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- Introduction to Matrices

Still Confused?

Try reviewing these fundamentals first.

Still Confused?

Try reviewing these fundamentals first.

Nope, I got it.

That's that last lesson.

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Get Started Now- Lesson: 116:26
- Lesson: 2a6:18
- Lesson: 2b9:01
- Lesson: 2c14:30
- Lesson: 2d5:25
- Lesson: 2e9:45

Now that we have learned how to represent a linear system as a matrix, we can now solve this matrix to solve the linear system! We use a method called "Gaussian elimination". This method involves a lot of matrix row operations. Our goal is to make it so that all entries in the bottom left of the matrix are 0. Once that is done, we take a look at the last row and convert it to a linear system. Then we solve for the variable. Then we look at the second last row, convert it to a linear system, and solve for the other variable. Rinse and repeat, and you will find all the variables which solve the linear system!

Note

Gaussian elimination (or row reduction) is a method used for solving linear systems. For example,

$x+y+z=3$
$x+2y+3z=0$
$x+3y+2z=3$

Can be represented as the matrix:

Using Gaussian elimination, we can turn this matrix into

(watch the intro video to learn how to do this!)

Now we can start solving for $x,y$ and $z$.

So in the third row, we see that $-3z=6$. So $z=-2$.

In the second row, we see that $2y+4z=-6$. Since we know that $z=-2$, then we can substitute it into the second row and solve for $y$. So,

$2y+4z=-6$→$2y+4(-2)=-6$

→$2y-8=-6$

→$2y=2$

→$y=1$

So now we know that $z=-2$, and $y=1$. Now let us take a look at the first row and solve for $x$.

$x+y+z=3$→$x+1-2=3$

→$x-1=3$

→$x=4$

Since we have solved for $x,y$ and $z$, then we have just solved the linear system.

Gaussian elimination (or row reduction) is a method used for solving linear systems. For example,

Can be represented as the matrix:

Using Gaussian elimination, we can turn this matrix into

Now we can start solving for $x,y$ and $z$.

So in the third row, we see that $-3z=6$. So $z=-2$.

In the second row, we see that $2y+4z=-6$. Since we know that $z=-2$, then we can substitute it into the second row and solve for $y$. So,

$2y+4z=-6$→$2y+4(-2)=-6$

→$2y-8=-6$

→$2y=2$

→$y=1$

So now we know that $z=-2$, and $y=1$. Now let us take a look at the first row and solve for $x$.

$x+y+z=3$→$x+1-2=3$

→$x-1=3$

→$x=4$

Since we have solved for $x,y$ and $z$, then we have just solved the linear system.

- 1.Gaussian elimination overview
- 2.
**Gaussian Elimination**

Solve the following linear systems:a)$x+2y=3$

$2x+3y=1$b)$x+4y+3z=1$

$x+2y+9z=1$

$x+6y+6z=1$c)$x+3y+3z=2$

$3x+9y+3z=3$

$3x+6y+6z=4$d)$4x-5y=-6$

$2x-2y=1$e)$x+3y+4z=4$

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

$3x+9y+6z=-6$

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