It’s Professor Dave, let’s look at inverse matrices.
We know about inverse functions, and the concept of an inverse can apply to a matrix as well.
With a function, we know how to swap X and Y and then solve for Y to find the inverse
function.
So how does this work for matrices?
Certainly it’s a different algorithm, so let’s learn about this now.
Although a matrix is different from a function, when denoting the inverse of a matrix, we
will use similar notation as we did for functions, where the inverse of matrix A will be A with
a negative one in superscript, just like a function F and its inverse.
We must realize that this is not an exponent, A inverse does not mean the reciprocal of
A, or one over A, because there is no such thing as matrix division.
But an inverse matrix is similar to the reciprocal of a number in another way.
We can recall that a number times its reciprocal equals one.
Similarly, a matrix times its inverse, or the inverse times the original matrix, will
yield an identity matrix, a square matrix with all zeroes except ones across its main
diagonal, which is kind of like the number one in matrix form.
So in that sense, there is a correlation.
Now, we know how to perform matrix multiplication, so how exactly can we structure an inverse
matrix so as to get an identity matrix when multiplying by the original matrix?
Let’s start with a simple two by two matrix with entries A, B, C, and D. The inverse of
this matrix will be one over the determinant of this matrix, times this new matrix, with
entries D, negative B, negative C, A. So A and D have swapped positions, B and C have
gone from positive to negative, and the resulting matrix is being divided by the determinant
of the original matrix.
To see why this is the case, let’s multiply this inverse, in this general form, by the
original matrix.
First, let’s just generate the two by two matrix that is the product of these two, without
worrying about this determinant.
For the first entry, we get DA plus negative BC, or keeping everything in alphabetical
order and simplifying the sign, AD minus BC.
For the second entry, we get BD minus BD.
Then moving down a row, we get negative AC plus AC, and for the last entry, negative
BC plus AD, or AD minus BC.
The upper right and lower left terms clearly become zero, leaving us with this.
Now we can divide by the determinant, like we originally said, meaning each entry in
the matrix will be divided by AD minus BC.
AD minus BC divided by AD minus BC is one, and the zeroes don’t change, so we are left
with this two by two identity matrix, just like we expected.
So that’s how you get the inverse of a two by two matrix.
You swap A and D, change the sign on B and C, and divide by the determinant of the original matrix.
Let’s try a concrete example to make sure we have this.
How about a matrix with entries four, three, three, two.
To get the inverse we find one over the determinant, and the determinant is eight minus nine, so
here we have one over negative one, and after that comes the two by two matrix with these
two entries swapped, and the signs of these two inverted.
That gives us two, negative three, negative three, four.
Incorporating this negative one out here is easy, that just gives us negative two, three,
three, negative four.
And that is the inverse of this original matrix.
Just to make sure this worked, let’s multiply them together and see what we get.
We do get the identity matrix, so we know our answer was correct.
So what do we need inverse matrices for?
Well as we said earlier, there is no such thing as matrix division, so we can use this
technique to solve equations with matrices where division would be necessary from an
algebraic standpoint.
If we have matrices X, A, and B, where X times A equals B, and we want to solve for X, we
can’t divide both sides by matrix A. Instead, we can multiply both sides by A inverse.
On the left side of the equation, A times A inverse will give us the identity matrix,
which we can essentially treat like the number one, and just get rid of it.
And on the right side of the equation we have a product of two matrices that we can compute,
which will be equal to X.
Of course, this assumes that the dimensions of these matrices are such that the product
can be calculated, otherwise this will not work.
Don’t forget that matrix multiplication is not commutative, so wherever we place A
inverse on one side, we have to place it the same way on the other side, so in this case,
as the rightmost entry on each side.
If the original equation was A times X equals B, we would have to place A inverse all the
way on the left, because A inverse times A should also equal the identity matrix, in
which case the right side must be A inverse times B, and not B times A inverse.
In addition, when taking the inverse of a product of matrices, like the quantity B times
A, inverse, we get A inverse times B inverse, in that order.
We should also note that not every matrix will have an inverse matrix.
If a matrix has a determinant that is equal to zero, its inverse will involve dividing
by zero, which is undefined, and that matrix will therefore not have an inverse.
Such a matrix is called a singular matrix.
Lastly, you may be wondering how we can find the inverse of any matrix other than a two
by two matrix.
Let’s look at a three by three matrix and see how this would work.
To find the inverse of this, we will first find something called the matrix of minors.
This will be a new three by three matrix, and each entry will be a particular determinant.
For the first entry, we locate the corresponding entry in the original matrix, and we block
out its row and column.
There is a two by two array of numbers left, and we find its determinant.
That determinant will become this first entry in the matrix of minors.
Then for the second entry, again we go to the corresponding entry in the original matrix,
block out its row and column, and find the determinant of what is left.
We continue doing this for every single entry until we have completed our matrix of minors.
Unfortunately, that was just step one.
Now for step two, we use our matrix of minors to generate a matrix of cofactors.
This isn’t too hard, we just have to apply a checkerboard of plus and minuses to get
the right sign on each term.
The corner entries and the central entry will remain as they are, and the other four entries
will have their signs inverted.
Then, step three is to find the adjugate, or the adjoint.
This involves reflecting all the entries across the diagonal.
The diagonal stays as it is, but the other terms are reflected, so these two swap, these
two corners swap, and these two swap.
Finally, the last step involves dividing by the determinant of the whole original three
by three matrix, which won’t seem so hard after doing that whole business with the matrix
of minors.
We just have this first entry times the determinant of this two by two matrix, minus the second
entry times the determinant of this two by two matrix, plus the third entry times the
determinant of this two by two matrix.
Whatever we get, we divide the last matrix we had by this value, meaning we divide every
entry in the matrix by that value, and that’s what we are left with for the inverse of the
original three by three matrix.
So to summarize the four steps we just took, first we found the matrix of minors, going
one entry at a time.
Then we applied the checkerboard pattern of signs to find the matrix of cofactors.
Then we transposed all entries across the diagonal to get the adjugate.
And lastly, we divided by the determinant of the original matrix to get the inverse
matrix.
This is essentially the same algorithm as we used for a two by two matrix, just many
more steps to achieve the same result.
So as we can see, even with just a three by three matrix, finding the inverse becomes
rather laborious.
If we were to extend this to a four by four matrix, it wouldn’t really be any harder,
we would just end up with an absurd amount of steps to follow.
For this reason, inverses of larger matrices are usually found using some kind of matrix
calculator, to avoid doing pages of arithmetic.
So that’s what we need to know about inverse matrices.
Now that we have covered the basics about matrices, their operations, determinants,
and inverses, we have many of the important algorithms out of the way, and we are ready
to examine some of the more abstract ideas in linear algebra.
But before we move forward, let’s check comprehension.