If 𝐴 is a non-singular square
matrix, is it correct that the dot product of the inverse of 𝐴 with the inverse of
𝐴 equals the identity matrix?
Before we answer this question, let
us begin by defining a few of the key terms.
First, let’s recall what we mean by
the word “non-singular.” A matrix is said to be non-singular
if it is invertible. In other words, it’s non-singular
if an inverse exists. Now because of the way the inverse
is calculated, a matrix is only invertible if its square and its determinant is
Next, we recall what we mean by the
identity matrix. That’s a square matrix whose
elements are zero with the exception of the leading diagonal, where the elements are
equal to one. For instance, the two-by-two
identity matrix is the matrix whose elements are one, zero, zero, one as shown.
So with this in mind, we now need
to establish whether if we have a non-singular square matrix 𝐴, the dot product of
the inverse of 𝐴 with itself is equal to the identity matrix. So we could attempt to prove
whether this is true for all possible non-singular square matrices. Before we do though, let’s look at
a simple two-by-two matrix 𝐴 equals one, two, three, four. We’re going to find the inverse of
this matrix 𝐴 and then find the dot product of the inverse of 𝐴 with itself and
see if it’s true for this.
To do so, we begin by finding the
determinant of 𝐴. And to find the determinant of 𝐴,
we multiply the top-left by the bottom-right element, and we subtract the product of
the top-right and bottom-left. So the determinant of 𝐴 is one
times four minus two times three. And that’s equal to negative
Now to find the inverse of 𝐴, we
multiply one over the determinant of 𝐴 by some two-by-two matrix. So we’re going to multiply one over
negative two by that two-by-two matrix. That is of course equivalent to
multiplying it by negative one-half. Then the two-by-two matrix that we
multiply this by is found by swapping the elements in the top left and bottom right
and changing the sign of the other two. So the inverse of 𝐴 is negative
one-half multiplied by the matrix with elements four, negative two, negative three,
We can then multiply each element
inside the matrix by negative one-half to find the inverse of 𝐴 to be the matrix
negative two, one, three over two, negative one-half. Now that we have the inverse of 𝐴,
let’s find the dot product of the inverse of 𝐴 with itself.
Let’s begin by finding the element
in the first row and the first column. To find this element, we find the
dot product of negative two, one with the common factor negative two, three over
two. And we might recall that to find
the dot product of this pair of matrices, we multiply negative two by negative two
and then add one multiplied by three over two. That’s four plus three over two,
which is equivalent to 11 over two.
And in fact, we don’t need to
continue finding the dot product of the inverse of 𝐴 with itself. We earlier looked at the two-by-two
identity matrix. And we saw that the first element
in the first row and first column should be one. Ours is 11 over two, which is quite
obviously not one. And so we know that this matrix
cannot be the identity matrix.
And so since it’s not true for the
matrix we defined to be equal to 𝐴, we cannot say that this is true for all
matrices. And so the answer is no. It is not correct that the dot
product of the inverse of 𝐴 with itself equals the identity matrix.