Question

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

x + y - z = -2
2x - y + 3z = 9
x - 4y - 2z = 1

Answer 1

x = 1 , y = - 1 , z = 2

Explanation:

`\begin{pmatrix}1&1&-1&-2\\ 2&-1&3&9\\ 1&-1&-2&1\end{pmatrix}\\\\\\= \begin{pmatrix}2&-1&3&9\\ 1&1&-1&-2\\ 1&-1&-2&1\end{pmatrix}`
`[ \ swap \ R_1 \ and \ R_2 \ ]`

`=\begin{pmatrix}2&-1&3&9\\ 0&(3)/(2)&-(5)/(2)&-(13)/(2)\\ 1&-4&-2&1\end{pmatrix}`
`[ \ R_2 = R_2 - (1)/(2) R_1 \ ]`

`=\begin{pmatrix}2&-1&3&9\\ 0&(3)/(2)&-(5)/(2)&-(13)/(2)\\ 0&-(7)/(2)&-(7)/(2)&-(7)/(2)\end{pmatrix}`
`[ \ R_3 = R_3 - (1)/(2) R_1 \ ]`

`=\begin{pmatrix}2&-1&3&9\\ 0&-(7)/(2)&-(7)/(2)&-(7)/(2)\\ 0&(3)/(2)&-(5)/(2)&-(13)/(2)\end{pmatrix}`
`[ \ swap \ R_2 \ and \ R_3 \ ]`

`=\begin{pmatrix}2&-1&3&9\\ 0&-(7)/(2)&-(7)/(2)&-(7)/(2)\\ 0&0&-4&-8\end{pmatrix}`
`[ \ R_3 = R_3 + (3)/(7)R_2 \ ]`

Therefore,

`-4z = - 8\\\\z = 2`

`-(7)/(2) y - (7)/(2)z = -(7)/(2)\\\\y + z = 1\\\\y + 2 = 1 \\\\y = 1 - 2 = - 1`

`2x - 1y + 3z = 9\\\\2x -1(-1) + 3(2) = 9\\\\2x + 1 + 6 = 9\\\\2x = 9 - 7\\\\2x = 2 \\\\x = 1`