Video Transcript
Consider the matrix 𝐴 which is equal to 𝐾, 𝑎, 𝑏, zero, 𝐿, 𝑐, zero, zero, 𝑀. Find its inverse given that it has the form 𝑋, 𝑝, 𝑞, zero, 𝑌, 𝑟, zero, zero, 𝑍, where 𝑋, 𝑌, 𝑍, 𝑝, 𝑞, and 𝑟 are expressions involving 𝐾, 𝐿, 𝑀, 𝑎, 𝑏, 𝑐 that you should find.
Well, the first thing we can notice from our matrix is that we have an upper triangular matrix. And we can see that because what we’ve got is a triangle shape here with the known elements and then with the bottom left-hand side, what we have are zeros. So this is called an upper triangular matrix. But why is this significant? Well, first of all, it means that the inverse would also take this form, which we can see that it does, because we’ve got the three zero elements in the bottom left-hand side. Okay, but also what it means is along the way, when we’re trying to work out the inverse, there’re a couple of steps which will make it a little bit easier.
Okay, so now let’s start to find the inverse. Well, there are four steps to this. Let’s remember what they are. So step one is finding the matrix of minors. For step two, what we do is find the matrix of cofactors. Then next, what we need to do is find the adjugate or adjoint matrix. And then finally, the fourth step, we multiply by one over the determinant of the original matrix. Okay, great! We’ve got our four steps, so let’s start to use them to solve our problem.
So as we see here, we’ve written our matrix of minors. But let’s remind ourselves exactly how we found our minors. Well, let’s take a look at our first element. We’ve got the minor 𝐿, 𝑐, zero, 𝑀. Well, if we look back to the corresponding element in our original matrix, we’ve got 𝐾. And then we delete the row and column that 𝐾 is in. We’re left with 𝐿, 𝑐, zero, 𝑀. And then what we have for the minor is the determinant of this two-by-two submatrix. So now what we need to do is just remind ourselves how we’d work out the determinant of a two-by-two matrix. Well, if we had 𝑎, 𝑏, 𝑐, 𝑑, then the determinant of this would be 𝑎𝑑 minus 𝑏𝑐. So we cross multiply and then subtract.
Well, so at this point, before we start working out our minors, then we can actually use one of the properties of our upper triangular matrix. And that is that if we look at the three minors here that I’ve circled some zeros in, we could see that there is zeros in both of our diagonals. So therefore, they’re just going to be equal to zero. And this is because these are areas that represent the bottom left-hand side of our original matrix, where we had the three zero elements. So therefore, we can fill in our zero elements straightaway with our matrix of minors. So then our first element is gonna be 𝐿𝑀, because it’s 𝐿 multiplied by 𝑀, minus 𝑐 multiplied by zero, so it’s just 𝐿 multiplied by 𝑀, 𝐿𝑀, minus zero.
And then if we move to the element just below, we’re gonna have 𝑎𝑀 because what we have is 𝑎 multiplied by 𝑀 minus 𝑏 multiplied by zero. Then 𝐾𝑀 for the next element. Then we’re gonna move on to the bottom row. Well, for the bottom-left element, we actually have two terms here because what we have is 𝑎𝑐 minus 𝑏𝐿. And then our last two elements are 𝐾𝑐 and 𝐾𝐿. Okay, great! So we’ve now found our matrix of minors, so step one.
So now for step two, to find the matrix of cofactors, this is nice and straight forward cause all we need to do is assign our elements some signs using the sign rule. And they are positive, negative, positive, negative, positive, negative, positive, negative, positive, in that order. Well, if we apply this, the only elements that it’s gonna affect, and that’s because we have the zero elements, are going to be negative 𝑎𝑀 and negative 𝐾𝑐. So now we’ve got our matrix of cofactors, which is step two complete.
So now for step three, what we want to do is find the adjugate or also known as the adjoint matrix. So to do that, what we do is swap our elements across our diagonal. So we’ve drawn on some arrows to show what we’re gonna do here. So now that’s step three complete cause we have our adjugate matrix, which is gonna be 𝐿𝑀, negative 𝑎𝑀, 𝑎𝑐 minus 𝑏𝐿, zero, 𝐾𝑀, negative 𝐾𝑐, zero, zero, 𝐾𝐿. And what we can see here is this has now taken the form of the upper triangular matrix, which is what we said that the final answer would need to take.
Well now, our final step is to multiply by one over the determinant of our original matrix 𝐴. So, now what we need to do is find the determinant. And if we’re gonna find the determinant of a three-by-three matrix, we’d usually use this method shown at the top right. However, we don’t need to do that with our problem. And here’s the reason why. Well, if we look at the method, what we’ve got is a value which is the element value from the first row of our matrix multiplied by a minor. Well, we’ve already calculated the minors because we did that earlier on when we found the matrix of minors. Well, therefore, what we need to do is just multiply the element of the first row from the original matrix by our minors and follow the pattern we have here.
However, to make it even easier, because we’ve got an upper triangular matrix, then we only need to do it with the first element because we have those two zeros in our first row of our matrix of minors. So, anything multiplied by zero will just give us zero. So what we’re gonna do now is clear a bit of space just to show what we’re doing. So therefore, to find the determinant of our matrix 𝐴, what we’re gonna do is multiply 𝐾 by 𝐿𝑀 because it’s the first element in our matrix by the first element in our matrix of minors. Well, this is gonna give us our determinant of 𝐴 as 𝐾𝐿𝑀.
So what we want to do is now complete our step four, multiply our adjugate matrix by one over the determinant of matrix 𝐴. So what we’re gonna get is that the inverse matrix is equal to one over 𝐾𝐿𝑀 multiplied by the matrix 𝐿𝑀, negative 𝑎𝑀, 𝑎𝑐 minus 𝑏𝐿, zero, 𝐾𝑀, negative 𝐾𝑐, zero, zero, 𝐾𝐿. So this is gonna give us the matrix 𝐿𝑀 over 𝐾𝐿𝑀, negative 𝑎𝑀 over 𝐾𝐿𝑀, 𝑎𝑐 minus 𝑏𝐿 over 𝐾𝐿𝑀, zero, 𝐾𝑀 over 𝐾𝐿𝑀, negative 𝐾𝑐 over 𝐾𝐿𝑀, zero, zero, 𝐾𝐿 over 𝐾𝐿𝑀.
Okay, so now let’s do some canceling. So now when we completed our canceling, we’ve completed our fourth step, which was multiplying by one over the determinant. So therefore, we can give our inverse as the matrix one over 𝐾, negative 𝑎 over 𝐾𝐿, 𝑎𝑐 minus 𝑏𝐿 over 𝐾𝐿𝑀, zero, one over 𝐿, negative 𝑐 over 𝐿𝑀, zero, zero, one over 𝑀. And this is exactly in the form that we want it because it’s in the form 𝑋, 𝑝, 𝑞, zero, 𝑌, 𝑟, zero, zero, 𝑍.
