Lesson Video: Matrix Operations Mathematics • First Year of Secondary School

Video Transcript

In this video, we will learn how to combine the operations of addition, subtraction, scalar multiplication, and transposing matrices. We will begin by recalling how we perform each of these operations.

If 𝐴 and 𝐵 are matrices of the same order or dimension, then they can be added or subtracted. If we let 𝐴 and 𝐵 be the two-by-two matrices as shown, we can simply add or subtract the matrices by adding or subtracting their corresponding components. The matrix 𝐴 plus 𝐵 will have the same order, two by two, and have the following four components. We can calculate the matrix 𝐴 minus 𝐵 in a similar way by subtracting the corresponding components. It is important to note that matrix 𝐴 plus matrix 𝐵 is equal to matrix 𝐵 plus matrix 𝐴. However, matrix 𝐴 minus matrix 𝐵 is not necessarily equal to matrix 𝐵 minus matrix 𝐴. This means that addition of matrices is commutative, whereas subtraction is not. The matrix 𝐴 minus 𝐵 is actually equal to negative one multiplied by matrix 𝐵 minus 𝐴.

Let’s now recall how we can multiply a matrix by a scalar. To multiply any matrix by a scalar, we multiply each element or component by the scalar. We can multiply a matrix of any order or size by a scalar. Let’s consider the two-by-two matrix as shown. We can multiply this matrix by the scalar 𝑘. This gives us the two-by-two matrix as shown. Let’s now recall what we mean by the transpose of a matrix. The transpose of a matrix is an operator which switches the rows and columns. When dealing with a square matrix, we simply flip the matrix over its leading or main diagonal. The transpose of a matrix is denoted with an exponent 𝑇. This is usually written as a capital but can sometimes be written as a lowercase 𝑡. When finding the transpose of a two-by-two matrix, the elements in the top-right and bottom-left corners swap.

Let’s now consider the three-by-two rectangular matrix 𝐵 as shown. It has elements 𝑢, 𝑣, 𝑤, 𝑥, 𝑦, 𝑧. As we’re switching the rows and columns, 𝐵 transpose is a two-by-three matrix. It has two rows and three columns with elements 𝑢, 𝑤, 𝑦, 𝑣, 𝑥, 𝑧. The first row of matrix 𝐵 is the first column of 𝐵 transpose. Likewise, the first column of matrix 𝐵 is the first row of 𝐵 transpose. This can be summarized as follows. A matrix with 𝑚 rows and 𝑛 columns will have a transpose with 𝑛 rows and 𝑚 columns. In this case, the original matrix will have order 𝑚 by 𝑛, whereas the matrix transpose will have order 𝑛 by 𝑚. This is different to perform in the operations of addition, subtraction, and scalar multiplication to rectangular matrices where the order remains the same.

We will now look at some examples where we need to perform a combination of these operations.

Given that matrix 𝐴 equals negative seven, five, negative four, negative two; matrix 𝐵 equals one, zero, seven, negative two, what is one-third multiplied by 𝐴 plus 𝐵?

We recall that we can add two matrices if they have the same order. In this case, both matrix 𝐴 and matrix 𝐵 are two-by-two square matrices. In order to add two matrices of the same order, we simply add their corresponding components or elements. Starting in the top-left corner, negative seven plus one is equal to negative six. Five plus zero is equal to five. Negative four plus seven is equal to three. And finally, negative two plus negative two is equal to negative four. The matrix 𝐴 plus 𝐵 is equal to negative six, five, three, negative four.

We are asked to calculate one-third of this, so we need to multiply the matrix by the scalar one-third. This can be done by multiplying each of the elements by one-third or dividing them by three. One-third multiplied by negative six is negative two. Repeating this process for the other elements gives us five-thirds, one, and negative four-thirds. If the matrices 𝐴 and 𝐵 are equal to negative seven, five, negative four, negative two and one, zero, seven, negative two, respectively, then one-third of 𝐴 plus 𝐵 is equal to negative two, five-thirds, one, negative four-thirds. Whilst we won’t prove it in this video, it is important to note that one-third multiplied by the matrix 𝐴 plus 𝐵 is equal to one-third of matrix 𝐴 plus one-third of matrix 𝐵.

In our next example, we will need to calculate the transpose of a two-by-two matrix.

The matrix six, negative four, negative three, two minus the transpose of the matrix five, negative three, negative four, one is equal to what. Is it (A) one, negative eight, negative six, one; (B) 11, zero, zero, three; (C) 𝐼; or (D) 𝑂?

Before starting this question, it is worth recalling what options (C) and (D) represent. 𝐼 is the identity matrix. This is a square matrix with ones on its leading or main diagonal and zeros elsewhere. The two-by-two identity matrix has elements one, zero, zero, one. Option (D) represents the zero matrix. All elements within this must equal zero. Therefore, the two-by-two zero matrix is equal to zero, zero, zero, zero.

