Lesson Video: Scalar Multiplication of Matrices Mathematics • First Year of Secondary School

Video Transcript

Scalar Multiplication of Matrices

In this video, we’re going to learn how to multiply any matrix by any scalar, that’s a real number, and we’ll see the properties and results that arise from this definition. Before we begin with scalar multiplications of matrices, we can recall we’ve seen the type of scalar multiplication before. We know how to use scalar multiplication and vectors. Consider the vector three, negative four. We know how we would multiply this by any real number. For example, we could multiply through by five. We then know to multiply a vector by five, we need to multiply it component-wise. In other words, we need to multiply each component through by five.

So multiplying each component by five, we get a new horizontal component of five times three and a new vertical component of five multiplied by negative four. This gives us the new vector with horizontal component 15 and vertical component negative 20. In other words, we’ve scaled our vector up by a factor of five. It turns out we can define exactly the same operation in terms of matrices. To start, let’s consider the matrix of order 𝑚 by 𝑛, which is given as follows. That is, the entry in row 𝑖 and column 𝑗 of this matrix is given by 𝑎 sub 𝑖𝑗. Just as we did with multiplying a vector by a scalar, we want to multiply every entry in our matrix by our scalar. So if we let 𝑘 be our scalar, we can define 𝑘 times 𝐴 as follows.

We want to multiply every single entry inside of our matrix by 𝑘. This is often referred to as scaling our matrix by a factor of 𝑘 because every single entry inside of our matrix is multiplied by 𝑘. And remember we can also represent this by what happens to every entry in our matrix. We can also represent this as the matrix whose entry in row 𝑖 and column 𝑗 is 𝑘 times 𝑎 𝑖𝑗. Let’s now see some examples of how we would calculate scalar multiplication of matrices.

Consider the matrix 𝐴. Find nine times 𝐴, where 𝐴 is equal to the one-by-two matrix two, negative one.

In this question, we’re given matrix 𝐴 and we need to find nine times 𝐴. Remember, nine is just a real number, so this is scalar multiplication of our matrix. So to answer this question, all we need to do is recall how we multiply a matrix by a scalar. We do this by multiplying every single entry inside of our matrix by our scalar. So we need to multiply the entry in the first row and first column by nine and the entry in the first row and second column by nine. This gives us the following matrix, and we can just evaluate this.

In the first row and first column, we have nine times two, which is of course just equal to 18. And in the first row and second column, we have nine multiplied by negative one. This is of course just equal to negative nine, and this gives us our final answer. Therefore, we were able to show if 𝐴 is equal to the one-by-two matrix two, negative one, then nine 𝐴 will be equal to the one-by-two matrix 18, negative nine.

Let’s now see an example which will show us how we can solve equations by using our new definition to multiply matrices by scalars.

Given that 𝑥 times the two-by-two matrix negative two, zero, negative three, negative five is equal to the two-by-two matrix 14, zero, 21, 35, find the value of 𝑥.

We’re given an equation involving two matrices and 𝑥, and we need to determine the value of 𝑥. First, we need to notice by looking at our equation, 𝑥 is a number. It’s not a matrix. This is because 𝑥 is given in lowercase. Another reason we can see this is we’re asked to find the value of 𝑥. We’re only asked to do this when 𝑥 represents a number. So in our equation, when we multiply our matrix by 𝐴, this is scalar multiplication of a matrix. So the first thing we’ll do is start by 𝑥 multiplied by our matrix and apply scalar multiplication to rewrite this in a different form.

To do this, we need to recall that when we multiply a matrix by a scalar, we multiply every single entry inside of our matrix by our scalar. In this case, we need to multiply all of our entries by 𝑥. By doing this, we get the two-by-two matrix of entries 𝑥 times negative two, 𝑥 times zero, 𝑥 times negative three, and 𝑥 times negative five. And of course, we can evaluate or simplify all of these entries. Doing this, we get the two-by-two matrix negative two 𝑥, zero, negative three 𝑥, negative five 𝑥.

But remember, in the question, we’re told that this matrix is exactly equal to the two-by-two matrix 14, zero, 21, 35. So we’ve now shown that these two matrices are equal. To find our value of 𝑥, we’re going to need to recall what it means for two matrices to be equal. Recall that we say that two matrices are equal if every single entry in both of our matrices are equal and they have the same order. Of course, both of these are two-by-two matrices. So all we need to do is check that all of their entries are equal.

Doing this, we get a series of equations. By equating the entries in row one and column one, we get 14 should be equal to negative two 𝑥. By equating the entries in row one and column two of our matrices, we get that zero should be equal to zero. By equating the entries in row two and column one of both of our matrices, we get that 21 should be equal to negative three 𝑥. And finally, by equating the entries in row two and column two of our matrices, we get that 35 should be equal to negative five 𝑥. This gives us a system of equations we need to solve. In fact, all of these are linear equations, so we could call this a system of linear equations.

