Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
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 hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.