Video: Finding the Product of Two Matrices

Given that 𝐴 is the matrix (βˆ’4, 2, 2, βˆ’4) and 𝐡 is the matrix (βˆ’3, βˆ’3, βˆ’1, 1), find 𝐴𝐡 and 𝐡𝐴.

04:20

Video Transcript

Given that 𝐴 is the matrix negative four, two, two, negative four and 𝐡 is the matrix negative three, negative three, negative one, one, find 𝐴𝐡 and 𝐡𝐴.

First we find 𝐴𝐡. What is 𝐴𝐡? Well it’s the product of the two matrices 𝐴 and 𝐡 in that order. So the first step is just to substitute in the values of 𝐴 and 𝐡 that were given in the question. 𝐴 is negative four, two, two, negative four and 𝐡 is negative three, negative three, negative one, one.

What do we get when we multiply these two matrices? Well the first thing to say is that we get another matrix, in fact another two-by-two matrix, so now we just have to fill in the entries of this matrix.

We’ll start with the first entry of the matrix, which is the entry in the first row and the first column of the matrix. We are in the first row, so we look at the first row of the matrix on the left, 𝐴. This is the row negative four, two, and this entry that we want to find is in the first column of 𝐴𝐡, so we look at the first column of the matrix on the right, 𝐡.

The entry of 𝐴𝐡 that we’re looking for is the dot product of the highlighted row with the highlighted column, so that is negative four times negative three plus two times negative one. So that’s the first entry of the first row of 𝐴𝐡.

Now let’s move on to the second entry of the first row. This entry is in the first row and second column of the matrix 𝐴𝐡. So it is the dot product of the first row of 𝐴 with the second column of 𝐡. So it is negative four times negative three plus two times one.

What’s the entry in the second row and first column of 𝐴𝐡? It’s the dot product of the second row of 𝐴 with the first column of 𝐡, two times negative three plus negative four times negative one. And finally the entry in the second row and second column of 𝐴𝐡 is the dot product of the second row of 𝐴 with the second column of 𝐡.

Now we just need to evaluate the entries. Negative four times negative three plus two times negative one is just 10. Negative four times negative three plus two times one is 14. Two times negative three plus negative four times negative one is negative two. And finally two times negative three plus negative four times one is negative 10.

Now we found 𝐴𝐡. We just have to find 𝐡𝐴. The only difference here is that the matrix 𝐡 is written before 𝐴, so we have negative three, negative three, negative one, one times negative four, two, two, negative four.

The entry in the first row and first column of 𝐡𝐴 is the dot product of the first row of 𝐡 with the first column of 𝐴. The entry in the first row and second column of 𝐡𝐴 is the dot product of the first row of 𝐡 with the second column of 𝐴. The entry in the second row and first column of 𝐡𝐴 is the dot product of the second row of 𝐡 with the first column of 𝐴. And finally the entry in the second row and second column of 𝐡𝐴 is the dot product of the second row of 𝐡 with the second column of 𝐴.

Evaluating all these entries, we get six, six, six, and negative six, and so our final answer is that 𝐴𝐡 is the matrix 10, 14, negative two, negative 10 and 𝐡𝐴 is the matrix six, six, six, negative six.

Notice that the values of 𝐴𝐡 and 𝐡𝐴 are different. This shows that, unlike the multiplication of natural numbers or real numbers or even complex numbers, matrix multiplication is not commutative. The commutative property does not hold. That isn’t to say that there can’t be two matrices 𝐴 and 𝐡 such that 𝐴𝐡 equals 𝐡𝐴; it just says that, in general, we can’t rely on this being true.

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