Answer :
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]
[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]