Answer
a. $\left[\begin{array}{lll}
1 & -1 & 1\\
0 & 2 & -1\\
2 & 3 & 0
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
8\\
-7\\
1
\end{array}\right]$
b. Solution set: $\{( 2$ , $-1$ , $5 ) \}$
Work Step by Step
$a.$
$A=\left[\begin{array}{lll}
1 & -1 & 1\\
0 & 2 & -1\\
2 & 3 & 0
\end{array}\right], \quad B=\left[\begin{array}{l}
8\\
-7\\
1
\end{array}\right]$
$AX=B$
$\left[\begin{array}{lll}
1 & -1 & 1\\
0 & 2 & -1\\
2 & 3 & 0
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
8\\
-7\\
1
\end{array}\right]$
$b.\qquad X=A^{-1}B$ ( given the matrix $A^{-1}$)
$\left[\begin{array}{lll}
3 & 3 & -1\\
-2 & -2 & 1\\
-4 & -5 & 2
\end{array}\right]\left[\begin{array}{l}
8\\
-7\\
1
\end{array}\right]=\left[\begin{array}{l}
3(8)+3(-7)-1(1)\\
-2(8)-2(-7)+1(1)\\
-4(8)-5(-7)+2(1)
\end{array}\right]=\left[\begin{array}{l}
2\\
-1\\
5
\end{array}\right]$
Solution set: $\{( 2$ , $-1$ , $5 ) \}$
