Question

Solve the system of equations by finding the reduced row-echelon form of the augmentee matrix for the system of equations..

2x+y+z=-3
3x-5y+3z=-4
5x-y+2z=-2

Answer 1

`x=1,y=-1,z=-4`

Explanation:

we are given system of equations as

2x+y+z=-3

3x-5y+3z=-4

5x-y+2z=-2

Firstly, we can write augmented matrix

`A=\begin{pmatrix}2&1&1&-3\\ 3&-5&3&-4\\ 5&-1&2&-2\end{pmatrix}`

now, we can change it into reduced- row echelon form

`\mathrm{Swap\:matrix\:rows:}\:R_1\:\leftrightarrow \:R_3`

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

`\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-(3)/(5)\cdot \:R_1`

`=\begin{pmatrix}5&-1&2&-2\\ 0&-(22)/(5)&(9)/(5)&-(14)/(5)\\ 2&1&1&-3\end{pmatrix}`

`\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3-(2)/(5)\cdot \:R_1`

`=\begin{pmatrix}5&-1&2&-2\\ 0&-(22)/(5)&(9)/(5)&-(14)/(5)\\ 0&(7)/(5)&(1)/(5)&-(11)/(5)\end{pmatrix}`

`\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_3\:\mathrm{\:by\:performing}\:R_3\:\leftarrow \:R_3+(7)/(22)\cdot \:R_2`

`=\begin{pmatrix}5&-1&2&-2\\ 0&-(22)/(5)&(9)/(5)&-(14)/(5)\\ 0&0&(17)/(22)&-(34)/(11)\end{pmatrix}`

`\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_3\:\leftarrow (22)/(17)\cdot \:R_3`

`=\begin{pmatrix}5&-1&2&-2\\ 0&-(22)/(5)&(9)/(5)&-(14)/(5)\\ 0&0&1&-4\end{pmatrix}`

`\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_2\:\mathrm{\:by\:performing}\:R_2\:\leftarrow \:R_2-(9)/(5)\cdot \:R_3`

`=\begin{pmatrix}5&-1&2&-2\\ 0&-(22)/(5)&0&(22)/(5)\\ 0&0&1&-4\end{pmatrix}`

`\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1-2\cdot \:R_3`

`=\begin{pmatrix}5&-1&0&6\\ 0&-(22)/(5)&0&(22)/(5)\\ 0&0&1&-4\end{pmatrix}`

`\mathrm{Multiply\:matrix\:row\:by\:constant:}\:R_2\:\leftarrow \:-(5)/(22)\cdot \:R_2`

`=\begin{pmatrix}5&-1&0&6\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}`

`\mathrm{Cancel\:leading\:coefficient\:in\:row\:}\:R_1\:\mathrm{\:by\:performing}\:R_1\:\leftarrow \:R_1+1\cdot \:R_2`

`=\begin{pmatrix}5&0&0&5\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}`

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

`=\begin{pmatrix}1&0&0&1\\ 0&1&0&-1\\ 0&0&1&-4\end{pmatrix}`

so, we will get solution as

`x=1,y=-1,z=-4`