Operations with Matrices
Many times we can treat matrices like we do numbers. In other words, we can add, subtract and multiply matrices, as long as they meet certain requirements. If these conditions are met, then there are very specific rules to follow to perform these operations.

Addition and subtraction of matrices can only be done if the dimensions are EXACTLY the same. Consider the following matrices:

A = B = F = G =
 C = D = H =

Addition and subtraction can only be performed with:
A and G
B and F
C and H
Once we determine if we can add or subtract two matrices based on their dimensions, we have to apply the correct procedure for adding or subtracting. Let’s start with A + G. We know that A is a matrix and G is also a matrix. So how do we go about adding them?

We simply add the corresponding entries together and put the answer in a new matrix. For example, we will add a11 to g11 and that answer will be the entry in the first row and first column of the answer matrix. We then just continue through the A and G matrices until all corresponding entries have been added.
A + G = + = =
Now let’s try B + F. Because they are the same dimensions, we can add these two matrices. We’ll use the same process as we did for A + G.
B + F = + = =
Now you should try a few examples to make sure that you understand this process.

Examples

Look at your answers for A + G and for G + A. Notice that they are the same even though we added in a different order. The same can be said for B + F and for F + B. Based on these observations, let’s look at H + C and see if we get the same answer as we did for C + H.
H + C = + = =
It turns out that addition of matrices is commutative, meaning that the order in which you add them does not matter. This is the only matrix operation that is commutative.

Now that we have a good idea of how addition works, let’s try subtraction. We’ll follow a very similar process as we did for addition. In other words, we’ll simply subtract corresponding entries in the two matrices. Look at the example below.
A - G = - = =
To see that the order in which you subtract matrices does matter (in other words, subtraction is NOT commutative), let’s look at G - A.
G - A = - = =
Even though A - G and G - A did not give us the same answers, it is interesting to note that they are opposites of each other.

Examples

Multiplication

Matrix multiplication comes in two distinct forms. One way is to multiply a matrix by a constant, this is also called scalar multiplication. This type of problem looks like 3A or -2B. Let’s look at each of these examples.

Recall from earlier in the lesson that

A = and B = .

To find 3A, we just multiply each element in A by 3. So,
3A = =
and
-2B = =
Notice that this is unique to multiplication. We were not able to add or subtract a constant to a matrix.

Examples

Scalar multiplication with matrices is not too difficult once you get in the routine of performing the process. You can practice more problems like those above on the accompanying worksheet.

A second type of multiplication is to multiply two matrices together and it is a little more involved.

Multiplication of matrices has different rules than addition and subtraction. For matrix multiplication, the columns of the first matrix MUST match with the rows of the second matrix.

A = and B = .
To multiply AB, we first have to make sure that the number of columns in A is the same as the number of rows in B. Matrix A has 2 columns and matrix B has 2 rows so we will be able to perform this operation. The dimension of the new matrix will be defined as:
Rows = number of rows in A
Columns = number of columns in B
An easier way to look at these dimensions is shown in the following figure:

Let’s do the multiplication and call our new matrix M (AB = M). This matrix will be a . We simply need to fill in the six entries in this matrix.
Remember we are working with the matrices A and B shown below.

A = and B = .

We begin by multiplying (4)(2) and adding that to (6)(–2). That is now the entry in m11.
Now multiply (4)( –3) and add that to (6)(0). This is the entry in m12.
Multiply (4)(1) and add that to (6)(5). This is the entry in m13.
We now have the top row of our answer matrix M.

We’ll follow the same process to fill in the second row of the answer matrix.
Entry for m21 = (1)(2) + (9)( –2)
Entry for m22 = (1)( –3) + (9)(0)
Entry for m23 = (1)(1) + (9)(5)
The answer matrix operations look like this:
M =
Matrix M simplifies to:
AB = M =
Notice that if you were to try to do BA, you could not because the dimensions would not match up. Always check your dimensions before beginning any multiplication!

Example
 Since K is a and L is a , we can do the multiplication. K = and L = The answer matrix, let’s call is N, will be a . So N will have four entries that we need to fill in. N = N11 is obtained by multiplying (2)(2) + (–3)( –2) + (1)(7). N = For N12 we get (2)( –3) + (–3)(0) + (1)(1). N = Complete this process and determine the final values for matrix N. What is your answer?

S Taylor

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