Answer
a. $\left[\begin{array}{lll}
1 & 2 & 5\\
2 & 3 & 8\\
-1 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
2\\
3\\
3
\end{array}\right]$
b. Solution set: $\{( -2$ , $-3$ , $2 ) \}$
Work Step by Step
$a.$
$A=\left[\begin{array}{lll}
1 & 2 & 5\\
2 & 3 & 8\\
-1 & 1 & 2
\end{array}\right], \quad B=\left[\begin{array}{l}
2\\
3\\
3
\end{array}\right]$
$AX=B$
$\left[\begin{array}{lll}
1 & 2 & 5\\
2 & 3 & 8\\
-1 & 1 & 2
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
2\\
3\\
3
\end{array}\right]$
$b.\qquad X=A^{-1}B$ ( given the matrix $A^{-1}$)
$X=\left[\begin{array}{lll}
2 & -1 & -1\\
12 & -7 & -2\\
-5 & 3 & 1
\end{array}\right]\left[\begin{array}{l}
2\\
3\\
3
\end{array}\right]=\left[\begin{array}{l}
2(2)-1(3)-1(3)\\
12(2)-7(3)-2(3)\\
-5(2)+3(3)+1(3)
\end{array}\right]=\left[\begin{array}{l}
-2\\
-3\\
2
\end{array}\right]$
Solution set: $\{( -2$ , $-3$ , $2 ) \}$