Answer
a) The linear system can be written as
$\left[ \begin{matrix}
2 & 6 & 6 \\
2 & 7 & 6 \\
2 & 7 & 7 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
8 \\
10 \\
9 \\
\end{matrix} \right]$
b) $ X=\left\{ \left( 1,2,-1 \right) \right\}$
Work Step by Step
(a)
Consider the given system of equations:
$\begin{align}
& 2x+6y+6z=8 \\
& 2x+7y+6z=10 \\
& 2x+7y+7z=9
\end{align}$
The linear system can be written as: $ AX=B $
$\left[ \begin{matrix}
2 & 6 & 6 \\
2 & 7 & 6 \\
2 & 7 & 7 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
8 \\
10 \\
9 \\
\end{matrix} \right]$
Where, $ A=\left[ \begin{matrix}
2 & 6 & 6 \\
2 & 7 & 6 \\
2 & 7 & 7 \\
\end{matrix} \right]$, $ X=\left[ \begin{align}
& x \\
& y \\
& z \\
\end{align} \right]$, $ B=\left[ \begin{align}
& 8 \\
& 10 \\
& 9 \\
\end{align} \right]$
(b)
Consider the given system of equations:
$\begin{align}
& 2x+6y+6z=8 \\
& 2x+7y+6z=10 \\
& 2x+7y+7z=9
\end{align}$
The linear system can be written as: $ AX=B $
$\left[ \begin{matrix}
2 & 6 & 6 \\
2 & 7 & 6 \\
2 & 7 & 7 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
8 \\
10 \\
9 \\
\end{matrix} \right]$
Where, $ A=\left[ \begin{matrix}
2 & 6 & 6 \\
2 & 7 & 6 \\
2 & 7 & 7 \\
\end{matrix} \right]$ $ X=\left[ \begin{align}
& x \\
& y \\
& z \\
\end{align} \right]$ $ B=\left[ \begin{align}
& 8 \\
& 10 \\
& 9 \\
\end{align} \right]$
Now, consider the coefficient matrix $ A=\left[ \begin{matrix}
2 & 6 & 6 \\
2 & 7 & 6 \\
2 & 7 & 7 \\
\end{matrix} \right]$
Use the inverse of the coefficient matrix.
${{\left[ A \right]}^{-1}}=\left[ \begin{matrix}
\frac{7}{2} & 0 & -3 \\
-1 & 1 & 0 \\
0 & -1 & 1 \\
\end{matrix} \right]$
Now, to find the values of the provided system:
Using formula $ X={{A}^{-1}}B $ we get, Where, ${{A}^{-1}}=\left[ \begin{matrix}
\frac{7}{2} & 0 & -3 \\
-1 & 1 & 0 \\
0 & -1 & 1 \\
\end{matrix} \right]$ $ B=\left[ \begin{align}
& 8 \\
& 10 \\
& 9 \\
\end{align} \right]$
Now, substitute the values in $ X={{A}^{-1}}B $ to get,
