Answer :
To solve the system of equations using either Cramer's Rule or Gaussian elimination, we need to manipulate and evaluate the given equations carefully. Here are the steps for solving the system using Gaussian elimination:
Given the system of equations:
1. [tex]\(3x + 2y = z + 1\)[/tex]
2. [tex]\(-5x + 2z = 8 - 5y\)[/tex]
3. [tex]\(x - 1 = y\)[/tex]
We can rewrite the system in the form of augmented matrix for Gaussian elimination. The augmented matrix will look like this after rewriting all equations in the form [tex]\( Ax = B \)[/tex].
1. Rewrite [tex]\(3x + 2y = z + 1\)[/tex] as [tex]\(3x + 2y - z = 1\)[/tex]
2. Rewrite [tex]\(-5x + 2z = 8 - 5y\)[/tex] as [tex]\(-5x + 5y + 2z = 8\)[/tex]
3. Rewrite [tex]\(x - 1 = y\)[/tex] as [tex]\(x - y = 1\)[/tex]
The augmented matrix representation of the above system is:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ -5 & 5 & 2 & | & 8 \\ 1 & -1 & 0 & | & 1 \end{pmatrix} \][/tex]
Step 1: Row Reduction
We will use Gaussian elimination (row operations) to transform this augmented matrix to row-echelon form.
The initial augmented matrix is:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ -5 & 5 & 2 & | & 8 \\ 1 & -1 & 0 & | & 1 \end{pmatrix} \][/tex]
Step 2: Eliminate the [tex]\(x\)[/tex]-term from rows 2 and 3
### Eliminate x from row 2:
Let's multiply Row 1 by [tex]\( \frac{5}{3} \)[/tex] and add to Row 2:
[tex]\[ \frac{5}{3} \times R1 + R2 \to R2 \][/tex]
[tex]\[ R1 = \begin{pmatrix} 3 & 2 & -1 & | & 1 \end{pmatrix} \Rightarrow \frac{5}{3}R1 = \begin{pmatrix} 5 & \frac{10}{3} & -\frac{5}{3} & | & \frac{5}{3} \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} -5 & 5 & 2 & | & 8 \end{pmatrix} \Rightarrow \begin{pmatrix} -5 & 5 & 2 & | & 8 \end{pmatrix} + \begin{pmatrix} 5 & \frac{10}{3} & -\frac{5}{3} & | & \frac{5}{3} \end{pmatrix} = \begin{pmatrix} 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \end{pmatrix} \][/tex]
### Eliminate x from row 3:
Let's multiply Row 1 by [tex]\(\frac{1}{3}\)[/tex] and subtract from Row 3:
[tex]\[ \frac{1}{3} \times R1 - R3 \to R3 \][/tex]
[tex]\[ R1 = \begin{pmatrix} 3 & 2 & -1 & | & 1 \end{pmatrix} \Rightarrow \frac{1}{3}R1 = \begin{pmatrix} 1 & \frac{2}{3} & -\frac{1}{3} & | & \frac{1}{3} \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 1 & -1 & 0 & | & 1 \end{pmatrix} \Rightarrow \begin{pmatrix} 1 & -1 & 0 & | & 1 \end{pmatrix} - \begin{pmatrix} 1 & \frac{2}{3} & -\frac{1}{3} & | & \frac{1}{3} \end{pmatrix} = \begin{pmatrix} 0 & -\frac{5}{3} & \frac{1}{3} & | & \frac{2}{3} \end{pmatrix} \][/tex]
