Sure, let's solve the system of linear equations using inverse matrices. The system of equations is given as:
[tex]\[
\begin{cases}
2x + 3y + z = -1 \\
3x + 3y + z = 1 \\
2x + 4y + z = -2
\end{cases}
\][/tex]
We can represent this system in matrix form [tex]\(AX = B\)[/tex], where:
[tex]\[
A = \begin{pmatrix}
2 & 3 & 1 \\
3 & 3 & 1 \\
2 & 4 & 1
\end{pmatrix}
\][/tex]
[tex]\[
X = \begin{pmatrix}
x \\
y \\
z
\end{pmatrix}
\][/tex]
[tex]\[
B = \begin{pmatrix}
-1 \\
1 \\
-2
\end{pmatrix}
\][/tex]
To find the solution vector [tex]\(X\)[/tex], we need to multiply the inverse of matrix [tex]\(A\)[/tex] with matrix [tex]\(B\)[/tex], i.e., [tex]\(X = A^{-1}B\)[/tex].
Now, we have the steps to find the solution:
1. Compute the inverse of matrix [tex]\(A\)[/tex], denoted as [tex]\(A^{-1}\)[/tex].
2. Multiply [tex]\(A^{-1}\)[/tex] with matrix [tex]\(B\)[/tex] to find [tex]\(X\)[/tex].
We know from our calculations that the result for [tex]\(X\)[/tex] is:
[tex]\[
X = \begin{pmatrix}
x \\
y \\
z
\end{pmatrix} = \begin{pmatrix}
2 \\
-1 \\
-2
\end{pmatrix}
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
(x, y, z) = (2, -1, -2)
\][/tex]
Therefore, the ordered triple that solves the system is:
[tex]\[
\boxed{(2, -1, -2)}
\][/tex]
