# 4. Multiplication of Matrices

**Important:** We can only multiply matrices if the number of columns in the first matrix is the same as the number of rows in the second matrix.

### Example 1

a) Multiplying a 2 × 3 matrix by a 3 × 4 matrix is possible and it gives a 2 × 4 matrix as the answer.

b) Multiplying a 7 × 1 matrix by a 1 × 2 matrix is okay; it gives a 7 × 2 matrix

c) A 4 × 3 matrix times a 2 × 3 matrix is NOT possible.

## How to Multiply 2 Matrices

We use letters first to see what is going on. We'll see a numbers example after.

As an example, let's take a general 2 × 3 matrix multiplied by a 3 × 2 matrix.

`[(a,b,c),(d,e,f)][(u,v),(w,x),(y,z)]`

The answer will be a 2 × 2 matrix.

We multiply and add the elements as follows. We work **across** the 1st row of the first matrix, multiplying **down** the 1st column of the second matrix, element by element. We **add** the resulting products. Our answer goes in position *a*_{11} (top left) of the answer matrix.

We do a similar process for the 1st row of the first matrix and the **2nd** column of the second matrix. The result is placed in position *a*_{12}.

Now for the **2nd** row of the first matrix and the **1st** column of the second matrix. The result is placed in position *a*_{21}.

Finally, we do the 2nd row of the first matrix and the 2nd column of the second matrix. The result is placed in position *a*_{22}.

So the result of multiplying our 2 matrices is as follows:

`[(a,b,c),(d,e,f)][(u,v),(w,x),(y,z)]` `=[(au+bw+cy,av+bx+cz),(du+ew+fy,dv+ex+fz)]`

Now let's see a number example.

### Phone users

**NOTE:** If you're on a phone, you can scroll any **wide matrices** on this page to the right or left to see the whole expression.

### Example 2

Multiply:

`((0,-1,2),(4,11,2))((3,-1),(1,2),(6,1))`

Answer

This is 2×3 times 3×2, which will give us a 2×2 answer.

`((0,-1,2),(4,11,2)) ((3,-1),(1,2),(6,1))`

`=((0xx3+ -1xx1 + 2xx6,0xx-1+ -1xx2 + 2xx1), (4xx3+11xx1+2xx6,4xx -1 + 11xx2 + 2xx1))`

` = ((0-1+12,0-2+2), (12+11+12,-4+22+2))`

` = ((11,0),(35,20)) `

Our answer is a 2×2 matrix.

## Multiplying 2 × 2 Matrices

The process is the same for any size matrix. We multiply **across** rows of the first matrix and **down** columns of the second matrix, element by element. We then add the products:

`((a,b),(c,d))((e,f),(g,h))` `=((ae+bg,af+bh),(ce+dg,cf+dh))`

In this case, we multiply a 2 × 2 matrix by a 2 × 2 matrix and we get a 2 × 2 matrix as the result.

### Example 3

Multiply:

`((8,9),(5,-1))((-2,3),(4,0))`

Answer

` ((8,9),(5,-1))((-2,3),(4,0)) `

`= ((8 xx -2+9xx4,8xx3+9xx0),(5xx-2+ -1xx4,5xx3 + -1xx0))`

` = ((-16+36,24+0),(-10+ -4,15 + 0)) `

` = ((20,24),(-14,15)) `

## Matrices and Systems of Simultaneous Linear Equations

We now see how to write a system of linear equations using matrix multiplication.

### Example 4

The system of equations

−3

x+y= 16

x− 3y= −4

can be written as:

`((-3,1),(6,-3))((x),(y))=((1),(-4))`

Matrices are ideal for computer-driven solutions of problems because computers easily form *arrays*. We can leave out the algebraic symbols. A computer only requires the first and last matrices to solve the system, as we will see in Matrices and Linear Equations.

## Note 1 - Notation

Care with **writing** matrix multiplication.

The following expressions have **different meanings:**

ABismatrix multiplication

A×Biscrossproduct, which returns avector

A*Bused in computer notation, but not on paper

A•Bdotproduct, which returns ascalar.

[See the Vector chapter for more information on vector and scalar quantities.]