Now our augmented matrix looks like this:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \\ 0 & -\frac{5}{3} & \frac{1}{3} & | & \frac{2}{3} \end{pmatrix} \][/tex]
Step 3: Continue Row Reduction
### Eliminate the y-term from row 3:
Let's multiply Row 2 by [tex]\(\frac{1}{25}\)[/tex] and add to Row 3:
[tex]\[ -\frac{5}{3}R2\to R3 \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \end{pmatrix} \Rightarrow \begin{pmatrix} 0 & \frac{5}{3} & \frac{1}{15} & | & \frac{29}{25} \end{pmatrix} \][/tex]
Now our augmented matrix looks like this:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & 25 & 1 & | & 29 \\ 0 & -5 & 1 & | & 2 \end{pmatrix} \][/tex]
Adding:
[tex]\[ \begin{pmatrix} 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \end{pmatrix} + \begin{pmatrix} 0 & -5 & 1 & | & 2 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 1 & | & \frac{65}{15} \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 0 & 1 & | & \frac{13}{2} \end{pmatrix} \][/tex]
So,
now our matrix looks like this:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & 25 & 1 & | & 29 \\ 0 & 0 & 1 & | & \frac{13}{2} \end{pmatrix} \][/tex]
Step 4: Back Substitution
Let's turn the augmented matrix into equations again to solve the remaining variables:
From the third row machine:
[tex]\[ z = \frac{13}{2} \][/tex]
Substitute [tex]\( z \)[/tex] into 2nd row:
[tex]\[ 25y + 1(\frac{32}{15}) = 29 \][/tex]
[tex]\[ y = \frac{9}{10} \][/tex]
Substitute [tex]\( y \)[/tex] into row 1:
[tex]\[ 3(0.19) + 2 * (0.9) - 0.65 = 1 \][/tex]
so we have:
[tex]\[ x = \frac{19}{10} \][/tex]
[tex]\[ y = \frac{9}{10} \][/tex]
[tex]\[ z = \frac{13}{2} \][/tex]
### Final solution
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & 25 & 1 & | & 29 \\ 0 & 0 & 1 & | & \frac{65}{15} \end{pmatrix} =m \][/tex]
Thus,
[tex]\[x = \frac{10}{25}\][/tex]
Given the system of equations:
1. [tex]\(3x + 2y = z + 1\)[/tex]
2. [tex]\(-5x + 2z = 8 - 5y\)[/tex]
3. [tex]\(x - 1 = y\)[/tex]
We can rewrite the system in the form of augmented matrix for Gaussian elimination. The augmented matrix will look like this after rewriting all equations in the form [tex]\( Ax = B \)[/tex].
1. Rewrite [tex]\(3x + 2y = z + 1\)[/tex] as [tex]\(3x + 2y - z = 1\)[/tex]
2. Rewrite [tex]\(-5x + 2z = 8 - 5y\)[/tex] as [tex]\(-5x + 5y + 2z = 8\)[/tex]
3. Rewrite [tex]\(x - 1 = y\)[/tex] as [tex]\(x - y = 1\)[/tex]
The augmented matrix representation of the above system is:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ -5 & 5 & 2 & | & 8 \\ 1 & -1 & 0 & | & 1 \end{pmatrix} \][/tex]
Step 1: Row Reduction
We will use Gaussian elimination (row operations) to transform this augmented matrix to row-echelon form.
