Video: Finding the Inverse of an Upper Triangular Matrix

Consider the matrix 𝐴 = [2, 1, 4 and 0, 5, 3 and 0, 0, 10]. Find its inverse, given that it has the form 𝐴⁻¹ = [π‘₯, 𝑝, π‘ž and 0, 𝑦, π‘Ÿ and 0, 0, 𝑧], where π‘₯, 𝑦, 𝑧, 𝑝, π‘ž, and π‘Ÿ are numbers you should find.

06:21

Video Transcript

Consider the matrix 𝐴 is equal to two, one, four, zero, five, three, zero, zero, 10. Find its inverse, given that it has the form the inverse of the matrix 𝐴 is equal to π‘₯, 𝑝, π‘ž, zero, 𝑦, π‘Ÿ, zero, zero, 𝑧, where π‘₯, 𝑦, 𝑧, 𝑝, π‘ž, and π‘Ÿ are numbers you should find.

Well, the first thing we’re gonna do here is clear a bit of space so that we can actually have room to do the working out. Okay, so we’ve now cleared some space. How we’re gonna work out what the inverse of our matrix is? Well, to find it, what we’re gonna do is go through a number of steps. Step one is finding the matrix of minors; step two, we need to find the cofactor matrix; then step three, we find the adjugate or the adjoint; and finally step four, we multiply by one over the determinant of our matrix. Okay, great! So we have our four-step approach, so let’s go ahead and start step one.

So what we have here is our matrix of minors 𝐴 sub 𝑀. Now let’s remind ourselves how we actually found out what the minors were for each element. Well, to find a minor, what we do is if we’re looking at the first element, so that’s the first element in the first column and row, then you delete the values or elements that are in the column and row that your element is in. And then you look at the four that are left over cause it’s a three-by-three matrix, and here, we would have five, three, zero, 10. Then the minor is the determinant of this two-by-two submatrix. And we’ve used that method and completed our matrix of minors.

Okay, great! We now need to find out what the values are for each of these elements. Well, to remind ourselves how we find the determinant of a two-by-two matrix, we have a look at this little example here. If we had the determinant the two-by-two matrix π‘Ž, 𝑏, 𝑐, 𝑑, then this would be equal to π‘Žπ‘‘ minus 𝑏𝑐. So what we do is we multiply diagonally and then subtract. So as an example, for the first element, what we’d have is five multiplied by 10 minus three multiplied by zero, which would be equal to 50.

Okay, we’ll just do the next element as one more example. Well, for the next element, we’d have zero multiplied by 10 minus three multiplied by zero, which is just zero. So that’s our next element filled in. And then we continue to use this method to fill in the rest of our elements. And when we’ve done this, we’ve got 50, zero, zero, 10, 20, zero, negative 17, six, 10. And this is our matrix of minors, so step one complete. So now we move on to step two, find the cofactor matrix.

Well, this step is in fact very straightforward because all we need to do is add some signs to our matrix of minors. And we do that using the sign rule, which tells us that the signs we need are gonna be positive, negative, positive; negative, positive, negative; positive, negative, positive. So if we put them in here, so we’ve got positive, negative, positive; negative, positive, negative; positive, negative, positive. So this just leaves us with our cofactor matrix, which is the matrix 50, zero, zero, negative 10, 20, zero, negative 17, negative six, 10. So great, that’s step two complete. We can move on to step three.

Well, for step three, what we need to do is find the adjugate or adjoint. So then to do this, what we do is take a look at our cofactor matrix, and we look at the diagonal that goes from top left to bottom right and keep this the same. And then we swap the other elements across this diagonal. And when we do this, we have the matrix 50, negative 10, negative 17, zero, 20, negative six, and zero, zero, 10. So we see that we swapped places. Okay, great! Step three complete.

So now for step four, what we need to do is multiply by one over the determinant. Well, we need to, first of all, find out what the determinant of our matrix is. Well, you might think, β€œWell, this is gonna be quite a long-winded process.” But in fact, it’s not; it’s actually gonna be very easy because we’ve already worked out the minors for each element of our matrix. But how is that gonna help?

Well, if we remind ourselves how we find the determinant of a three-by-three matrix β€” I’ll just quickly clear some space in the bottom right here to do this. Well, if we have the matrix π‘Ž, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, β„Ž, 𝑖, then the determinant of this matrix is equal to π‘Ž multiplied by the minor which is 𝑒, 𝑓, β„Ž, 𝑖 β€” or the two-by-two submatrix 𝑒, 𝑓, β„Ž, 𝑖 β€” the determinant of that minus 𝑏 multiplied by the determinant of 𝑑, 𝑓, 𝑔, 𝑖 plus 𝑐 multiplied by the determinant of 𝑑, 𝑒, 𝑔, β„Ž.

Well, as we already have the value of our minors or the determinants of our two-by-two submatrices, then all we have to do is multiply this number by our π‘Ž, 𝑏, and 𝑐. So what we’re gonna get is the determinant of 𝐴 is gonna be equal to two multiplied by 50, and we get that because two is the first element of our matrix 𝐴 and 50 is the first element of our matrix of minors, minus one multiplied by zero plus four multiplied by zero. So this is gonna give us 100, which is great. So we now know that the determinant of our matrix 𝐴 is 100.

So now for step four, what we need to do is multiply the adjugate by one over this amount. So therefore, we know that the inverse of our matrix is gonna be equal to one over 100 multiplied by the matrix 50, negative 10, negative 17, zero, 20, negative six, zero, zero, 10. And when we do this, we’ll get the matrix a half, negative one over 10, negative 17 over 100, zero, fifth, negative three over 50, zero, zero, one over 10. And what this is is the inverse of the matrix two, one, four, zero, five, three, zero, zero, 10, where the values of π‘₯, 𝑝, π‘ž, 𝑦, π‘Ÿ, 𝑧 are a half, negative one-tenth, negative 17 over 100, a fifth, negative three over 50, and one over 10, respectively.

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