Firstly, we can see that the second equation is true for all values of 𝑥. Next, we can just solve all three of the remaining linear equations. We’ll divide the first one through by negative two, the second one through by negative three, and the third one through by negative five. And by doing this, we can see that all of these are solved by 𝑥 being equal to negative seven. This means we were able to show that the value of 𝑥 should be negative seven. However, it’s also worth pointing out we can confirm this answer either by substituting the value of 𝑥 is equal to negative seven into our matrix or substituting 𝑥 is equal to negative seven into our original equation. We could then verify that this gives us the correct solution to the equation.

Therefore, given that 𝑥 times the two-by-two matrix negative two, zero, negative three, negative five was equal to the two-by-two matrix 14, zero, 21, 35, we were able to show that the value of 𝑥 must be equal to negative seven.

Let’s now see an example of a useful result which comes from our definition of scalar matrix multiplication.

If 𝐴 is equal to the one-by-three matrix eight, negative three, one, what is the value of zero 𝐴.

We’re given a matrix 𝐴 and we’re asked to evaluate zero multiplied by 𝐴. Of course, zero is a number, so this is scalar multiplication of our matrix. The first thing we’re going to need to recall is how we multiply a matrix by a scalar. We recall scalar multiplication of a matrix means we multiply every single entry by our scalar. In this case, we’re going to need to multiply every entry by zero. Doing this, we get the one-by-three matrix with entry in row one, column one zero times eight; row one, column two zero times negative three; and row one, column three zero times one. And of course we can evaluate the expressions of all of our entries. We know that zero times eight is equal to zero, zero times negative three is equal to zero, and zero times one is equal to zero. So this gives us the one-by-three matrix with every entry equal to zero, which is our final answer.

Therefore, we were able to show if 𝐴 is equal to the one-by-three matrix eight, negative three, one, then zero 𝐴 will be equal to the one-by-three zero matrix.

We can notice a very useful result from this question. We know that any number multiplied by zero is equal to zero. So, in fact, it didn’t matter what our matrix was. It always would’ve given a matrix with every entry equal to zero. So let’s confirm this result. If we have a matrix of order 𝑚 by 𝑛, and we’ll call this matrix 𝐴, and we’ll call the entry in row 𝑖 and column 𝑗 of matrix 𝐴 𝑎 𝑖𝑗, then if we multiply our matrix by the scalar zero, every entry inside of our matrix should be zero. In other words, this should be equal to the 𝑚-by-𝑛 zero matrix, represented by zero sub 𝑚𝑛.

And in fact, we can prove this result. We could do this by writing the matrix 𝐴 out in matrix notation. However, we already know the entry in row 𝑖 and column 𝑗. So when we multiply our matrix 𝐴 by the scalar zero, we multiply every single entry by zero. In other words, the entry in row 𝑖 and column 𝑗 of this matrix is zero times 𝑎 𝑖𝑗. However, we know for any number, zero times that number will always be equal to zero. In other words, for all 𝑖 and all 𝑗, zero times 𝑎 𝑖𝑗 is equal to zero. So every entry inside of our matrix is equal to zero. In other words, this is equal to the zero matrix of order 𝑚 by 𝑛.

And it’s important to remember we need to keep the order of our matrix because scalar multiplication does not change the order of our matrix. But this is not the only useful result we can get from this definition of scalar multiplication. Another question we can ask is, what is one multiplied by a matrix 𝐴? Remember, when we multiply a matrix by a scalar, we multiply every single entry in that matrix by the scalar. So, in this case, we’re multiplying every entry inside of our matrix by one. Of course, this isn’t going to change the value of any of our entries. So this should be equal to 𝐴.

And in fact, we can prove this using a very similar method to what we did above. When we multiply our matrix 𝐴 by the scalar one, we’re multiplying every single entry in matrix 𝐴 by one. In other words, the entry in row 𝑖 and column 𝑗 is going to be equal to one times 𝑎 𝑖𝑗 because we know the entry in row 𝑖 and column 𝑗 of matrix 𝐴 is 𝑎 𝑖𝑗. And of course, one multiplied by any number is equal to that same number. So one times 𝑎 𝑖𝑗 is going to be equal to 𝑎 𝑖𝑗 for any values of 𝑖 and 𝑗. Therefore, the entries inside of our matrix don’t change. The entry in row 𝑖 and column 𝑗 is just 𝑎 𝑖𝑗. Therefore, we’ve shown for any matrix 𝐴, one 𝐴 is equal to 𝐴.

