Discover how IDNLearn.com can help you find the answers you need quickly and easily. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.
Sagot :
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]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
