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 𝑥, 𝑝, 𝑞, zero, 𝑦, 𝑟, zero,
zero, 𝑧, where 𝑥, 𝑦, 𝑧, 𝑝, 𝑞, and 𝑟 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 𝑎, 𝑏, 𝑐, 𝑑, then this would be equal to 𝑎𝑑 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, “Well, 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 — I’ll just quickly clear some space
in the bottom right here to do this. Well, if we have the matrix 𝑎, 𝑏,
𝑐, 𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖, then the determinant of this matrix is equal to 𝑎
multiplied by the minor which is 𝑒, 𝑓, ℎ, 𝑖 — or the two-by-two submatrix 𝑒, 𝑓,
ℎ, 𝑖 — 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, ℎ.
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 𝑎, 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
𝑥, 𝑝, 𝑞, 𝑦, 𝑟, 𝑧 are a half, negative one-tenth, negative 17 over 100, a
fifth, negative three over 50, and one over 10, respectively.