Video Transcript
Write the matrix three, negative eight, negative one, negative nine in the form 𝑎 multiplied by the matrix one, zero, zero, zero plus 𝑏 multiplied by the matrix zero, one, zero, zero plus 𝑐 multiplied by the matrix zero, zero, one, zero plus 𝑑 multiplied by the matrix zero, zero, zero, one, where 𝑎, 𝑏, 𝑐, and 𝑑 are real numbers that you should find.
So what we can see here is that we’ve got scalars multiplied by matrices. So if we were to multiply a matrix by a scalar, what we do is multiply each of the elements inside the matrix by our scalar value. So, for instance, it’d be 𝑎 multiplied by and then each of the elements one, zero, zero, and zero in the first matrix. Now what we could do is multiply each of our matrices out and then solve to find out what our 𝑎, 𝑏, 𝑐, and 𝑑 are.
However, with this particular question, there’s an easier way of doing this. And that’s because, on inspection, we can see that each of our matrices only carries one element that has a value that isn’t zero. And in fact, it’s one in each of these cases. If we look at the first matrix, it’s the top-left element; second matrix, top-right element; third matrix, bottom-left element; and fourth matrix, bottom-right element.
So what I’m going to do is, just as an example, have a look at the top-left elements, so the first element in each of our matrices. So what we can see is that if we multiplied it by the scalar and then added them together, we’d have 𝑎, because it’s 𝑎 multiplied by one which is just 𝑎, plus zero 𝑏, cause it’s 𝑏 multiplied by zero, plus zero 𝑐 plus zero 𝑑. And what this would be is equal to the corresponding matrix in the matrix we’re looking for, which is the matrix three, negative eight, negative one, negative nine. So therefore, we’d say that this is all equal to three. However, as the terms involving 𝑏, 𝑐, and 𝑑 are all zero terms, we can disregard these. So we’ll just get 𝑎 is equal to three.
Well, now if we look at the top-right element, the only matrix with a value in that element, a value of one, is the matrix with 𝑏 as a scalar. So therefore, we don’t have to go through the whole process of multiplying out. What we can say is that 𝑏 is gonna be equal to negative eight. And that’s because negative eight is the corresponding element in the matrix three, negative eight, negative one, negative nine. And then using the same rationale, we can see that 𝑐 is gonna be equal to negative one and 𝑑 is equal to negative nine.
So great, have we solved the problem? Well, no, because we found 𝑎, 𝑏, 𝑐, and 𝑑; however, what we’re asked to do is write the matrix three, negative eight, negative one, negative nine in the form and then the form that we’ve got laid out in front of us, which is 𝑎 multiplied by the matrix one, zero, zero, zero plus 𝑏 multiplied by the matrix zero, one, zero, zero plus 𝑐 multiplied by the matrix zero, zero, one, zero plus 𝑑 multiplied by the matrix zero, zero, zero, one.
So then, if we substitute in our values, we get three multiplied by the matrix one, zero, zero, zero minus eight multiplied by the matrix zero, one, zero, zero and then minus — well, we would have one but we don’t need to write that — the matrix zero, zero, one, zero minus nine multiplied by the matrix zero, zero, zero, one.
