Question

Hello, I need some assistance with this precalculus question, please?HW Q12

Answer 1

A. The solution to the system of equations is (2, -4, 3)

Explanation:

Given the system of linear equations:

`\begin{gathered} x-4y+5z=33 \\ 2x+y+z=3 \\ -4x+4y-3z=-33 \end{gathered}`

The row matrix for the system is given below:

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ {2} & {1} & {1} & {3} \\ {-4} & {4} & {-3} & {-33} \\ {} & {} & {} & {}\end{bmatrix}`

We perform the row operations below to solve for x, y and z.

Step 1: Multiply R2 by 1/2:

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 1 & 0.5 & 0.5 & 1.5 \\ {-4} & {4} & {-3} & {-33} \\ {} & {} & {} & {}\end{bmatrix}`

Step 2: Multiply R3 by -1/4:

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 1 & 0.5 & 0.5 & 1.5 \\ 1 & -1 & {(3)/(4)} & {(33)/(4)} \\ {} & {} & {} & {}\end{bmatrix}`

Step 3: Carry out the operation: R2-R1-->R2

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 4.5 & -4.5 & -(63)/(2) \\ 1 & -1 & {(3)/(4)} & {(33)/(4)} \\ {} & {} & {} & {}\end{bmatrix}`

Step 4: Carry out the operation: R3-R1-->R3

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 4.5 & -4.5 & -(63)/(2) \\ 0 & 3 & -(17)/(4) & -{(99)/(4)} \\ {} & {} & {} & {}\end{bmatrix}`

Step 5: Multiply R2 by 2/9.

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 3 & -(17)/(4) & -{(99)/(4)} \\ {} & {} & {} & {}\end{bmatrix}`

Step 6: Multiply R3 by 1/3:

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 1 & -(17)/(12) & -{(33)/(4)} \\ {} & {} & {} & {}\end{bmatrix}`

Step 7: Subtract R2 from R3

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 0 & -(5)/(12) & -{(5)/(4)} \\ {} & {} & {} & {}\end{bmatrix}`

Step 8: Multiply R3 by -12/5.

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & -1 & -7 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}`

Step 9: Add R2 to R3 on R2:

`\begin{bmatrix}{1} & {-4} & {5} & {33} \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}`

Step 10: Perform the operation: R1-5R3 on Row 1.

`\begin{bmatrix}{1} & {-4} & 0 & 18 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}`

Step 11: Perform the operation: R1+4R2 on Row 1.

`\begin{bmatrix}{1} & 0 & 0 & 2 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 3 \\ {} & {} & {} & {}\end{bmatrix}`

We see that the identity 3 x 3 matrix is formed on the left.

Therefore, the solution to the system of equations is:

`(x,y,z)=(2,-4,3)`

Option A is correct.