# Find the solution of the following system using Gauss elimination. x - 5y + z = 16 5y - 4z = 1 ...

## Question:

Find the solution of the following system using Gauss elimination.

x - 5y + z = 16

5y - 4z = 1

x + y + 5z = -18

## Linear System Solution:

A linear system of equations can be solved by the process of Gauss's elimination. The essence of the process is to perform appropriate row operations on the matrix of equations coefficients augmented with the free term vector column to bring it to the upper triangular form and then continue operations until it is reduced to the identity matrix. The transformed free term vector will then produce a vector of solutions of the system.

Performing consecutive row operations on augmented matrix, we get:

1) Subtract first row from the third one:

{eq}\left ( \begin{matrix} 1&-5&1&16 \\ 0&5&-4&1 \\ 1&1&5&-18 \end{matrix} \right) \rightarrow \left ( \begin{matrix} 1&-5&1&16 \\ 0&5&-4&1 \\ 0&3&2&-17 \end{matrix} \right) {/eq}

2) Add second row to the first one:

{eq}\left ( \begin{matrix} 1&-5&1&16 \\ 0&5&-4&1 \\ 0&3&2&-17 \end{matrix} \right) \rightarrow \left ( \begin{matrix} 1&0&-3&17 \\ 0&5&-4&1 \\ 0&3&2&-17 \end{matrix} \right) {/eq}

3) Multiply the second row by {eq}\dfrac {3}{5} {/eq} and subtract from the third row, then divide the third row by 22 and multiply by 5:

{eq}\left ( \begin{matrix} 1&0&-3&17 \\ 0&5&-4&1 \\ 0&3&2&-17 \end{matrix} \right) \rightarrow \left ( \begin{matrix} 1&0&-3&17 \\ 0&5&-4&1 \\ 0&0&\dfrac {22}{5}&-\dfrac {88}{5} \end{matrix} \right) \rightarrow \left ( \begin{matrix} 1&0&-3&17 \\ 0&5&-4&1 \\ 0&0& 1&-4 \end{matrix} \right) {/eq}

4) Multiply the third row by 4 and add to the second row, then divide the second row by five:

{eq}\left ( \begin{matrix} 1&0&-3&17 \\ 0&5&-4&1 \\ 0&0& 1&-4 \end{matrix} \right)\rightarrow \left ( \begin{matrix} 1&0&-3&17 \\ 0&5&0&-15 \\ 0&0& 1&-4 \end{matrix} \right) \rightarrow \left ( \begin{matrix} 1&0&-3&17 \\ 0&1&0&-3 \\ 0&0& 1&-4 \end{matrix} \right) {/eq}

5) Multiply the third row by 3 and add to the first row:

{eq}\left ( \begin{matrix} 1&0&-3&17 \\ 0&1&0&-3 \\ 0&0& 1&-4 \end{matrix} \right) \rightarrow \left ( \begin{matrix} 1&0&0&5 \\ 0&1&0&-3 \\ 0&0& 1&-4 \end{matrix} \right) {/eq}

Hence, the solution is:

{eq}x = 5 \\ y = -3 \\ z = -4 {/eq}