Certainly! To solve the system of linear equations:
[tex]\[
\left\{\begin{array}{l}
6x + y - z = 14 \\
4x + 2y - z = 13 \\
x + 2y + 4z = 12
\end{array}\right.
\][/tex]
we seek to find the values of [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] that satisfy all three equations simultaneously.
### Step-by-Step Solution:
Step 1: Write the system of equations in matrix form.
We represent the system as [tex]\(A\mathbf{x} = \mathbf{B}\)[/tex], where [tex]\(A\)[/tex] is the matrix of coefficients, [tex]\(\mathbf{x}\)[/tex] is the column vector of variables, and [tex]\(\mathbf{B}\)[/tex] is the column vector of constants.
The coefficient matrix [tex]\(A\)[/tex]:
[tex]\[
A = \begin{pmatrix}
6 & 1 & -1 \\
4 & 2 & -1 \\
1 & 2 & 4 \\
\end{pmatrix}
\][/tex]
The constants vector [tex]\(\mathbf{B}\)[/tex]:
[tex]\[
\mathbf{B} = \begin{pmatrix}
14 \\
13 \\
12 \\
\end{pmatrix}
\][/tex]
The variables vector [tex]\(\mathbf{x}\)[/tex]:
[tex]\[
\mathbf{x} = \begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}
\][/tex]
Thus, the system can be written as:
[tex]\[
\begin{pmatrix}
6 & 1 & -1 \\
4 & 2 & -1 \\
1 & 2 & 4 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z \\
\end{pmatrix}
=
\begin{pmatrix}
14 \\
13 \\
12 \\
\end{pmatrix}
\][/tex]
Step 2: Solve the matrix equation.
There are various methods to solve this system of linear equations, such as Gaussian elimination, Cramer's rule, or using matrix inversion. Here's the resulting solution:
[tex]\[
x \approx 2.0
\][/tex]
[tex]\[
y \approx 3.0
\][/tex]
[tex]\[
z \approx 1.0
\][/tex]
(Note: Due to the limitations of numerical precision in calculations, the values for [tex]\(y\)[/tex] and [tex]\(z\)[/tex] will have tiny numerical errors close to zero.)
### Final Answer:
[tex]\[
x = 2.0, \quad y = 3.0, \quad z = 1.0
\][/tex]
These values satisfy all three equations in the system.
