Solve the following system of equations:

[tex]\[
\left\{\begin{array}{l}
6x + y - z = 14 \\
4x + 2y - z = 13 \\
x + 2y + 4z = 12
\end{array}\right.
\][/tex]



Answer :

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.