Answer
please see step-by-step
Work Step by Step
Use matrix row operations to get ls down the main diagonal from upper left to lower right, and Os below the $1\mathrm{s}$.
Transition first to second matrix:
we want 0s below the 1 at $a_{11}$
In the second row, 0 and 5 are obtained by
multiplying the number above with -2 and aading to the number in row 2,
($-2$)R1+R3,
$-2(1)+2=0$
$-2(-1)+3=5$
apply the same transformation to the rest of row 2:
$-2(1)+(-1)=-3$
$-2(8)+(-2)=-18$
Row 3 is replaced with (-3)R1+R3,
$-3(1)+3=0$
$-3(-1)+(-2)=1$
$-3(1)+(-9)=-12$
$-3(8)+9=-15$
The second matrix is
$\left[\begin{array}{lllll}
1 & -1 & 1 & | & 8\\
0 & 5 & [-3] & | & [-18]\\
0 & 1 & [-12] & | & [-15]
\end{array}\right]$
Next step: we want a 1 at $a_{22}$
- Replace R2 with $\displaystyle \frac{1}{5}R_{2}\qquad(*)$
So the third matrix is
$\left[\begin{array}{lllll}
1 & -1 & 1 & | & 8\\
0 & 1 & [-3/5] & | & [-18/5]\\
0 & 1 & [-12] & | & [-15]
\end{array}\right]$
$(*)$ another row operation is possible, $R_{2}-4R_{3}\rightarrow R_{2}$
(to place a 1 on the diagonal at $a_{22}),$
$\left[\begin{array}{lllll}
1 & -1 & 1 & | & 8\\
0 & 1 & [45] & | & [42]\\
0 & 1 & [-12] & | & [-15]
\end{array}\right]$