And that’s another useful result we’re going to need going forward. We want to ask the question, what is negative one 𝐴 equal to? Of course we know how to do this. We multiply every single entry inside the matrix 𝐴 by negative one. So this seems to suggest that negative one times 𝐴 is going to be equal to negative 𝐴. And to fully express this, we’re going to need to recall exactly what we mean by negative 𝐴. The easiest way to do this is to think what happens if you subtract a matrix from itself.

Remember, when you subtract matrices, you do it component-wise. So when we calculate the matrix 𝐴 minus itself, every entry is going to be subtracted from itself. Every entry is going to be equal to zero. And of course, it keeps the dimension 𝑚𝑛. So perhaps a better way to write this equation would be to add our matrix 𝐴 to both sides of the equation. This would give us the equivalent statement 𝐴 plus negative one times 𝐴 is equal to 𝐴 minus 𝐴. And we know exactly what 𝐴 minus 𝐴 is equal to. It’s the zero matrix of order 𝑚 by 𝑛. And we could prove this in exactly the same way we did above.

First, to evaluate negative one multiplied by 𝐴, we need to use scalar multiplication. We multiply every entry inside of our matrix by negative one. So in row 𝑖, column 𝑗, we’ll have 𝑎 𝑖𝑗 plus negative one times 𝑎 𝑖𝑗. And of course we can then simplify this. Negative one multiplied by 𝑎 𝑖𝑗 is going to be equal to negative 𝑎 𝑖𝑗 for any values of 𝑖 and 𝑗. But remember, when we add two matrices together, we do this component-wise. So in row 𝑖, column 𝑗, we’re going to get 𝑎 𝑖𝑗 minus 𝑎 𝑖𝑗, and the number minus itself is equal to zero. So every entry inside of our matrix is going to be zero. This is going to be equal to the 𝑚-by-𝑛 zero matrix.

And there’s one final result which we can get from this definition of scalar multiplication. Before we do this, let’s clear a little bit of space. This result is going to be very similar to our first result. However, this time, instead of multiplying a matrix by the scalar zero, we’re instead going to multiply a zero matrix by any scalar. If we let 𝑘 be any number, then we can consider what happens when we multiply the 𝑚-by-𝑛 zero matrix by 𝑘. Of course, multiplying by the scalar 𝑘 means we multiply every single entry inside of the zero matrix by 𝑘.

But all of the entries inside of the zero matrix are zero. So we’re just going to get 𝑘 multiplied by zero for all of our entries. This is just going to be equal to the 𝑚-by-𝑛 zero matrix. And we could prove this by using a very similar method to what we did before. When we multiply a matrix by a scalar, we need to multiply every single entry inside of our matrix by the scalar. So in this case, we’re going to get in row 𝑖, column 𝑗 𝑘 multiplied by the entry in row 𝑖 and column 𝑗 of the zero matrix. However, every single entry in the zero matrix is zero. So in row 𝑖, column 𝑗, we get 𝑘 times zero which is equal to zero. So this is just equal to the 𝑚-by-𝑛 zero matrix.

Let’s now see how we can use these results to help us answer questions.

Let 𝑍 be a two-by-three matrix whose entries are all zero. If 𝐴 is any two-by-three matrix, which of the following is equivalent to five 𝐴 minus three 𝑍? Option (A) two times 𝑍 times 𝐴, option (B) negative two times 𝐴 times 𝑍, option (C) negative three 𝑍, option (D) five 𝐴, or option (E) two 𝐴.

In this question, we’re given an expression in terms of two matrices, and we need to determine which of five options is this expression equal to. And in fact, there’s a lot of different ways of answering this question. However, there’s one important thing we do need to notice. Both matrix 𝑍 and matrix 𝐴 are two-by-three matrices. This means the number of rows on one matrix and the number of columns on the other matrix are never equal. And when this happens, this means we can’t multiply 𝐴 by 𝑍, and we can’t multiply 𝑍 by 𝐴. So options (A) and (B) can’t be correct.

Let’s now see what we can do with the expression we’re given. Let’s start by recalling that 𝑍 is a two-by-three matrix, where every entry is equal to zero. Now we could write this out in terms of matrices. However, we can also write this as the two-by-three zero matrix, zero, two, three. Let’s now write out matrix 𝐴 and the zero matrix in full. To do this, we need to recall a matrix of order two by three will have two rows and three columns. We’ll also call the entry in matrix 𝐴 in row 𝑖, column 𝑗 𝑎 𝑖𝑗. This gives us the following expression.

