Question

Consider the system of linear equations 2x+3y−1z=2

x+2y+z=3
−x−y+3z=1
a. Write the system of the equations above in an augmented matrix [A∣B] b. Solve the system using Gauss Elimination Method.

Answer 1

`(x,y,z)=(-5,4,0)`

Explanation:

Use Gauss Elimination Method

`\left[\begin{array}{cccc}2&3&-1&2\\1&2&1&3\\-1&-1&3&1\end{array}\right] \\\\\\\left[\begin{array}{cccc}1&(3)/(2)&-(1)/(2)&1\\1&2&1&3\\-1&-1&3&1\end{array}\right] \leftarrow (1)/(2)R_1\\\\\\\left[\begin{array}{cccc}1&(3)/(2)&-(1)/(2)&1\\0&-(1)/(2)&-(3)/(2)&-2\\-1&-1&3&1\end{array}\right] \leftarrow R_1-R_2\\\\\\\left[\begin{array}{cccc}1&(3)/(2)&-(1)/(2)&1\\0&-(1)/(2)&-(3)/(2)&-2\\0&(1)/(2)&(5)/(2)&2\end{array}\right] \leftarrow R_3+R_1`

`\left[\begin{array}{cccc}1&(3)/(2)&-(1)/(2)&1\\0&1&3&4\\0&(1)/(2)&(5)/(2)&2\end{array}\right] \leftarrow -2R_2\\\\\\\left[\begin{array}{cccc}1&(3)/(2)&-(1)/(2)&1\\0&1&3&4\\0&0&2&0\end{array}\right] \leftarrow 2R_3-R_2\\\\\\\left[\begin{array}{cccc}1&(3)/(2)&-(1)/(2)&1\\0&1&3&4\\0&0&1&0\end{array}\right] \leftarrow (1)/(2)R_3`

Write augmented matrix as a system of equations

`x+(3)/(2)y-(1)/(2)z=1\\y+3z=4\\z=0\\\\y+3z=4\\y+3(0)=4\\y=4\\\\x+(3)/(2)y-(1)/(2)z=1\\x+(3)/(2)(4)-(1)/(2)(0)=1\\x+6=1\\x=-5`

Therefore, the solution to the system is
`(x,y,z)=(-5,4,0)`.