To solve the system of equations using the matrix tool, we start by writing the system in the form of [tex]\[A\mathbf{x} = \mathbf{b}\][/tex] where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(\mathbf{x}\)[/tex] is the vector of variables, and [tex]\(\mathbf{b}\)[/tex] is the constants vector.
The given system of linear equations is:
[tex]\[
\begin{array}{rcl}
4x + 11y &=& -5 \\
4x + 8y &=& -8
\end{array}
\][/tex]
First, identify the coefficient matrix [tex]\(A\)[/tex], the variables vector [tex]\(\mathbf{x}\)[/tex], and the constants vector [tex]\(\mathbf{b}\)[/tex]:
[tex]\[
\begin{pmatrix}
4 & 11 \\
4 & 8
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
-5 \\
-8
\end{pmatrix}
\][/tex]
Here,
[tex]\[
A = \begin{pmatrix}
4 & 11 \\
4 & 8
\end{pmatrix},
\quad
\mathbf{x} = \begin{pmatrix}
x \\
y
\end{pmatrix},
\quad
\mathbf{b} = \begin{pmatrix}
-5 \\
-8
\end{pmatrix}
\][/tex]
We can use the matrix method to find [tex]\(\mathbf{x} = A^{-1}\mathbf{b}\)[/tex], where [tex]\(A^{-1}\)[/tex] is the inverse of the coefficient matrix [tex]\(A\)[/tex].
To find the inverse of [tex]\(A\)[/tex], first calculate the determinant of [tex]\(A\)[/tex]:
[tex]\[
\text{det}(A) = (4 \cdot 8) - (4 \cdot 11) = 32 - 44 = -12
\][/tex]
Since the determinant is non-zero, the inverse of [tex]\(A\)[/tex] exists. Now, calculate the inverse of the matrix [tex]\(A\)[/tex]:
[tex]\[
A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}
=
\frac{1}{-12}
\begin{pmatrix}
8 & -11 \\
-4 & 4
\end{pmatrix}
=
\begin{pmatrix}
-\frac{2}{3} & \frac{11}{12} \\
\frac{1}{3} & -\frac{1}{3}
\end{pmatrix}
\][/tex]
Now, multiply [tex]\(A^{-1}\)[/tex] by [tex]\(\mathbf{b}\)[/tex]:
[tex]\[
\mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix}
-\frac{2}{3} & \frac{11}{12} \\
\frac{1}{3} & -\frac{1}{3}
\end{pmatrix}
\begin{pmatrix}
-5 \\
-8
\end{pmatrix}
\][/tex]
Perform the matrix multiplication:
[tex]\[
\begin{pmatrix}
-\frac{2}{3} & \frac{11}{12} \\
\frac{1}{3} & -\frac{1}{3}
\end{pmatrix}
\begin{pmatrix}
-5 \\
-8
\end{pmatrix}
=
\begin{pmatrix}
(-\frac{2}{3}) \cdot (-5) + \frac{11}{12} \cdot (-8) \\
(\frac{1}{3}) \cdot (-5) + (-\frac{1}{3}) \cdot (-8)
\end{pmatrix}
\][/tex]
Calculate each element step-by-step:
[tex]\[
\begin{pmatrix}
\frac{10}{3} - \frac{88}{12} \\
-\frac{5}{3} + \frac{8}{3}
\end{pmatrix}
=
\begin{pmatrix}
\frac{10}{3} - \frac{22}{3} \\
\frac{3}{3}
\end{pmatrix}
=
\begin{pmatrix}
-\frac{12}{3} \\
1
\end{pmatrix}
=
\begin{pmatrix}
-4 \\
1
\end{pmatrix}
\][/tex]
Thus, the solution to the given system of equations is the ordered pair [tex]\((-4, 1)\)[/tex].
Therefore, the solution is:
[tex]\[
\boxed{(-4, 1)}
\][/tex]
