Answer :
To solve the system of linear equations using matrices, we'll break it down into several clear steps. Let’s start with the given matrix equation:
[tex]\[ \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix} \][/tex]
This can be interpreted as the following system of linear equations:
1. [tex]\(2x + 3y = 5\)[/tex]
2. [tex]\(x + 2y = 4\)[/tex]
### Step 1: Write the augmented matrix
We start by writing the augmented matrix for the given system of equations:
[tex]\[ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 1 & 2 & 4 \end{array}\right] \][/tex]
### Step 2: Perform row operations to solve
To solve the system using matrix operations, we can use methods like Gaussian elimination. Our goal is to transform this augmented matrix into a form where we can easily read off the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
#### (a) Ensure the leading entry of the first row is 1:
Although it's already a non-zero pivot (2), we'll keep it as it is and perform row operations directly.
#### (b) Make the leading coefficient in the second row of the first column 0:
We can subtract an appropriate multiple of the first row from the second row.
Row 2 [tex]\( \leftarrow \)[/tex] Row 2 - (1/2) × Row 1
[tex]\[ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 0 & \left(2 - \frac{3}{2}\right) & \left(4 - \frac{5}{2}\right) \end{array}\right] \implies \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 0 & \frac{1}{2} & \frac{3}{2} \end{array}\right] \][/tex]
#### (c) Scale the second row to make the pivot equal to 1:
Multiply the second row by 2:
[tex]\[ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 0 & 1 & 3 \end{array}\right] \][/tex]
#### (d) Make the remaining element in the first row of the second column 0:
Row 1 [tex]\( \leftarrow \)[/tex] Row 1 - 3 × Row 2
[tex]\[ \left[\begin{array}{cc|c} 2 & 0 & -4 \\ 0 & 1 & 3 \end{array}\right] \][/tex]
Next, scale Row 1 by 1/2 to make the leading coefficient in Row 1 equal to 1:
[tex]\[ \left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 3 \end{array}\right] \][/tex]
### Step 3: Interpret the results
From the final matrix, we have the following equations:
1. [tex]\(x = -2\)[/tex]
2. [tex]\(y = 3\)[/tex]
### Final Solution
Thus, the solution to the system of equations is:
[tex]\[ x = -2, \; y = 3 \][/tex]
[tex]\[ \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix} \][/tex]
This can be interpreted as the following system of linear equations:
1. [tex]\(2x + 3y = 5\)[/tex]
2. [tex]\(x + 2y = 4\)[/tex]
### Step 1: Write the augmented matrix
We start by writing the augmented matrix for the given system of equations:
[tex]\[ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 1 & 2 & 4 \end{array}\right] \][/tex]
### Step 2: Perform row operations to solve
To solve the system using matrix operations, we can use methods like Gaussian elimination. Our goal is to transform this augmented matrix into a form where we can easily read off the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
#### (a) Ensure the leading entry of the first row is 1:
Although it's already a non-zero pivot (2), we'll keep it as it is and perform row operations directly.
#### (b) Make the leading coefficient in the second row of the first column 0:
We can subtract an appropriate multiple of the first row from the second row.
Row 2 [tex]\( \leftarrow \)[/tex] Row 2 - (1/2) × Row 1
[tex]\[ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 0 & \left(2 - \frac{3}{2}\right) & \left(4 - \frac{5}{2}\right) \end{array}\right] \implies \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 0 & \frac{1}{2} & \frac{3}{2} \end{array}\right] \][/tex]
#### (c) Scale the second row to make the pivot equal to 1:
Multiply the second row by 2:
[tex]\[ \left[\begin{array}{cc|c} 2 & 3 & 5 \\ 0 & 1 & 3 \end{array}\right] \][/tex]
#### (d) Make the remaining element in the first row of the second column 0:
Row 1 [tex]\( \leftarrow \)[/tex] Row 1 - 3 × Row 2
[tex]\[ \left[\begin{array}{cc|c} 2 & 0 & -4 \\ 0 & 1 & 3 \end{array}\right] \][/tex]
Next, scale Row 1 by 1/2 to make the leading coefficient in Row 1 equal to 1:
[tex]\[ \left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 3 \end{array}\right] \][/tex]
### Step 3: Interpret the results
From the final matrix, we have the following equations:
1. [tex]\(x = -2\)[/tex]
2. [tex]\(y = 3\)[/tex]
### Final Solution
Thus, the solution to the system of equations is:
[tex]\[ x = -2, \; y = 3 \][/tex]
