Question 3: Solve Linear System by Elimination

Solve the following system of linear equations using Gaussian elimination:
$\begin{aligned} x - 4 z & = 8 \\ - 2 x - y + 8 z & = - 25 \\ 2 x + 3 y - 4 z & = 40 \end{aligned}$

(a) Perform Gaussian elimination to reduce the augmented matrix to row echelon form.
Maintain the order of variables. Enter the resulting augmented matrix below.

[0.5] [0]

Acceptable answers include:

$(\begin{array}{r} 1 & 0 & - 4 & 8 \\ 0 & - 1 & 0 & - 9 \\ 0 & 0 & 4 & - 3 \end{array})$ and $(\begin{array}{c} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 9 \\ 0 & 0 & 1 & - \frac{3}{4} \end{array})$

Any other equivalent row echelon form would be accepted.

Step 1 of 2

(a) Let's write the matrix equation $A \cdot x = b$ :

$(\begin{array}{r} 1 & 0 & - 4 \\ - 2 & - 1 & 8 \\ 2 & 3 & - 4 \end{array}) \cdot (\begin{array}{c} x \\ y \\ z \end{array}) = (\begin{array}{r} 8 \\ - 25 \\ 40 \end{array})$

Form the augmented matrix $[ A | b ]$ :

$(\begin{array}{r} 1 & 0 & - 4 & | & 8 \\ - 2 & - 1 & 8 & | & - 25 \\ 2 & 3 & - 4 & | & 40 \end{array})$

Step 2 of 2

Perform Gaussian elimination to reduce the augmented matrix to the row echelon form.

The original augmented matrix is $(\begin{array}{r} 1 & 0 & - 4 & 8 \\ - 2 & - 1 & 8 & - 25 \\ 2 & 3 & - 4 & 40 \end{array})$ .

$R_{2} \to 2 R_{1} + R_{2}$

$(\begin{array}{r} 1 & 0 & - 4 & 8 \\ 0 & - 1 & 0 & - 9 \\ 2 & 3 & - 4 & 40 \end{array})$

$R_{3} \to R_{3} - 2 R_{1}$

$(\begin{array}{r} 1 & 0 & - 4 & 8 \\ 0 & - 1 & 0 & - 9 \\ 0 & 3 & 4 & 24 \end{array})$

$R_{3} \to 3 R_{2} + R_{3}$

$(\begin{array}{r} 1 & 0 & - 4 & 8 \\ 0 & - 1 & 0 & - 9 \\ 0 & 0 & 4 & - 3 \end{array})$

Any equivalent row echelon form of the augmented matrix will be accepted.

(b) Using back substitution on the row echelon form obtained in (a), determine the values of the variables $x$ , $y$ and $z$ . Enter their values as an ordered list ${x, y, z}$ in the answer field.

[0.5] [0]

${5, 9, - \frac{3}{4}}$

Step 1 of 5

(b) Now we have the simplified set of equations:

$\begin{aligned} x - 4 z & = 8 \\ - y & = - 9 \\ 4 z & = - 3 \end{aligned}$

Step 2 of 5

We obtain $z$ by solving $4 z = - 3$ :

$z = - \frac{3}{4}$

Step 3 of 5

Substitute $z = - \frac{3}{4}$ in $- y = - 9$ and solve for $y$ :

$\begin{aligned} - y & = - 9 \\ y & = 9 \end{aligned}$

Step 4 of 5

Substitute $y = 9$ and $z = - \frac{3}{4}$ in $- y = - 9$ and solve for $x$ :

$\begin{aligned} x + 3 & = 8 \\ x & = 5 \end{aligned}$

Step 5 of 5

So the answer to be entered is ${5, 9, - \frac{3}{4}}$ .