Answer:
A. The solution to the system of equations is (2, -4, 3)
Explanation:
Given the system of linear equations:
[tex]\begin{gathered} x-4y+5z=33 \\ 2x+y+z=3 \\ -4x+4y-3z=-33 \end{gathered}[/tex]The row matrix for the system is given below:
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ {2} & {1} & {1} & {3} \\ {-4} & {4} & {-3} & {-33} \\ {} & {} & {} & {}\end{bmatrix}[/tex]We perform the row operations below to solve for x, y and z.
Step 1: Multiply R2 by 1/2:
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 1 & 0.5 & 0.5 & 1.5 \\ {-4} & {4} & {-3} & {-33} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 2: Multiply R3 by -1/4:
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 1 & 0.5 & 0.5 & 1.5 \\ 1 & -1 & {\frac{3}{4}} & {\frac{33}{4}} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 3: Carry out the operation: R2-R1-->R2
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 4.5 & -4.5 & -\frac{63}{2} \\ 1 & -1 & {\frac{3}{4}} & {\frac{33}{4}} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 4: Carry out the operation: R3-R1-->R3
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 4.5 & -4.5 & -\frac{63}{2} \\ 0 & 3 & -\frac{17}{4} & -{\frac{99}{4}} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 5: Multiply R2 by 2/9.
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 3 & -\frac{17}{4} & -{\frac{99}{4}} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 6: Multiply R3 by 1/3:
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 1 & -\frac{17}{12} & -{\frac{33}{4}} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 7: Subtract R2 from R3
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 0 & -\frac{5}{12} & -{\frac{5}{4}} \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 8: Multiply R3 by -12/5.
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 9: Add R2 to R3 on R2:
[tex]\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 10: Perform the operation: R1-5R3 on Row 1.
[tex]\begin{bmatrix}{1} & {-4} & 0 & 18 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}[/tex]Step 11: Perform the operation: R1+4R2 on Row 1.
[tex]\begin{bmatrix}{1} & 0 & 0 & 2 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}[/tex]We see that the identity 3 x 3 matrix is formed on the left.
Therefore, the solution to the system of equations is:
[tex](x,y,z)=(2,-4,3)[/tex]Option A is correct.
