Answer :
Sure, let's solve this system of linear equations using the Gaussian elimination method:
[tex]\[ \begin{array}{rr} x + 2y + 2z &= 10 \\ -x + 3y + 5z &= 13 \\ x + y - 4z &= -13 \end{array} \][/tex]
First, we construct the augmented matrix for the system:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ -1 & 3 & 5 & | & 13 \\ 1 & 1 & -4 & | & -13 \end{pmatrix} \][/tex]
Step 1: Eliminate \(x\) from the second and third rows
Add the first row to the second row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 5 & 7 & | & 23 \\ 1 & 1 & -4 & | & -13 \end{pmatrix} \][/tex]
Subtract the first row from the third row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 5 & 7 & | & 23 \\ 0 & -1 & -6 & | & -23 \end{pmatrix} \][/tex]
Step 2: Eliminate \(y\) from the third row
Multiply the second row by \( \frac{1}{5} \):
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & -1 & -6 & | & -23 \end{pmatrix} \][/tex]
Add the second row to the third row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & 0 & -\frac{23}{5} & | & \frac{23}{5} - 23 \end{pmatrix} \][/tex]
Simplifying the third row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & 0 & -\frac{23}{5} & | & -\frac{92}{5} \end{pmatrix} \][/tex]
Step 3: Solve for \(z\)
From the third row:
[tex]\[ -\frac{23}{5}z = -\frac{92}{5} \implies z = 4 \][/tex]
Step 4: Back-substitute to find \(y\) and \(x\)
Substitute \(z = 4\) into the second row:
[tex]\[ 0 + y + \frac{7}{5}(4) = \frac{23}{5} \implies y + \frac{28}{5} = \frac{23}{5} \implies y = \frac{23}{5} - \frac{28}{5} = -1 \][/tex]
Substitute \( y = -1 \) and \( z = 4 \) into the first row:
[tex]\[ x + 2(-1) + 2(4) = 10 \implies x - 2 + 8 = 10 \implies x + 6 = 10 \implies x = 4 \][/tex]
Therefore, the solution is \((4, -1, 4)\).
So, the correct choice is:
A. There is one solution. The solution is [tex]\(\{(4, -1, 4)\}\)[/tex].
[tex]\[ \begin{array}{rr} x + 2y + 2z &= 10 \\ -x + 3y + 5z &= 13 \\ x + y - 4z &= -13 \end{array} \][/tex]
First, we construct the augmented matrix for the system:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ -1 & 3 & 5 & | & 13 \\ 1 & 1 & -4 & | & -13 \end{pmatrix} \][/tex]
Step 1: Eliminate \(x\) from the second and third rows
Add the first row to the second row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 5 & 7 & | & 23 \\ 1 & 1 & -4 & | & -13 \end{pmatrix} \][/tex]
Subtract the first row from the third row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 5 & 7 & | & 23 \\ 0 & -1 & -6 & | & -23 \end{pmatrix} \][/tex]
Step 2: Eliminate \(y\) from the third row
Multiply the second row by \( \frac{1}{5} \):
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & -1 & -6 & | & -23 \end{pmatrix} \][/tex]
Add the second row to the third row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & 0 & -\frac{23}{5} & | & \frac{23}{5} - 23 \end{pmatrix} \][/tex]
Simplifying the third row:
[tex]\[ \begin{pmatrix} 1 & 2 & 2 & | & 10 \\ 0 & 1 & \frac{7}{5} & | & \frac{23}{5} \\ 0 & 0 & -\frac{23}{5} & | & -\frac{92}{5} \end{pmatrix} \][/tex]
Step 3: Solve for \(z\)
From the third row:
[tex]\[ -\frac{23}{5}z = -\frac{92}{5} \implies z = 4 \][/tex]
Step 4: Back-substitute to find \(y\) and \(x\)
Substitute \(z = 4\) into the second row:
[tex]\[ 0 + y + \frac{7}{5}(4) = \frac{23}{5} \implies y + \frac{28}{5} = \frac{23}{5} \implies y = \frac{23}{5} - \frac{28}{5} = -1 \][/tex]
Substitute \( y = -1 \) and \( z = 4 \) into the first row:
[tex]\[ x + 2(-1) + 2(4) = 10 \implies x - 2 + 8 = 10 \implies x + 6 = 10 \implies x = 4 \][/tex]
Therefore, the solution is \((4, -1, 4)\).
So, the correct choice is:
A. There is one solution. The solution is [tex]\(\{(4, -1, 4)\}\)[/tex].