We can now simplify this expression either by using our definition for scalar multiplication of a matrix or by using the fact that for any number 𝑘, 𝑘 multiplied by the 𝑚-by-𝑛 zero matrix is just equal to the 𝑚-by-𝑛 zero matrix. This gives us that three 𝑍 is just equal to the two-by-three zero matrix. Now we see we’re subtracting the zero matrix from our matrix five 𝐴. There’s a few different things we could do. For example, we could use our definition of scalar multiplication to bring five inside of our matrix. Then we could use our definition of matrix subtraction to answer this question.

However, this is not necessary because we’re subtracting the zero matrix of the same order. And when we do this, we subtract zero from every entry inside of our matrix. Of course, subtracting zero is not going to change any of the values, so this is just equal to five 𝐴. And this gives us that option (D) is the correct answer. It’s also worth pointing out we could check that option (C) and option (E) are not correct in all scenarios. For example, if we set 𝐴 equal to the two-by-three matrix where all entries are one, then we’ve already shown that five 𝐴 minus three 𝑍 should be equal to five 𝐴. And, of course, five 𝐴 is every entry in 𝐴 multiplied by five. It’s the two-by-three matrix where every entry is five.

This is, of course, not equal to two 𝐴, since this would be the two-by-three matrix where every entry is two, and it’s also not equal to the matrix negative three, 𝑍 since we’ve already shown that this will be equal to the two-by-three zero matrix. Therefore, we were able to show the only correct option is option (D); five 𝐴 minus three 𝑍 will be equal to five 𝐴.

Let’s now look at one last property that we can get from scalar multiplication of matrices. We’ll first show this property by looking at it in action.

Let 𝐴 and 𝐵 be the following two matrices. We want to calculate the following two expressions. First, we want to calculate three multiplied by the sum of 𝐴 and 𝐵. We then want to calculate three 𝐴 plus three 𝐵.

In this case, we’re checking whether we can distribute three over our parentheses in this case. And if we can do this in general, this would be called a distributive property.

Let’s start with the first expression. Let’s find three times 𝐴 plus 𝐵. Remember, we first need to calculate the expression inside of our parentheses. So we’ll start by writing our matrices 𝐴 and 𝐵 out in full. And remember, when we’re adding two matrices of the same order together, we just add their entries together. So in row one, column one, we’ll get negative four plus three which we can calculate as negative one. In row one, column two, we’ll get six plus seven which is equal to 13. And we can do the same to find the rest of the entries. We get negative five, 14, 13, and one.

And remember, we need to multiply this matrix by three. So we’re multiplying this matrix by the scalar three. Remember to do this, we multiply every entry inside our matrix by three. For the entry in row one, column one of our matrix, we have three multiplied by negative one, which we can calculate as negative three. For the entry in row one, column two, we have three multiplied by 13, which we can calculate as 39. And we can do the same to find the rest of the entries. We get negative 15, 42, 39, and three.

Let’s now see what happens if we calculate three 𝐴 plus three 𝐵. Once again, we’ll start by writing our matrices 𝐴 and 𝐵 out in full. This time, we first need to use our scalar multiplication to multiply both of our matrices by three. Remember, we do this by multiplying every single entry inside of our matrix by three. We’ll start by calculating three 𝐴. Multiplying every entry by three and simplifying, we get the two-by-three matrix negative 12, 18, zero, 24, 21, six. We can then do the same to find three 𝐵; it’s equal to the following two-by-three matrix.

Finally, all we need to do is add these two together. Remember, we do this by adding all the entries together. By adding the entries in row one and column one together, we get negative 12 plus nine, which is equal to negative three. By adding the entries in row one, column two together, we get 39. And we can do the same to find the rest of the entries. We get the following two-by-three matrix, which we can see is exactly equal to the one we found before. In other words, three times 𝐴 plus 𝐵 is equal to three 𝐴 plus three 𝐵. And in fact, this is true in general. For any number 𝑘 and 𝑚-by-𝑛 matrices 𝐴 and 𝐵, we have 𝑘 times 𝐴 plus 𝐵 is equal to 𝑘𝐴 plus 𝑘𝐵. And this is called the distributive property because we can distribute 𝑘 over our parentheses.

Let’s now go over the key points of this video. First, we showed that multiplying a matrix 𝐴 by a scalar 𝑘 means multiplying every single entry of 𝐴 by 𝑘. And in fact, it’s worth pointing out 𝑘 can be any number. It could be complex. However, we’ve only used real numbers in this video. And we can write this as 𝑘𝐴. Finally, we showed five useful properties for 𝑚-by-𝑛 matrices 𝐴 and 𝐵. The first four of these are identities we directly derive from the definition of scalar multiplication and matrix addition. Finally, we showed the scalar multiplication is distributive when combined with matrix addition.