Lesson Video: Properties of Matrix Multiplication Mathematics • 10th Grade

Video Transcript

In this video, we will learn how to identify the properties of the multiplication of matrices and compare them to the properties of multiplication of numbers. We will begin by recalling how we multiply two matrices and define the properties they must have for us to be able to do so.

We can only multiply two matrices if the number of columns in the first matrix is equal to the number of rows in the second matrix, for example, a three-by-two matrix and a two-by-five matrix. There are two columns in the first matrix and two rows in the second. The resultant matrix will have three rows and five columns. This is the number of rows of the first matrix and the number of columns of the second.

In general, we can multiply an 𝑚-by-𝑛 matrix with an 𝑛-by-𝑝 matrix, which results in an 𝑚-by-𝑝 matrix. This can be demonstrated as shown. Matrix 𝐴 has elements from 𝑎 one one to 𝑎 𝑚𝑛, and matrix 𝐵 has elements from 𝑏 one one to 𝑏 𝑛𝑝. Multiplying matrix 𝐴 by matrix 𝐵 gives us matrix 𝐶, with elements from 𝑐 one one to 𝑐 𝑚𝑝. Each of the elements in matrix 𝐶 can be calculated using the following formula. The general term 𝑐 𝑖𝑗 is equal to the sum from 𝑘 equals one to 𝑘 equals 𝑛 of 𝑎 𝑖𝑘 multiplied by 𝑏 𝑘𝑗, where 𝑎 𝑖𝑘 and 𝑏 𝑘𝑗 are the general terms in the matrix 𝐴 and 𝐵. This is equal to the sum of 𝑎 𝑖 one multiplied by 𝑏 one 𝑗 and so on up to 𝑎 𝑖𝑛 multiplied by 𝑏 𝑛𝑗.

We will now consider how this works in a practical example.

Given that matrix 𝐴 is equal to negative four, two, two, negative four and matrix 𝐵 is equal to negative three, negative three, negative one, one, find 𝐴𝐵 and 𝐵𝐴.

We recall that we can only multiply two matrices if the number of columns of the first matrix is equal to the number of rows in the second matrix. As both of our matrices are two by two, this will hold for 𝐴𝐵 and 𝐵𝐴. Let’s begin by multiplying matrix 𝐴 by matrix 𝐵. When multiplying two matrices, we multiply the elements of each row in the first matrix by the elements of each column in the second matrix.

The first element in matrix 𝐴𝐵 will be equal to negative four multiplied by negative three plus two multiplied by negative one. This is equal to 10, as negative four multiplied by negative three is 12 and two multiplied by negative one is negative two. The top-right element in matrix 𝐴𝐵 will be equal to negative four multiplied by negative three plus two multiplied by one. This is equal to 14.

We now repeat this process by multiplying the numbers in the second row of matrix 𝐴 by the columns in matrix 𝐵. Two multiplied by negative three plus negative four multiplied by negative one is equal to negative two. And two multiplied by negative three plus negative four multiplied by one is equal to negative 10. The matrix 𝐴𝐵 is equal to 10, 14, negative two, negative 10.

We now need to repeat this process for matrix 𝐵𝐴. Negative three multiplied by negative four plus negative three multiplied by two is equal to six. Repeating this for the other rows and columns gives us values of six, six, and negative six. Matrix 𝐵𝐴 is therefore equal to six, six, six, negative six.

We notice that matrix 𝐴𝐵 is not equal to matrix 𝐵𝐴. This leads us to a general rule when dealing with the multiplication of matrices. As 𝐴𝐵 is not equal to 𝐵𝐴, matrix multiplication is not commutative. This is different to the multiplication of numbers, as multiplying any two numbers is commutative.

In our next question, we will look at a specific example when 𝐴𝐵 is equal to 𝐵𝐴.

State whether the following statement is true or false. If 𝐴 and 𝐵 are both two-by-two matrices, then 𝐴𝐵 is never the same as 𝐵𝐴.

In order to prove that a statement is false, we simply need to find one example where the statement is not true. We are told that both of our matrices are two by two. And we will let the elements of matrix 𝐴 be 𝑎, 𝑏, 𝑐, 𝑑. Whilst we could let the elements of matrix 𝐵 have any values, in this case, we will let matrix 𝐵 be the identity matrix: one, zero, zero, one. We know that the identity matrix has ones on its leading diagonal and zeros everywhere else.

