Question

Solve the following system of equations using either Gaussian elimination or a matrix. Write your answer as a point and explain how you would check your work:

2x - y + 4z = 33
x + 2y - 3z = -26
-5 x - 3y +5z = 23

Answer 1

x = 5, y = -11 and z = 3

As a point it is (5, -11, 3)

You can check your work by plugging these values into each of the three equations and seeing if they satisfy each equation

Explanation:

Using Gaussian elimination

Write a matrix with the coefficients and solutions

`\begin{bmatrix}2&-1&4& | &33\\ 1&2&-3& |&-26\\ -5&-3&5&|&23\end{bmatrix}`

Reduce matrix to row echelon form:

`\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}`

Steps

`\begin{pmatrix}2&-1&4&33\\ 1&2&-3&-26\\ -5&-3&5&23\end{pmatrix}`

Swap matrix rows
`R_1\:\leftrightarrow \:R_3`

`=\begin{pmatrix}-5&-3&5&23\\ 1&2&-3&-26\\ 2&-1&4&33\end{pmatrix}`

Cancel leading coefficient in row
`R_2` by performing
`R_2\:\leftarrow \:R_2+(1)/(5)\cdot \:R_1`

`=\begin{pmatrix}-5&-3&5&23\\ 0&(7)/(5)&-2&-(107)/(5)\\ 2&-1&4&33\end{pmatrix}`

Cancel leading coefficient in row
`R_3` by performing
`R_3\:\leftarrow \:R_3+(2)/(5)\cdot \:R_1`

`=\begin{pmatrix}-5&-3&5&23\\ 0&(7)/(5)&-2&-(107)/(5)\\ 0&-(11)/(5)&6&(211)/(5)\end{pmatrix}`

Swap matrix rows:
`R_2\:\leftrightarrow \:R_3`

`=\begin{pmatrix}-5&-3&5&23\\ 0&-(11)/(5)&6&(211)/(5)\\ 0&(7)/(5)&-2&-(107)/(5)\end{pmatrix}`

Cancel leading coefficient in row
`R_3` by performing
`R_3\:\leftarrow \:R_3+(7)/(11)\cdot \:R_2`

`=\begin{pmatrix}-5&-3&5&23\\ 0&-(11)/(5)&6&(211)/(5)\\ 0&0&(20)/(11)&(60)/(11)\end{pmatrix}`

Multiply matrix row by constant:
`\:R_3\:\leftarrow (11)/(20)\cdot \:R_3`

`=\begin{pmatrix}-5&-3&5&23\\ 0&-(11)/(5)&6&(211)/(5)\\ 0&0&1&3\end{pmatrix}`

Cancel leading coefficient in row
`\:R_2\:` by performing
`R_2\:\leftarrow \:R_2-6\cdot \:R_3`

`=\begin{pmatrix}-5&-3&5&23\\ 0&-(11)/(5)&0&(121)/(5)\\ 0&0&1&3\end{pmatrix}`

Cancel leading coefficient in row
`R_1` by performing
`R_1\:\leftarrow \:R_1-5\cdot \:R_3`

`=\begin{pmatrix}-5&-3&0&8\\ 0&-(11)/(5)&0&(121)/(5)\\ 0&0&1&3\end{pmatrix}`

Multiply matrix row by constant:
`R_2\:\leftarrow \:-(5)/(11)\cdot \:R_2`

`=\begin{pmatrix}-5&-3&0&8\\ 0&1&0&-11\\ 0&0&1&3\end{pmatrix}`

Cancel leading coefficient in row
`R_1` by performing
`R_1\:\leftarrow \:R_1+3\cdot \:R_2`

`\begin{pmatrix}-5&0&0&-25\\ 0&1&0&-11\\ 0&0&1&3\end{pmatrix}`

Multiply matrix row by constant :
`R_1\:\leftarrow \:-(1)/(5)\cdot \:R_1`

`=\begin{pmatrix}1&0&0&5\\ 0&1&0&-11\\ 0&0&1&3\end{pmatrix}`

This means x = 5, y = -11 and z = 3

Other row transformation sequences are possible but as you long as you can get a matrix to row echelon form you can determine what the solution set is