Answer
a. $\left[\begin{array}{lll}
1 & -6 & 3\\
2 & -7 & 3\\
4 & -12 & 5
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
11\\
14\\
25
\end{array}\right]$
b. Solution set: $\{( 2$ , $-1$ , $1 ) \}$
Work Step by Step
$a.$
$A=\left[\begin{array}{lll}
1 & -6 & 3\\
2 & -7 & 3\\
4 & -12 & 5
\end{array}\right], \quad B=\left[\begin{array}{l}
11\\
14\\
25
\end{array}\right]$
$AX=B$
$\left[\begin{array}{lll}
1 & -6 & 3\\
2 & -7 & 3\\
4 & -12 & 5
\end{array}\right]\left[\begin{array}{l}
x\\
y\\
z
\end{array}\right]=\left[\begin{array}{l}
11\\
14\\
25
\end{array}\right]$
$b.\qquad X=A^{-1}B$ ( given the matrix $A^{-1}$)
$\left[\begin{array}{lll}
1 & -6 & 3\\
2 & -7 & 3\\
4 & -12 & 5
\end{array}\right]\left[\begin{array}{l}
11\\
14\\
25
\end{array}\right]=\left[\begin{array}{l}
1(11)-6(14)+3(25)\\
2(11)-7(14)+3(25)\\
4(11)-12(14)+5(25)
\end{array}\right]=\left[\begin{array}{l}
2\\
-1\\
1
\end{array}\right]$
Solution set: $\{( 2$ , $-1$ , $1 ) \}$
