Answer
a) The linear system can be written as $\left[ \begin{matrix}
1 & 2 & 5 \\
2 & 3 & 8 \\
-1 & 1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
2 \\
3 \\
3 \\
\end{matrix} \right]$
b) the value of the inverse matrix is $ X=\left\{ \left( -2,-3,2 \right) \right\}$
Work Step by Step
(a)
Consider the given system of equations:
$\begin{align}
& x+2y+5z=2 \\
& 2x+3y+8z=3 \\
& -x+y+2z=3
\end{align}$
The linear system can be written as: $ AX=B $
$\left[ \begin{matrix}
1 & 2 & 5 \\
2 & 3 & 8 \\
-1 & 1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
2 \\
3 \\
3 \\
\end{matrix} \right]$
Where, $ A=\left[ \begin{matrix}
1 & 2 & 5 \\
2 & 3 & 8 \\
-1 & 1 & 2 \\
\end{matrix} \right]$ $ X=\left[ \begin{align}
& x \\
& y \\
& z \\
\end{align} \right]$ $ B=\left[ \begin{align}
& 2 \\
& 3 \\
& 3 \\
\end{align} \right]$
(b)
Consider the given system of equations:
$\begin{align}
& x+2y+5z=2 \\
& 2x+3y+8z=3 \\
& -x+y+2z=3
\end{align}$
The linear system can be written as: $ AX=B $
$\left[ \begin{matrix}
1 & 2 & 5 \\
2 & 3 & 8 \\
-1 & 1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
2 \\
3 \\
3 \\
\end{matrix} \right]$
Where, $ A=\left[ \begin{matrix}
1 & 2 & 5 \\
2 & 3 & 8 \\
-1 & 1 & 2 \\
\end{matrix} \right]$ $ X=\left[ \begin{align}
& x \\
& y \\
& z \\
\end{align} \right]$ $ B=\left[ \begin{align}
& 2 \\
& 3 \\
& 3 \\
\end{align} \right]$
Now, consider the coefficient matrix $ A=\left[ \begin{matrix}
1 & 2 & 5 \\
2 & 3 & 8 \\
-1 & 1 & 2 \\
\end{matrix} \right]$
Use the inverse of the coefficient matrix
${{\left[ A \right]}^{-1}}=\left[ \begin{matrix}
2 & -1 & -1 \\
12 & -7 & -2 \\
-5 & 3 & 1 \\
\end{matrix} \right]$
Now, to find the values of the provided system:
We will use the formula $ X={{A}^{-1}}B $
Where, ${{A}^{-1}}=\left[ \begin{matrix}
2 & -1 & -1 \\
12 & -7 & -2 \\
-5 & 3 & 1 \\
\end{matrix} \right]$ $ B=\left[ \begin{align}
& 2 \\
& 3 \\
& 3 \\
\end{align} \right]$
Now, substitute the values in $ X={{A}^{-1}}B $
