Answer
a) The required matrix equation is:
$\left[ \begin{matrix}
1 & -1 & 2 \\
0 & 1 & -1 \\
1 & 0 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
12 \\
-5 \\
10 \\
\end{matrix} \right]$
b) The solution set is: $\left\{ 4,-2,3 \right\}$
Work Step by Step
(a)
Note that the coefficient matrix $ A $ contains all coefficients of the variables $ x,y\text{ and }z $.
The matrix $ X $ contains all the variable and the matrix $ B $ contains all the constant terms of the provided linear system. Therefore, the matrix equation is given by:
$\left[ \begin{matrix}
1 & -1 & 2 \\
0 & 1 & -1 \\
1 & 0 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
12 \\
-5 \\
10 \\
\end{matrix} \right]$
Hence, the required matrix equation is: $\left[ \begin{matrix}
1 & -1 & 2 \\
0 & 1 & -1 \\
1 & 0 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
x \\
y \\
z \\
\end{matrix} \right]=\left[ \begin{matrix}
12 \\
-5 \\
10 \\
\end{matrix} \right]$
(b)
Consider the given system $ AX=B $
The solution of this system is $ X={{A}^{-1}}B $
Since it is provided that inverse of the matrix $ A=\left[ \begin{matrix}
1 & 1 & 2 \\
0 & 1 & 3 \\
3 & 0 & -2 \\
\end{matrix} \right]$ is $\left[ \begin{matrix}
-2 & 2 & 1 \\
9 & -8 & -3 \\
-3 & 3 & 1 \\
\end{matrix} \right]$.
Therefore, $\begin{align}
& X={{A}^{-1}}B \\
& =\left[ \begin{matrix}
2 & 2 & -1 \\
-1 & 0 & 1 \\
-1 & -1 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
12 \\
-5 \\
10 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2\left( 12 \right)+2\left( -5 \right)-1\left( 10 \right) \\
-1\left( 12 \right)+0\left( -5 \right)+1\left( 10 \right) \\
-1\left( 12 \right)-1\left( -5 \right)+1\left( 10 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4 \\
-2 \\
3 \\
\end{matrix} \right]
\end{align}$
This implies that $ x=4,y=-2\text{ and }z=3$.