The initial augmented matrix is:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ -5 & 5 & 2 & | & 8 \\ 1 & -1 & 0 & | & 1 \end{pmatrix} \][/tex]
Step 2: Eliminate the [tex]\(x\)[/tex]-term from rows 2 and 3
### Eliminate x from row 2:
Let's multiply Row 1 by [tex]\( \frac{5}{3} \)[/tex] and add to Row 2:
[tex]\[ \frac{5}{3} \times R1 + R2 \to R2 \][/tex]
[tex]\[ R1 = \begin{pmatrix} 3 & 2 & -1 & | & 1 \end{pmatrix} \Rightarrow \frac{5}{3}R1 = \begin{pmatrix} 5 & \frac{10}{3} & -\frac{5}{3} & | & \frac{5}{3} \end{pmatrix} \][/tex]
[tex]\[ R2 = \begin{pmatrix} -5 & 5 & 2 & | & 8 \end{pmatrix} \Rightarrow \begin{pmatrix} -5 & 5 & 2 & | & 8 \end{pmatrix} + \begin{pmatrix} 5 & \frac{10}{3} & -\frac{5}{3} & | & \frac{5}{3} \end{pmatrix} = \begin{pmatrix} 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \end{pmatrix} \][/tex]
### Eliminate x from row 3:
Let's multiply Row 1 by [tex]\(\frac{1}{3}\)[/tex] and subtract from Row 3:
[tex]\[ \frac{1}{3} \times R1 - R3 \to R3 \][/tex]
[tex]\[ R1 = \begin{pmatrix} 3 & 2 & -1 & | & 1 \end{pmatrix} \Rightarrow \frac{1}{3}R1 = \begin{pmatrix} 1 & \frac{2}{3} & -\frac{1}{3} & | & \frac{1}{3} \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 1 & -1 & 0 & | & 1 \end{pmatrix} \Rightarrow \begin{pmatrix} 1 & -1 & 0 & | & 1 \end{pmatrix} - \begin{pmatrix} 1 & \frac{2}{3} & -\frac{1}{3} & | & \frac{1}{3} \end{pmatrix} = \begin{pmatrix} 0 & -\frac{5}{3} & \frac{1}{3} & | & \frac{2}{3} \end{pmatrix} \][/tex]
Now our augmented matrix looks like this:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \\ 0 & -\frac{5}{3} & \frac{1}{3} & | & \frac{2}{3} \end{pmatrix} \][/tex]
Step 3: Continue Row Reduction
### Eliminate the y-term from row 3:
Let's multiply Row 2 by [tex]\(\frac{1}{25}\)[/tex] and add to Row 3:
[tex]\[ -\frac{5}{3}R2\to R3 \][/tex]
[tex]\[ R2 = \begin{pmatrix} 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \end{pmatrix} \Rightarrow \begin{pmatrix} 0 & \frac{5}{3} & \frac{1}{15} & | & \frac{29}{25} \end{pmatrix} \][/tex]
Now our augmented matrix looks like this:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & 25 & 1 & | & 29 \\ 0 & -5 & 1 & | & 2 \end{pmatrix} \][/tex]
Adding:
[tex]\[ \begin{pmatrix} 0 & \frac{25}{3} & \frac{1}{3} & | & \frac{29}{3} \end{pmatrix} + \begin{pmatrix} 0 & -5 & 1 & | & 2 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 1 & | & \frac{65}{15} \end{pmatrix} \][/tex]
[tex]\[ R3 = \begin{pmatrix} 0 & 0 & 1 & | & \frac{13}{2} \end{pmatrix} \][/tex]
So,
now our matrix looks like this:
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & 25 & 1 & | & 29 \\ 0 & 0 & 1 & | & \frac{13}{2} \end{pmatrix} \][/tex]
Step 4: Back Substitution
Let's turn the augmented matrix into equations again to solve the remaining variables:
From the third row machine:
[tex]\[ z = \frac{13}{2} \][/tex]
Substitute [tex]\( z \)[/tex] into 2nd row:
[tex]\[ 25y + 1(\frac{32}{15}) = 29 \][/tex]
[tex]\[ y = \frac{9}{10} \][/tex]
Substitute [tex]\( y \)[/tex] into row 1:
[tex]\[ 3(0.19) + 2 * (0.9) - 0.65 = 1 \][/tex]
so we have:
[tex]\[ x = \frac{19}{10} \][/tex]
[tex]\[ y = \frac{9}{10} \][/tex]
[tex]\[ z = \frac{13}{2} \][/tex]
### Final solution
[tex]\[ \begin{pmatrix} 3 & 2 & -1 & | & 1 \\ 0 & 25 & 1 & | & 29 \\ 0 & 0 & 1 & | & \frac{65}{15} \end{pmatrix} =m \][/tex]
Thus,
[tex]\[x = \frac{10}{25}\][/tex]
