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