## Note 2 - Commutativity of Matrix Multiplication

Does `AB = BA`?

Let's see if it is true using an example.

### Example 5

If

`A=((0,-1,2),(4,11,2))`

and

`B=((3,-1),(1,2),(6,1))`

find *AB* and *BA.*

Answer

We performed *AB* above, and the answer was:

`AB = ((0,-1,2),(4,11,2)) ((3,-1),(1,2),(6,1))`

` = ( (11,0),(35,20) )`

Now *BA* is (3 × 2)(2 × 3) which will give 3 × 3:

`BA= ((3,-1),(1,2),(6,1))((0,-1,2),(4,11,2))`

` = ((0-4,-3-11,6-2),(0+8,-1+22,2+4),(0+4,-6+11,12+2))`

` = ((-4,-14,4),(8,21,6),(4,5,14)) `

So in this case, *AB* does NOT equal *BA.*

In fact, for most matrices, you cannot reverse the order of multiplication and get the same result.

In general, when multiplying matrices, the commutative law doesn't hold, i.e. *AB* ≠ *BA*. There are two common exceptions to this:

- The identity matrix:
*IA*=*AI*=*A*. - The
**inverse**of a matrix:*A*^{-1}*A*=*AA*^{-1}=*I.*

In the next section we learn how to find the inverse of a matrix.

### Example 6 - Multiplying by the Identity Matrix

Given that

`A=((-3,1,6),(3,-1,0),(4,2,5))`

find *AI*.

Answer

`AI = ((-3,1,6),(3,-1,0),(4,2,5)) ((1,0,0),(0,1,0),(0,0,1))`

`=((-3+0+0,0+1+0,0+0+6),(3+0+0,0+ -1+0,0+0+0),(4+0+0,0+2+0,0+0+5))`

` =((-3,1,6),(3,-1,0),(4,2,5))`

` =A`

We see that multiplying by the identity matrix does not change the value of the original matrix.

That is,

AI = A

## Exercises

1. If possible, find *BA* and *AB*.

`A=((-2,1,7),(3,-1,0),(0,2,-1))`

`B=(4\ \ -1\ \ \ 5)`

Answer

`BA=(4\ \ -1\ \ \ 5)((-2,1,7),(3,-1,0),(0,2,-1))`

`=( -8+(-3)+0\ \ \ 4+1+10\ \ \ 28+0+(-5))`

`=(-11\ \ 15\ \ 23)`

*AB* is not possible. (3 × 3) × (1 × 3).

2. Determine if *B* = *A*^{-1}, given:

`A=((3,-4),(5,-7))`

`B=((7,4),(5,3))`

Answer

If *B* = *A*^{-1}, then `AB = I`.

`AB=((3,-4),(5,-7))((7,4),(5,3))`

`=((21-20,12-12),(35-35,20-21))`

`=((1,0),(0,-1))`

` !=I`

So *B* is NOT the inverse of *A.*

3. In studying the motion of electrons, one of the Pauli spin matrices is

`s=((0,-j),(j,0))`

where

`j=sqrt(-1)`

Show that *s*^{2} = *I.*

[If you have never seen *j* before, go to the section on complex numbers].

Answer

`s^2=( (0,-j),(j,0))((0,-j),(j,0))`

`=(( 0-j^2,0+0), (0+0,-j^2+0))`

`= ((1,0),(0,1))`

`=I`

4. Evaluate the following matrix multiplication which is used in directing the motion of a robotic mechanism.

`( (cos\ 60° ,-sin\ 60° ,0),(sin\ 60°, cos\ 60°,0),(0,0,1))((2),(4),(0))`

Answer

`( (cos\ 60° ,-sin\ 60° ,0),(sin\ 60°, cos\ 60°,0),(0,0,1))((2),(4),(0))`

`=((2(0.5)-4(0.866)+0),(2(0.866)+4(0.5)+0),(0+0+0))`

`=((-2.464),(3.732),(0))`

The interpretation of this is that the robot arm moves from
position (2, 4, 0) to position (-2.46, 3.73, 0). That is, it
moves in the *x-y* plane, but its height remains at *z* = 0*.* The 3 × 3 matrix containing sin and
cos values tells it how many degrees to move.