array}\right]→\left[\begin{array}{cccccc}1& 0& 0& 3& -3& 1\\ 0& 1& 0& -2& 2& -1\\ 0& 0& 1& -4& 5& -2\end{array}\right]$

We can see that the reduced row echelon form of $\left[A|I_3\right]$contains the identity matrix $I_3$on the left of the vertical bar and $A^{-1}$on the right of the vertical bar.

$A^{-1}=\left[\begin{array}{ccc}3& -3& 1\\ -2& 2& -1\\ -4& 5& -2\end{array}\right]$.