Find the multiplicative inverse of the matrix negative five, zero, zero, zero, negative five, zero, zero, zero, negative five.
Well, the multiplicative inverse of a matrix is in fact the matrix, so the inverse matrix, that when multiplied by the original matrix will give us 𝐼, where 𝐼 is the identity matrix. And the identity matrix is a matrix where all the elements are zero apart from the diagonal from the top left to the bottom right, where the elements are one.
Now, because we’re looking at a matrix here which is in fact a diagonal matrix, that is, the only elements that aren’t zero are diagonal, the process of finding our inverse is gonna be a lot more simple. But let’s look at the steps anyway. Well, the four steps we have when finding the inverse of a matrix are step one, find the matrix of minors. Step two, find the cofactor matrix, then step three, the adjugate or adjoint matrix. And step four, finally, we multiply by one over the determinant of the original matrix.
So, if we call our original matrix 𝐴 in the question, then let’s have a look at finding the matrix of minors. So now, we’ve got our matrix of minors. Let’s remind ourselves how we found each of the minors. Well, if we take a look at our first element, we can see that the minor here is created by removing the row and column that the first element is in. And then, it’s the two-by-two submatrix which is left behind, which is negative five, zero, zero, negative five. Then, we find the determinant of this.
Okay, so we’ve got our matrix of minors. But what we want to do now is calculate the values. However, as I said earlier, we’ve got a diagonal matrix. So, this next step is a lot easier. So, if we quickly remind ourselves how we find the determinant of a two-by-two matrix, what we do is we have the matrix here 𝑎, 𝑏, 𝑐, 𝑑. The determinant is equal to 𝑎𝑑 minus 𝑏𝑐. So, we cross multiply and then we subtract. Well, because we’re looking at a diagonal matrix, we can see that all of the minors we have, apart from the three from the top left to the bottom right, have zeroes in both their diagonals. So therefore, they’re just gonna be equal to zero because zero multiplied by anything is zero.
And also, all the diagonal minors are identical. So, we just need to work one out. So, what we have is negative five multiplied by negative five, which is 25, then minus zero multiplied by zero, which is just zero. So, they’re all 25. So great, we’ve got our matrix of minors.
So now, the next step is nice and easy because all we need to do for the cofactor matrix is adding the signs that we get from the sign rule for our cofactor matrix. And they are positive, negative, positive, negative, positive, negative, positive, negative, positive. And, in fact, if we look at our matrix of minors, this won’t effect it at all because the only values are our diagonals and they are all positive anyway.
And because they are all positive anyway in our sign rule, it means that their signs won’t change. So therefore, our cofactor matrix will just be the same as our matrix of minors. Okay, great. So, that’s step two complete.
Now, step three, again, is gonna be very simple. And this is because of the special type of matrix we have, which I already said was a diagonal matrix. Well, to find the adjugate matrix in step three, all we would do is swap the elements across our diagonal as shown by the arrows. However, these are all the same; these are all zero. So, once again, the adjugate matrix is gonna be just the same as our matrix of cofactors. So now, all we need to do is move on to step four, which is multiply by one over our determinant of 𝐴.
Well, it’s worth noting at this point that whenever we have this type of matrix, we can always skip out steps two and three because they won’t change the matrix that we’re looking at.
So, what we need to do at this point is remind ourselves how we find the determinant of a three-by-three matrix. So, if we’ve got the matrix 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖, this is equal to 𝑎 multiplied by the minor or the determinant of the two-by- two submatrix 𝑒, 𝑓, ℎ, 𝑖 then minus 𝑏 multiplied by 𝑑, 𝑓, 𝑔, 𝑖 plus 𝑐 multiplied by the determinant of the two-by-two submatrix 𝑑, 𝑒, 𝑔, ℎ.
So, actually, this would usually be a bit of a long-winded process. But in this question, it won’t be at all because we’ve already found our minors. And then, all we need to do is multiply these by each of the elements in the first row of our original matrix. But what makes this even easier, because this special type of matrix we’re looking at, is the fact that the second and third elements in our first row are both zero in our matrix of minors and our original matrix. So therefore, we can disregard these.
So, therefore our determinant is just gonna be equal to negative five multiplied by 25, which is equal to negative 125. So therefore, the inverse of the matrix is gonna be negative one over 125 because that’s one over the determinant of our matrix multiplied by the matrix 25, zero, zero, zero, 25, zero, zero, zero, 25, which could also be written as negative one-fifth, zero, zero, zero, negative one-fifth, zero, zero, zero, negative one-fifth.
And what we can notice by this is, in fact, all we had to do to create this matrix was look at negative five and think, “Well, what are we gonna have to multiply negative five by to get one?” Well, negative five multiplied by negative one over five gives us one. So therefore, this would’ve given us our inverse or multiplicative inverse matrix. So, this could’ve been a shortcut to the same result.