To solve the system of linear equations using matrices, let's break down the given matrix equation:
[tex]\[
\begin{pmatrix}
3 & 10 \\
1 & 3
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
6 \\
1
\end{pmatrix}
\][/tex]
This matrix equation corresponds to the following system of linear equations:
1. [tex]\(3x + 10y = 6\)[/tex]
2. [tex]\(1x + 3y = 1\)[/tex]
We can solve this system step-by-step:
### Step 1: Setting Up the Augmented Matrix
We start by setting up the augmented matrix from the system of equations:
[tex]\[
\left[\begin{array}{ccc}
3 & 10 & 6 \\
1 & 3 & 1
\end{array}\right]
\][/tex]
### Step 2: Performing Row Operations to Achieve Row Echelon Form
We'll use Gaussian elimination to transform the augmented matrix into its row echelon form.
First, use the second row to eliminate the [tex]\(x\)[/tex]-coefficient in the first row. Multiply the second row by 3 and subtract it from the first row:
[tex]\[
\begin{aligned}
R_1 - 3R_2 & \rightarrow R_1 \\
3 & 10 & 6 \\
- 3(1 & 3 & 1) \\
\end{aligned}
\][/tex]
[tex]\[
\left[\begin{array}{ccc}
3 & 10 & 6 \\
1 & 3 & 1
\end{array}\right]
\Rightarrow
\left[\begin{array}{ccc}
0 & 1 & 3 \\
1 & 3 & 1
\end{array}\right]
\][/tex]
### Step 3: Back Substitution
At this point, the augmented matrix is in row echelon form, and we can proceed to back substitution.
From the second row:
[tex]\[
y = 3
\][/tex]
Next, substitute [tex]\(y = 3\)[/tex] back into the first equation:
[tex]\[
3x + 10(3) = 6 \\
3x + 30 = 6 \\
3x = 6 - 30 \\
3x = -24 \\
x = -8
\][/tex]
### Conclusion
Thus, the solution to the system of linear equations is:
[tex]\[
x = -8, \quad y = 3
\][/tex]
So, the final solution for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] is:
[tex]\(
x = -8, \quad y = 3
\)[/tex]
Therefore, the solution to the system of equations is:
[tex]\(
(-8, 3)
\)[/tex]