To calculate matrix 𝐴𝐵, we need to multiply 𝑎, 𝑏, 𝑐, 𝑑 by one, zero, zero, one. When multiplying matrices, we multiply the elements of each row in the first matrix by each column in the second matrix. 𝑎 multiplied by one is equal to 𝑎, and 𝑏 multiplied by zero is zero. Therefore, the first element in the matrix 𝐴𝐵 is 𝑎. Repeating this for the other rows and columns, we get the elements 𝑏, 𝑐, and 𝑑. Matrix 𝐴𝐵 is equal to 𝑎, 𝑏, 𝑐, 𝑑, which is equal to matrix 𝐴.

We will now repeat this method when multiplying matrix 𝐵, the identity matrix, by matrix 𝐴. Once again, this gives us the elements 𝑎, 𝑏, 𝑐, 𝑑. We have therefore found an example where the matrix 𝐴𝐵 is the same as the matrix 𝐵𝐴. This leads us to a general rule. When we multiply any matrix by the identity matrix, it is the same as multiplying the identity matrix by this matrix. In both cases, the original matrix remains the same. 𝐴𝐼 is equal to 𝐼𝐴, which is equal to the matrix 𝐴.

We can actually go one stage further when looking at the commutative property of matrices. We will now let matrix 𝐵 have the elements 𝑒, 𝑓, 𝑔, ℎ. Multiplying the matrices 𝐴𝐵 and 𝐵𝐴, we get the following two-by-two matrices. At first glance, matrix 𝐴𝐵 and 𝐵𝐴 appear to have nothing in common. However, we do notice that the top-left element contains 𝑎𝑒 or 𝑒𝑎 and the bottom-right element contains 𝑑ℎ or ℎ𝑑. The elements 𝑎, 𝑑, 𝑒, and ℎ are the elements on the leading diagonals of matrices 𝐴 and 𝐵, respectively. We can see that if all the other products were equal to zero, the two matrices would be the same.

Let’s consider what happens if 𝑏, 𝑐, 𝑓, and 𝑔 are all equal to zero. Matrices 𝐴𝐵 and 𝐵𝐴 are both equal to 𝑎𝑒, zero, zero, 𝑑ℎ. This is an example of a diagonal matrix, as all the elements apart from those on the leading diagonal are equal to zero. This leads us to another general rule of matrix multiplication. If 𝐴 and 𝐵 are both diagonal matrices, then the two matrices are commutative. 𝐴𝐵 is equal to 𝐵𝐴.

In our next question, we will demonstrate how we can distribute matrix multiplication over addition.

Given three matrices 𝐴, 𝐵, and 𝐶, which of the following is equal to 𝐴 multiplied by 𝐵 plus 𝐶? Is it (A) 𝐴𝐵 plus 𝐶, (B) 𝐴𝐵 plus 𝐴𝐶, (C) 𝐵𝐴 plus 𝐶𝐴, (D) 𝐵𝐴 plus 𝐶, or (E) 𝐵 plus 𝐴𝐶?

In order to answer this question, we need to use the distributive property of matrices. We can distribute matrices in a similar way to how we distribute real numbers. Multiplying matrix 𝐴 by matrix 𝐵 plus 𝐶 is equal to matrix 𝐴𝐵 plus matrix 𝐴𝐶. It is important to note though that if the parentheses came first, we were multiplying 𝐵 plus 𝐶 by 𝐴, then our answer would be 𝐵𝐴 plus 𝐶𝐴. If the matrix 𝐴 is distributed from the left side, we must ensure that the product in the resulting sum has 𝐴 on the left. In the same way, if matrix 𝐴 is distributed from the right side, each product in the resulting sum must have 𝐴 on the right. We can therefore see that the correct answer is option (B). 𝐴 multiplied by 𝐵 plus 𝐶 is equal to 𝐴𝐵 plus 𝐴𝐶.

It is important to remember that when performing matrix addition and matrix multiplication, the order of each matrix is key. In order to add matrix 𝐵 and 𝐶, they must have the same order. To perform matrix multiplication, the number of columns in matrix 𝐴 must be equal to the number of rows in matrix 𝐵 and 𝐶.

Our final question will include an application of the distributive property.