Our question here also contains the operation transpose. To calculate the transpose of a matrix, we switch the rows and the columns. When dealing with a two-by-two square matrix, we flip the matrix over its leading or main diagonal. In this example, the numbers negative three and negative four will swap places. The transpose of the matrix five, negative three, negative four, one is five, negative four, negative three, one. We need to subtract this from the matrix six, negative four, negative three, two.

We recall that when subtracting two matrices, they must be of the same order, and we simply subtract the corresponding elements or components. Six minus five is equal to one. Negative four minus negative four is the same as negative four plus four, which equals zero. Likewise, negative three minus negative three is equal to zero. And two minus one is equal to one. The matrix six, negative four, negative three, two minus the transpose of the matrix five, negative three, negative four, one is equal to one, zero, zero, one. As previously mentioned, this is the identity matrix. Therefore, the correct answer is option (C).

In our next example, we need to find the missing matrix in the equation.

If negative three 𝑋 plus the matrix negative three, zero, nine, 12 is equal to the zero matrix, then 𝑋 equals what. Is it (A) one, zero, negative three, negative four; (B) negative one, zero, three, four; (C) three, zero, negative nine, negative 12; or (D) negative three, zero, nine, 12?

We recall that the zero matrix has all elements equal to zero. Therefore, the two-by-two zero matrix is equal to zero, zero, zero, zero. There are lots of ways of solving this problem. We could divide through by three or negative three. However, in this example, we will let the matrix 𝑋 be the two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑. In order to multiply any matrix by a scalar, we simply multiply each of the elements or components by that scalar. This means that the matrix negative three 𝑋 has elements negative three 𝑎, negative three 𝑏, negative three 𝑐, negative three 𝑑. Adding the matrix negative three, zero, nine, 12 to this will give us the zero matrix.

We can then set up four linear equations by comparing the corresponding components or elements. Firstly, negative three 𝑎 plus negative three is equal to zero. We can add three to both sides such that negative three 𝑎 is equal to three. Dividing through by negative three gives us 𝑎 is equal to negative one. Next, we can look at the top-right elements. This gives us negative three 𝑏 plus zero is equal to zero. As negative three 𝑏 is equal to zero, dividing both sides by negative three gives us 𝑏 is equal to zero. We can repeat this process for the bottom row, giving us values of 𝑐 and 𝑑 equal to three and four. The matrix 𝑋 is therefore equal to negative one, zero, three, four. We can therefore conclude that the correct answer is option (B).

We will now consider one final example where we look at rectangular matrices.

Given that matrix 𝑋 equals negative three, negative two, one, five, negative eight, negative eight; matrix 𝑌 equals negative one, eight, negative nine, negative nine, seven, negative two; and matrix 𝑍 equals three, negative eight, negative seven, zero, negative eight, five, what is the matrix three 𝑋 plus 𝑌 minus three 𝑍?

We recall that in order to multiply any matrix by a scalar, we simply multiply each of the elements or components by that scalar. This means that the matrix three 𝑋 is equal to negative nine, negative six, three, 15, negative 24, negative 24. In the same way, the matrix three 𝑍 is equal to nine, negative 24, negative 21, zero, negative 24, 15. We need to add matrix 𝑌 to three 𝑋 and then subtract three 𝑍. We can only add and subtract matrices when they are of the same order. In this question, all three matrices have three rows and two columns. Therefore, their order is three by two.

We now need to add and subtract the corresponding elements or components. Negative nine plus negative one is negative 10. And subtracting nine from this gives us negative 19. In the top right, we have negative six plus eight, which is equal to two. And then subtracting negative 24 from this gives us 26. Repeating this method for our other four elements gives us 15, six, seven, and negative 41. If 𝑋, 𝑌, and 𝑍 are the matrices given, then the matrix three 𝑋 plus 𝑌 minus three 𝑍 is equal to negative 19, 26, 15, six, seven, negative 41. Note that this matrix has the same order as our original matrices.

We will now summarize the key points from this video. Matrix addition and subtraction is only well defined between matrices of the same order. This means that we can only add or subtract matrices if they are of the same size. Matrix addition and subtraction is completed entry by entry. This means that we add or subtract the corresponding components or elements. Multiplying a matrix by a scalar is also completed entry by entry. We can multiply a matrix of any order by any scalar. For any matrix 𝐴 of order 𝑚 by 𝑛, the matrix transpose is of order 𝑛 by 𝑚. This means that we can perform the transpose of any matrix whether it is square or rectangular. The matrix transpose simply switches the rows and columns. When dealing with a square matrix, we can flip the matrix over its leading or main diagonal.