Question

Use Gauss-Jordan elimination to solve the following linear system.

–4x + 4y + 5z = 9
–2x + 2y + z = 3
–4x + 5y = 4

A. (3,1,–4)
B. (–1,0,1)
C. (–5,–1,0)
D. (1,2,–4) "SOME HELP PLEASE"!!!!!!!!!!!!!!!

Answer 1

By taking a linear expression and looking at -4(x)+4(y)+5(z)=9
we can say that -1=x, 0=y, and 1=z
-4(-1)=4, 4(0)=0, 5(1)=5 So 4+5=9
-2(-1)=2, 2(0)=0, z=1 So 2+1=3
-4(-1)=4, 5(0)=0, So 4=4

Answer 2

The correct answer is B.
`(-1,0,1)`

Explanation:

The given linear system is

`-4x+4y+5z=9`

`-2x+2y+z=3`

`-4x+5y=4`

The augmented matrix is

`\left[\begin{array}{cccc}-4&4&5&|9\\-2&2&1&|3\\-4&5&0&|4\end{array}\right]`

To solve this by Gauss-Jordan elimination means that, we need to reduce to reduced row echelon form.

`-(1)/(4)R_1\rightarrow R_1`

`\left[\begin{array}{cccc}1&-1&-(5)/(4) &\:\:\:\:\:\:|-(9)/(4) \\-2&2&1&|3\\-4&5&0&|4\end{array}\right]`

`R_2+2R_1\rightarrow R_2`

`R_3+4R_1\rightarrow R_3`

`\left[\begin{array}{cccc}1&-1&-(5)/(4) &|-(9)/(4) \\0&0&-(3)/(2) &|-(3)/(2) \\0&1&-5&|-5\end{array}\right]`

`R_2\leftrightarrow R_3`

`\left[\begin{array}{cccc}1&-1&-(5)/(4) &|-(9)/(4) \\0&1&-5 &|-5 \\0&0&-(3)/(2)&|-(3)/(2)\end{array}\right]`

`R_1+R_2\rightarrow R_1`

`\left[\begin{array}{cccc}1&0&-(25)/(4) &|-(29)/(4) \\0&1&-5 &|-5 \\0&0&-(3)/(2)&|-(3)/(2)\end{array}\right]`

`-(2)/(3)R_3\rightarrow R_3`

`\left[\begin{array}{cccc}1&0&-(25)/(4) &|-(29)/(4) \\0&1&-5 &|\:\:-5 \\0&0&1&|\:\:\:\:\:\:\:1\end{array}\right]`

`R_2+5R_3\rightarrow R_2`

`R_1+(25)/(4)R_3\rightarrow R_1`

`\left[\begin{array}{cccc}1&0&0&|-1 \\0&1&0&|\:\:\:\:\:\:0\\0&0&1&|\:\:\:\:\:\:1\end{array}\right]`

The matrix is now in the reduced row echelon form.

This gives the solution to be,

`(-1,0,1)`