Answer
a. $\left[\begin{array}{lll}
2 & 6 & 6\\
2 & 7 & 6\\
2 & 7 & 7
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
8\\
10\\
9
\end{array}\right]$
b. Solution set: $\{( 1$ , $2$ , $-1 ) \}$
Work Step by Step
$a.$
$A=\left[\begin{array}{lll}
2 & 6 & 6\\
2 & 7 & 6\\
2 & 7 & 7
\end{array}\right],\qquad B=\left[\begin{array}{l}
8\\
10\\
9
\end{array}\right]$
$AX=B$
$\left[\begin{array}{lll}
2 & 6 & 6\\
2 & 7 & 6\\
2 & 7 & 7
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
8\\
10\\
9
\end{array}\right]$
$b.\qquad X=A^{-1}B$
Find the inverse, $A^{-1}:$
$\left[\begin{array}{lllllll}
2 & 6 & 6 & | & 1 & 0 & 0\\
2 & 7 & 6 & | & 0 & 1 & 0\\
2 & 7 & 7 & | & 0 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
\times\frac{1}{2}\\
-R_{1}\\
-R_{1}
\end{array}\right.$
$\left[\begin{array}{lllllll}
1 & 3 & 3 & | & 1/2 & 0 & 0\\
0 & 1 & 0 & | & -1 & 1 & 0\\
0 & 1 & 1 & | & -1 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
-R_{2}.
\end{array}\right.$
$\left[\begin{array}{lllllll}
1 & 3 & 3 & | & 1/2 & 0 & 0\\
0 & 1 & 0 & | & -1 & 1 & 0\\
0 & 0 & 1 & | & 0 & -1 & 1
\end{array}\right]\left\{\begin{array}{l}
-3R_{2}\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllllll}
1 & 0 & 3 & | & 7/2 & -3 & 0\\
0 & 1 & 0 & | & -1 & 1 & 0\\
0 & 0 & 1 & | & 0 & -1 & 1
\end{array}\right]\left\{\begin{array}{l}
-3R_{3}\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{lllllll}
1 & 0 & 0 & | & 7/2 & 0 & -3\\
0 & 1 & 0 & | & -1 & 1 & 0\\
0 & 0 & 1 & | & 0 & -1 & 1
\end{array}\right]$
$A^{-1}=\left[\begin{array}{lll}
7/2 & 0 & -3\\
-1 & 1 & 0\\
0 & -1 & 1
\end{array}\right]$
$X=A^{-1}B$
$X=\left[\begin{array}{lll}
7/2 & 0 & -3\\
-1 & 1 & 0\\
0 & -1 & 1
\end{array}\right]\left[\begin{array}{l}
8\\
10\\
9
\end{array}\right]=\left[\begin{array}{l}
28+0-27\\
-8+10+0\\
0-10+9
\end{array}\right]$
$X=\left[\begin{array}{l}
1\\
2\\
-1
\end{array}\right]$
Solution set: $\{( 1$ , $2$ , $-1 ) \}$
