Answer to Question #258113 in Linear Algebra for Monte

Question

Solve the following system of linear equations for 𝑥, 𝑦, 𝑧, 𝑡: −𝑦+3𝑧+2𝑥+4𝑡 = 9

𝑥−2𝑧+7𝑡 =11 𝑧−3𝑦+3𝑥+5𝑡 = 8 𝑦 + 2𝑥 + 4𝑧 + 4𝑡 = 10

Answer 1

$$\begin{matrix} 2x-y+3z+4t=9 \\ x-2z+7t=11\\ 3x-3y+z+5t=8\\ 2x+y+4z+4t=10 \end{matrix}$$

Augmented matrix

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 1 & 0 & -2 & 7 & & 11 \\ 3 & -3 & 1 & 5 & & 8 \\ 2 & 1 & 4 & 4 & & 10 \\ \end{pmatrix}$$

$R_2=R_2-R_1/2$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 3 & -3 & 1 & 5 & & 8 \\ 2 & 1 & 4 & 4 & & 10 \\ \end{pmatrix}$$

$R_3=R_3-3R_1/2$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & -3/2 & -7/2 & -1 & & -11/2 \\ 2 & 1 & 4 & 4 & & 10 \\ \end{pmatrix}$$

$R_4=R_4-R_1$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & -3/2 & -7/2 & -1 & & -11/2 \\ 0 & 2 & 1 & 0 & & 1 \\ \end{pmatrix}$$

$R_3=R_3+3R_2$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 2 & 1 & 0 & & 1 \\ \end{pmatrix}$$

$R_4=R_4-4R_2$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 0 & 15 & -20 & & -25 \\ \end{pmatrix}$$

$R_4=R_4+15R_3/14$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 0 & 0 & -5 & & -10 \\ \end{pmatrix}$$

$R_4=R_4/(-5)$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 14 & & 14 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_3=R_3-14R_4$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & -14 & 0 & & -14 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_3=R_3/(-14)$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 5 & & 13/2 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_2=R_2-5R_4$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & -7/2 & 0 & & -7/2 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_2=R_2+7R_3/2$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1/2 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_2=2R_2$

$$\begin{pmatrix} 2 & -1 & 3 & 4 & & 9 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_1=R_1-4R_4$

$$\begin{pmatrix} 2 & -1 & 3 & 0 & & 1 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_1=R_1-3R_3$

$$\begin{pmatrix} 2 & -1 & 0 & 0 & & -2 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_1=R_1+R_2$

$$\begin{pmatrix} 2 & 0 & 0 & 0 & & -2 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

$R_1=R_1/2$

$$\begin{pmatrix} 1 & 0 & 0 & 0 & & -1 \\ 0 & 1 & 0 & 0 & & 0 \\ 0 & 0 & 1 & 0 & & 1 \\ 0 & 0 & 0 & 1 & & 2 \\ \end{pmatrix}$$

The solution is

$$x=-1, y=0, z=1, t=2$$

$$(x,y,z,t)=(-1,0,1,2)$$