Question

Solve the system of equations using matrices use the Gaussian elimination method with back substitutionWhat’s the solution set? ( 3 separate numbers?

Answer 1

We need to start by rearranging the equations into the form: ax+by+cz=d

Equation 1:

`\begin{gathered} x+3y=z+15 \\ \text{Subtract z from both sides} \\ x+3y-z=z+15-z \\ x+3y-z=15 \end{gathered}`

Equation 2:

`\begin{gathered} x=-3+y-z \\ \text{Subtract y from both sides} \\ x-y=-3+y-z-y \\ x-y=-3-z \\ \text{Add z to both sides} \\ x-y+z=-3-z+z \\ x-y+z=-3 \end{gathered}`

Equation 3:

`x+y+7z=13`

Now, rewrite the system as an augmented matrix:

`\begin{bmatrix}{1} & {3} & -1 & {15} \\ {1} & {-1} & +1 & {-3} \\ {1} & {1} & +7 & {13} \\ {} & {} & {} & {}\end{bmatrix}`

1. Subtract the first row from the second row, and the result will be the second row. It is written as:

`R_2-R_1\to R_2`

The new matrix is:

`\begin{bmatrix}{1} & {3} & -1 & {15} \\ {0} & {-4} & +2 & {-18} \\ {1} & {1} & +7 & {13} \\ {} & {} & {} & {}\end{bmatrix}`

The next operation is:

`R_3-R_1\to R_3`

The matrix is:

`\begin{bmatrix}{1} & {3} & -1 & {15} \\ {0} & {-4} & +2 & {-18} \\ {0} & {-2} & +8 & {-2} \\ {} & {} & {} & {}\end{bmatrix}`

Next step:

`\begin{gathered} R_3-(1)/(2)R_2\to R_3 \\ \begin{bmatrix}{1} & {3} & -1 & {15} \\ {0} & {-4} & +2 & {-18} \\ {0} & {0} & +7 & {7} \\ {} & {} & {} & {}\end{bmatrix} \end{gathered}`

Next step:

`\begin{gathered} (1)/(7)R_3\to R_3 \\ \begin{bmatrix}{1} & {3} & -1 & {15} \\ {0} & {-4} & +2 & {-18} \\ {0} & {0} & +1 & {1} \\ {} & {} & {} & {}\end{bmatrix} \end{gathered}`

Then, we have the following system of equations:

`\begin{gathered} x+3y-z=15\text{ Equation 1} \\ -4y+2z=-18\text{ Equation 2} \\ z=1\text{ Equation 3} \end{gathered}`

Replace equation 3 into equation 2 and solve for y:

`\begin{gathered} -4y+2\cdot1=-18 \\ -4y+2=-18 \\ -4y=-18-2 \\ -4y=-20 \\ y=(-20)/(-4) \\ y=5 \end{gathered}`

Replace y and z into equation 1 and solve for x:

`\begin{gathered} x+3\cdot5-1=15 \\ x+15-1=15 \\ x=15-15+1 \\ x=1 \end{gathered}`

The solution set is { 1 , 5 , 1 }