Question Video: Finding the Inverse of an Upper Triangular Matrix Mathematics • 10th Grade

Consider the matrix 𝐴 = [1, 2, 3, and 0, 1, 4, and 0, 0, 1]. Find its inverse, given that it has the form 𝐴⁻¹ = [1, 𝑝, 𝑞, and 0, 1, 𝑟, and 0, 0, 1], where 𝑝, 𝑞, and 𝑟 are numbers that you should find.

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Video Transcript

Consider the matrix 𝐴 is equal to one, two, three, zero, one, four, zero, zero, one. Find its inverse given that it has the form one, 𝑝, 𝑞, zero, one, 𝑟, zero, zero, one, where 𝑝, 𝑞, and 𝑟 are numbers that you should find.

Well, the first thing we can notice about our matrix 𝐴 is that it is an upper triangular matrix. We know this because we look at the elements in the bottom left-hand side, they’re all three zero. So, therefore, the elements that are not zero form a triangle in the top right-hand side. And what we know about an upper triangular matrix is that the inverse of said matrix will also be an upper triangular matrix. And if we can check the form of our inverse, we can see that this is the case, because once again, we have the bottom left three elements as zero.

What we’ll also notice when we’re looking to find the inverse is that a lot of steps that we’ll usually complete will be a lot easier because of this particular matrix we have. But we’ll look at that as we go through. So, first of all, if we’re gonna find the inverse, step one is to find the matrix of minors. Step two, we then find the cofactor matrix. Then, for step three, we find the adjugate, also known as the adjoint matrix. And then, finally, for step four, what we do is multiply by one over the determinant of our original matrix. Okay, great. So, let’s go through each of these steps. And as I already said, what we’ll be able to do is make some of them a lot more simple because of the fact that we’re looking at upper triangular matrix.

So now, our matrix of minors is gonna be the matrix we got here. And if we remind ourselves how we’d find the matrix of minors, well, let’s look at the first element. Well, what we’ve got in the original matrix is one as our first element. Then, what we want to do is remove every value or every element that’s in the same column and row with it. And then, what we’re left with is a two-by-two submatrix one, four, zero, one. And what this is is our minor. And so, our minor is the determinant of this two-by-two matrix.

So now, what we can quickly do is remind ourselves how we’d find the determinant of a two-by-two matrix. Well, if we’ve got the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑, this will be equal to 𝑎𝑑 minus 𝑏𝑐. So, what we do is cross multiply and then subtract. Well, here, this next step, what we’d usually do is work out each of the values, so each of our minors. However, we can straightaway put in three values, and this is because we know that we’re dealing with an upper triangular matrix. These three values are zero, zero, and zero, and they’re in the top right-hand side of our matrix. And we can do that because without even having to work out our minors, first of all, we can see that both diagonals in our minors have a zero in. So, therefore, zero multiplied by anything is zero. So, the results are just going to be zero.

It’s also worth noting that when we’re having a look at the matrix of minors for a matrix that is an upper triangular matrix, then we always have these three zeros in these positions, which are opposite to where they are in the matrix itself. And then, if we want to fill in the first element, we’d have one multiplied by one which is just one minus four multiplied by zero which is zero, which is just gonna give us a result as one. And then, if we take a look at the element below, we’ve got two multiplied by one minus three multiplied by zero, which gives us two. And then, we carry out this method until we fill out the rest of our matrix. So, our matrix of minors is one, zero, zero, two, one, zero, five, four, one. So, that’s step one complete.

Now, we’re gonna move on to step two and find the cofactor matrix. Well, in fact, this is the easiest one because all we need to do is apply our sign rule to help us work out what it’s going to be. Our sign rule for the cofactor matrix just tells us what the signs of each element is going to be. So, we just apply these signs to the elements we had in our matrix of minors. And when we do that, what we get is our cofactor matrix. And that is one, zero, zero, negative two, one, zero, five, negative four, one. Okay, great. That’s step two complete.

So now, let’s move on to step three. Let’s find the adjugate matrix. So, all we need to do to find the adjugate matrix is swap the elements across our diagonal that goes from top left to bottom right. So, if we do that, we’ll then get our adjugate matrix. So, what this is gonna give us is an adjugate matrix one, negative two, five, zero, one, negative four, zero, zero, one. And what we can see is that this is, in fact, an upper triangular matrix. So, actually, we’re moving towards what we’re looking for in our inverse because we’d already identified that our inverse is gonna be an upper triangular matrix.

So now, let’s complete step four. And step four is our final step. And all we need to do is multiply by one over the determinant of our original matrix. Well, if we want to find the determinant of a three-by-three matrix and if we have the matrix 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖, this will be equal to 𝑎 multiplied by the two-by-two submatrix determinant 𝑒, 𝑓, ℎ, 𝑖 minus 𝑏 multiplied by the two-by-two submatrix determinant 𝑑, 𝑓, 𝑔, 𝑖 plus 𝑐 multiplied by the determinant 𝑑, 𝑒, 𝑔, ℎ. Well, this would usually be a bit of a lengthy process. However, because we’ve already found our matrix of minors, this means we’ve already calculated our determinants of our two-by-two submatrices.

So, therefore, to find out our determinant, what we’re gonna need to do is multiply the first row, each of the terms by the corresponding terms in the first row of our matrix of minors. However, this gets even quicker because we’re dealing with an upper triangular matrix. Well, this is because the determinant is just going to be one multiplied by one. And that’s because we don’t have to even bother trying to work out the next two values because as we can see the next two values in the top row of our matrix of minors are zeros. So, therefore, we can see that our determinant is just going to be equal to one.

Well, if our determinant is just equal to one, this’s gonna make step four rather easy because our inverse is just gonna be equal to one multiplied by our adjugate matrix, so one multiplied by the matrix one, negative two, five, zero, one, negative four, zero, zero, one.

So, therefore, what we can say is that the inverse of the matrix one, two, three, zero, one, four, zero, zero, one in the form of the matrix one, 𝑝, 𝑞, zero, one, 𝑟, zero, zero, one is the matrix one, negative two, five, zero, one, negative four, zero, zero, one, where 𝑝, 𝑞, and 𝑟 are negative two, five, and negative four, respectively.

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