To solve the system of linear equations represented by the augmented matrix
[tex]\[
\begin{pmatrix}
1 & -3 & -1 & 400 \\
-3 & 10 & 3 & 300 \\
0 & 6 & 1 & 150
\end{pmatrix}
\][/tex]
we can follow these steps:
1. Extract the coefficient matrix [tex]\(A\)[/tex] and the constant matrix [tex]\(B\)[/tex]:
The augmented matrix can be split into a coefficient matrix [tex]\(A\)[/tex] and a constants column vector [tex]\(B\)[/tex].
[tex]\[
A = \begin{pmatrix}
1 & -3 & -1 \\
-3 & 10 & 3 \\
0 & 6 & 1
\end{pmatrix}
\][/tex]
[tex]\[
B = \begin{pmatrix}
400 \\
300 \\
150
\end{pmatrix}
\][/tex]
2. Solve the system of equations [tex]\(A \mathbf{x} = B\)[/tex]:
To find the solution vector [tex]\( \mathbf{x} \)[/tex], we need to solve the matrix equation [tex]\(A \mathbf{x} = B\)[/tex]. This can be done using a variety of mathematical methods such as Gaussian elimination, matrix inversion, or more practically using linear algebra techniques.
Given the information provided, the solution to this system is:
[tex]\[
\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -3950 \\ 1500 \\ -8850 \end{pmatrix}
\][/tex]
So, we have:
[tex]\[
x_1 = -3950, \quad x_2 = 1500, \quad x_3 = -8850
\][/tex]
This means that the solution to the system of equations is:
- [tex]\( x_1 = -3950 \)[/tex]
- [tex]\( x_2 = 1500 \)[/tex]
- [tex]\( x_3 = -8850 \)[/tex]
Therefore, the values that satisfy all three equations are:
[tex]\[
\begin{cases}
1 \cdot (-3950) - 3 \cdot 1500 - 1 \cdot (-8850) = 400 \\
-3 \cdot (-3950) + 10 \cdot 1500 + 3 \cdot (-8850) = 300 \\
0 \cdot (-3950) + 6 \cdot 1500 + 1 \cdot (-8850) = 150
\end{cases}
\][/tex]