Suppose matrix 𝐴 is equal to one, negative three, negative four, two; matrix 𝐵 is equal to two, zero, one, negative one; and matrix 𝐶 is equal to zero, one, negative three, zero. There are four parts to this question. Find matrix 𝐴𝐵. Find matrix 𝐴𝐶. Find 𝐴 multiplied by two 𝐵 plus seven 𝐶. And express 𝐴 multiplied by two 𝐵 plus seven 𝐶 in terms of 𝐴𝐵 and 𝐴𝐶.

In order to multiply matrix 𝐴 by matrix 𝐵, we need to multiply all of the elements in the rows of matrix 𝐴 by the columns of matrix 𝐵. One multiplied by two plus negative three multiplied by one is equal to negative one. Repeating this for the other rows and columns gives us the elements three, negative six, and negative two. Matrix 𝐴𝐵 is equal to negative one, three, negative six, negative two.

To work out matrix 𝐴𝐶, we multiply one, negative three, negative four, two by zero, one, negative three, zero. This gives us the elements nine, one, negative six, and negative four. This is the matrix 𝐴𝐶.

In the third part of our question, we begin by multiplying matrix 𝐵 by the scalar or constant two and matrix 𝐶 by the scalar seven. When multiplying a matrix by a scalar, we simply multiply each of the elements by that scalar. This means that two 𝐵 is equal to four, zero, two, negative two. In the same way, seven 𝐶 is equal to zero, seven, negative 21, zero.

Next, we need to add these two matrices. We do this by adding the elements in corresponding positions in each matrix. Four plus zero is equal to four. Repeating this for the other elements gives us the matrix four, seven, negative 19, negative two.

Finally, we need to multiply this matrix by matrix 𝐴. The order here is important. We must multiply matrix 𝐴 by the matrix four, seven, negative 19, negative two. This gives us the elements 61, 13, negative 54, and negative 32. 𝐴 multiplied by two 𝐵 plus seven 𝐶 is equal to 61, 13, negative 54, negative 32.

In the final part of this question, we can use the distributive property of matrix multiplication. We can multiply matrix 𝐴 by two 𝐵 and then add matrix 𝐴 multiplied by seven 𝐶. This gives us one, negative three, negative four, two multiplied by four, zero, two, negative two plus one, negative three, negative four, two multiplied by zero, seven, negative 21, zero. The first product gives us negative two, six, negative 12, negative four. The second product gives us 63, seven, negative 42, negative 28.

We might be tempted to simply add these matrices. However, we were asked to give our answer in terms of 𝐴𝐵 and 𝐴𝐶. We notice that our first matrix negative two, six, negative 12, negative four is two times matrix 𝐴𝐵. We also notice that the second matrix 63, seven, negative 42, negative 28 is seven times matrix 𝐴𝐶. This means that matrix 𝐴 multiplied by two 𝐵 plus seven 𝐶 is equal to two multiplied by matrix 𝐴𝐵 plus seven multiplied by matrix 𝐴𝐶.

We will now summarize the key points from this video. We saw in our first example that matrix multiplication is generally not commutative. Matrix 𝐴𝐵 is not equal to matrix 𝐵𝐴. There were a couple of exceptions to this though. Multiplying a matrix by the identity matrix gives the original matrix. This can be done in either order. 𝐴 multiplied by 𝐼 is equal to 𝐼 multiplied by 𝐴, which is equal to matrix 𝐴.

We also saw that if 𝐴 and 𝐵 are both diagonal matrices with the same order, then 𝐴𝐵 is equal to 𝐵𝐴. Two diagonal matrices of the same order are commutative. We also saw that matrix multiplication is distributive with respect to matrix addition. That is, 𝐴 multiplied by 𝐵 plus 𝐶 is equal to 𝐴𝐵 plus 𝐴𝐶. It is important to note here that as matrix 𝐴 is in front of the parentheses, it will be the first matrix in each of the products. This is not the same as matrix 𝐵𝐴 plus matrix 𝐶𝐴.

The order of each matrix is also important. To perform matrix addition, both matrices must have the same order. In order to perform matrix multiplication, the number of columns of the first matrix must be equal to the number of rows in the second matrix. If the orders don’t have these properties, then the matrix addition and matrix multiplication cannot be defined. These properties have some similarities and some differences between the properties of multiplication of numbers.