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

Answer:

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

Step-by-step 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.

$-\frac{1}{4}R_1\rightarrow R_1$

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

$R_2+2R_1\rightarrow R_2$

$R_3+4R_1\rightarrow R_3$

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

$R_2\leftrightarrow R_3$

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

$R_1+R_2\rightarrow R_1$

$\left[\begin{array}{cccc}1&0&-\frac{25}{4} &|-\frac{29}{4} \\0&1&-5 &|-5 \\0&0&-\frac{3}{2}&|-\frac{3}{2}\end{array}\right]$

$-\frac{2}{3}R_3\rightarrow R_3$

$\left[\begin{array}{cccc}1&0&-\frac{25}{4} &|-\frac{29}{4} \\0&1&-5 &|\:\:-5 \\0&0&1&|\:\:\:\:\:\:\:1\end{array}\right]$

$R_2+5R_3\rightarrow R_2$

$R_1+\frac{25}{4}R_3\rightarrow R_1$

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

The matrix is now in the reduced row echelon form.

This gives the solution to be,

$(-1,0,1)$