Answer
a. $\left[\begin{array}{llll}
2 & 0 & 1 & 1\\
3 & 0 & 0 & 1\\
-1 & 1 & -2 & 1\\
4 & -1 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
w\\
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
6\\
9\\
4\\
6
\end{array}\right]$
b. Solution set: $\{( 2$ , $1$ , $-1$ , $3 ) \}$
Work Step by Step
$a.$
$A=\left[\begin{array}{llll}
2 & 0 & 1 & 1\\
3 & 0 & 0 & 1\\
-1 & 1 & -2 & 1\\
4 & -1 & 1 & 0
\end{array}\right], \quad B=\left[\begin{array}{l}
6\\
9\\
4\\
6
\end{array}\right]$
$AX=B$
$\left[\begin{array}{llll}
2 & 0 & 1 & 1\\
3 & 0 & 0 & 1\\
-1 & 1 & -2 & 1\\
4 & -1 & 1 & 0
\end{array}\right]\left[\begin{array}{l}
w\\
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
6\\
9\\
4\\
6
\end{array}\right]$
$b.\qquad X=A^{-1}B$ ( given the matrix $A^{-1}$)
$\left[\begin{array}{llll}
-1 & 2 & -1 & -1\\
-4 & 9 & -5 & -6\\
0 & 1 & -1 & -1\\
3 & -5 & 3 & 3
\end{array}\right]\left[\begin{array}{l}
6\\
9\\
4\\
6
\end{array}\right]=$
$=\left[\begin{array}{l}
-1(6)+2(9)-1(4)-1(6)\\
-4(6)+9(9)-5(4)-6(6)\\
0(6)+1(9)-1(4)-1(6)\\
3(6)-5(9)+3(4)+3(6)
\end{array}\right]=\left[\begin{array}{l}
2\\
1\\
-1\\
3
\end{array}\right]$
Solution set: $\{( 2$ , $1$ , $-1$ , $3 ) \}$
