Solving Systems of Linear Equations Using Matrices

Hi there! This page is only going to make sense when you know a little about Systems of Linear Equations and Matrices, so please go and learn about those if you don't know them already.

The Example

One of the last examples on Systems of Linear Equations was this one:

Example: Solve

x + y + z = 6
2y + 5z = −4
2x + 5y − z = 27

We went on to solve it using "elimination", but we can also solve it using Matrices!

Using Matrices makes life easier because we can use a computer program (such as the Matrix Calculator) to do all the "number crunching".

But first we need to write the question in Matrix form.

In Matrix Form?

OK. A Matrix is an array of numbers:

Three by three matrix containing numbers 1, 1, 1, 0, 2, 5, 2, 5, and negative 1
A Matrix

Well, think about the equations:

x	+	y	+	z	=	6
		2y	+	5z	=	−4
2x	+	5y	−	z	=	27

They could be turned into a table of numbers like this:

1	1	1	=	6
0	2	5	=	−4
2	5	−1	=	27

We could even separate the numbers before and after the "=" into:

1	1	1		6
0	2	5	and	−4
2	5	−1		27

Now it looks like we have 2 Matrices.

In fact we have a third one, which is [x y z]:

Matrix equation AX equals B with coefficient matrix, variable column vector, and constant vector

Why does [x y z] go there? Because when we Multiply Matrices we use the "Dot Product" like this:

Dot product of the first row of matrix A and the variable column vector X

Which is the first of our original equations above (you might like to check that). Here it is for the second line.

matrix dot product

Try the third line for yourself.

The Matrix Solution

We can shorten this:

Matrix equation AX equals B with coefficient matrix, variable column vector, and constant vector

to this:

AX = B

where

A is the 3x3 matrix of x, y and z coefficients
X is x, y and z, and
B is 6, −4 and 27

Then (as shown on the Inverse of a Matrix page) the solution is this:

X = A^-1B

What does that mean?

It means that we can find the X matrix (the values of x, y and z) by multiplying the inverse of the A matrix by the B matrix.

It has to be done in that order, as order of multiplication matters with matrices.

So let's go ahead and do that.

First, we need to find the inverse of the A matrix (assuming it exists!)

Using the Matrix Calculator we get this:

Inverse of matrix A shown as 1 over negative 11 multiplied by a 3 by 3 matrix

(I left the 1/determinant outside the matrix to keep the numbers simpler)

If the determinant of Matrix A is zero, the inverse does not exist. This means the system of equations either has no solution or infinitely many solutions.

Then multiply A^-1 by B (we can use the Matrix Calculator again):

systems linear equations matrix [x,y,z] equals solution

And we are done! The solution is:

x = 5
y = 3
z = −2

Just like on the Systems of Linear Equations page.

Quite neat and elegant, and the human does the thinking while the computer does the calculating.

Just For Fun ... Do It Again!

For fun, let's do this all again, but put matrix "X" first.

It is good to see this different method, as many people think the solution above is so neat it must be the only way.

So we will solve it like this:

XA = B

And because of the way matrices are multiplied we need to set up the matrices differently now. The rows and columns have to be switched over ("transposed"):

dot product example

And XA = B looks like this:

systems linear equations matrix

The Matrix Solution

Then (also shown on the Inverse of a Matrix page) the solution is this:

X = BA^-1

This is what we get for A^-1:

matrix inverse

In fact it is just like the Inverse we got before, but Transposed (rows and columns swapped over).

Next we multiply B by A^-1:

systems linear equations matrix solution

And the solution is the same:

x = 5, y = 3 and z = −2

It didn't look as neat as the previous solution, but it does show us that there is more than one way to set up and solve matrix equations. Just be careful about the rows and columns!

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