# Gaussian Elimination

## Key Questions

The goals of Gaussian elimination are to get $1$s in the main diagonal and $0$s in every position below the $1$s,

Then you can use back substitution to solve for one variable at a time.

#### Explanation:

EXAMPLE:

Use Gaussian elimination to solve the following system of equations.

$x + 2 y + 3 z = - 7$
$2 x - 3 y - 5 z = 9$
$- 6 z - 8 y + z = - 22$

Solution:

Set up an augmented matrix of the form.

$\left(\begin{matrix}1 & 2 & 3 & | & - 7 \\ 2 & 3 & - 5 & | & 9 \\ - 6 & - 8 & 1 & | & 22\end{matrix}\right)$

Goal 1. Get a 1 in the upper left hand corner.

Goal 2a: Get a zero under the 1 in the first column.

Multiply Row 1 by $- 2$ to get

$\left(\left(- 2 , - 4 , - 6 , | , 14\right)\right)$

Add the result to Row 2 and place the result in Row 2.

We signify the operations as -2R_2+R_1→R_2.

((1,2,3,|,-7),(2,3,-5,|,9),(-6,-8,1,|,22)) stackrel(-2R_1+R_2→R_2)(→) ((1,2,3,|,-7),(0,-7,-11,|,23),(-6,-8,1,|,22))

Goal 2b: Get another zero in the first column.

To do this, we need the operation 6R_1+R_3→R_3.

((1,2,3,|,-7),(0,-7,-11,|,23),(-6,-8,1,|,22)) stackrel(6R_2+R_3→R_3)(→) ((1,2,3,|,-7),(0,-7,-11,|,23),(0,4,19,|,-64))

Goal 2c. Get the remaining zero.

Multiply Row 2 by $- \frac{1}{7}$.

((1,2,3,|,-7),(0,-7,-11,|,23),(0,4,19,|,-64)) stackrel(-(1/7)R_2 → R_2)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,4,19,|,-64))

Now use the operation -4R_2+R_3 →R_3.

((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,4,19,|,-64)) stackrel(-4R_2+R_3 →R_3)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,89/7,|,-356/7))

Multiply the third row by $\frac{7}{89}$.

((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,89/7,|,-356/7)) stackrel(7/89R_3 →R_3)(→) ((1,2,3,|,-7),(0,1,11/7,|,-23/7),(0,0,1,|,-4))

Goal 3. Use back substitution to get the values of $x$, $y$, and $z$.

Goal 3a. Calculate $z$.

$z = - 4$

Goal 3b. Calculate $y$.

$y + \frac{11}{7} z = - \frac{23}{7}$
$y - \frac{44}{7} = - \frac{23}{7}$
$y = \frac{44}{7} - \frac{23}{7} = \frac{21}{7}$

$y = 3$

Goal 3c. Calculate x.

$x + 2 y + 3 z = - 7$
$x + 6 - 12 = - 7$
$x - 6 = - 7$

$x = 1$

The solution is $x = 1 , y = 3 , z = - 4$

• A determinant of a square matrix is different from Gaussian elimination...so I will address both topics lightly for you!

The determinant of a 2x2 matrix is found by subtracting the products of the diagonals like: $1 \cdot 5 - 3 \cdot 2$ = 5 - 6 = -1.

For a larger square matrix like a 3x3, there are different methods. Link to Purple math for one method.

The TI-nspire calculator (as well as other calculators and online services) can do a determinant quickly for you: Gaussian elimination is a method of solving a system of linear equations. First, the system is written in "augmented" matrix form. Then, legal row operations are used to transform the matrix into a specific form that leads the student to answers for the variables.

Ex: 3x + 4y = 10
-x + 5y = 3

The command "ref" on the TI-nspire means "row echelon form", which takes the matrix down to a stage where the last variable is solved for, and the first coefficient is "1". In this example, y = 1, and $1 x + \frac{4}{3} y = \frac{10}{3}$. Substitute y = 1 and solve for x: $x + \frac{4}{3} = \frac{10}{3}$
$x = \frac{6}{3} \mathmr{and} 2$.

Without showing you all of the steps (row operations), you probably don't have the feel for how to do this yourself! Here is another LINK to Purple Math to see what they say about Gaussian elimination.

• A matrix only has an inverse if it is a square matrix (like 2x2 or 3x3...) and its determinant is not equal to 0.

First, to find a determinant by hand, we can look at a 2x2: In my calculator, you see the abbreviation of determinant is "det".
The calculator knows to expect a square matrix inside the parentheses, otherwise this command would not be possible.

For a 2x2, you can see the product of the first diagonal subtracted by the product of the second diagonal. Simple.

Then, you take the reciprocal of that answer (-34), and multiply that as a scalar multiple on a new matrix where you switch the positions of the 3 and -2 (first diagonal), and change signs on the second diagonal (7 and 4). In this example, some of the fractions were reduced.

A 3x3 matrix is not as easy, and I would usually suggest using a calculator like i did here: I hope this was helpful. Ask another question if you are interested in more about inverse matrices!