Video: Operations on Matrices Involving Equality of Matrices

Let 𝐴 = [1, 2 and 3, 4] and 𝐡 = [1, 2 and 1, π‘˜]. Is it possible to choose a value for π‘˜ so that 𝐴𝐡 = 𝐡𝐴? If so, what is this value?

03:32

Video Transcript

Let 𝐴 equal the matrix one, two, three, four and 𝐡 equal the matrix one, two, one, π‘˜. Is it possible to choose a value for π‘˜ so that 𝐴𝐡 is equal to 𝐡𝐴? If so, what is this value?

So the first thing we need to do is remind ourselves how we multiply two two-by-two matrices. So let’s imagine we’ve got the matrices π‘Ž, 𝑏, 𝑐, 𝑑 and 𝑒, 𝑓, 𝑔, β„Ž. Then the first element of the result, which will be a two-by-two matrix, is gonna be π‘Ž multiplied by 𝑒, so the first element in the first row of the first matrix multiplied by the first element in the first column of the second matrix, then plus 𝑏𝑔 because that’s the second element in the first row of the first matrix multiplied by the second element in the first column of the second matrix.

And then if we move along to the next element, what we get is π‘Žπ‘“ plus π‘β„Ž. And that’s because what we’ve got is the first element in the first row of the first matrix multiplied by the first element in the second column of the second matrix plus the second element in the first row of the first matrix multiplied by the second element in the second column of the second matrix. And then, if we carry on this pattern, we’ll have 𝑐𝑒 plus 𝑑𝑔. And then for our final element, we’ll have 𝑐𝑓 plus π‘‘β„Ž.

Okay, great. So we remind ourselves how we multiply our matrices. So now what we’re going to do is work out what 𝐴𝐡 and 𝐡𝐴 are. So we’re gonna start with 𝐴𝐡. So it’s the matrix one, two, three, four multiplied by the matrix one, two, one, π‘˜. So for the first element, we’re gonna have one multiplied by one plus two multiplied by one. Then for the next element, we’ve got one multiplied by two plus two multiplied by π‘˜. Then we’ll have three multiplied by one plus four multiplied by one. And then finally, three multiplied by two plus four multiplied by π‘˜. So this is gonna give us the matrix three, two plus two π‘˜, seven, six plus four π‘˜. So great, we’ve found out what 𝐴𝐡 is.

So now let’s have a look at 𝐡𝐴. And then for 𝐡𝐴, what we’re gonna have is matrix one, two, one, π‘˜ multiplied by the matrix one, two, three, four. And this is gonna be equal to one multiplied by one plus two multiplied by three, one multiplied by two plus two multiplied by four, one multiplied by one plus π‘˜ multiplied by three, then finally one multiplied by two plus π‘˜ multiplied by four. And what this is gonna give us is the matrix seven, 10, one plus three π‘˜, two plus four π‘˜.

Well, we can see straightaway that 𝐴𝐡 cannot be equal to 𝐡𝐴. And that’s because if we have a look at the matrices that we’ve got here, our first element in 𝐴𝐡 is three and the first element in 𝐡𝐴 is seven. And even if we tried with the other elements, if we take a look at the next element along, we’ve got two plus two π‘˜ and 10. Well, if we equated these, we get two plus two π‘˜ equals 10. So two π‘˜ would be equal to eight. And π‘˜ would be equal to four. Then we substituted back in this value for π‘˜, we’d get all different values for the different elements.

So therefore, we can say that in answer to the question, β€œIs it possible to choose a value for π‘˜ so that 𝐴𝐡 equals 𝐡𝐴?” we can say that there is no possible choice for π‘˜ because 𝐴𝐡 cannot be equal to 𝐡𝐴.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.