Video Transcript
True or False: If 𝐴 is any given
square matrix of order 𝑛, then 𝐴 multiplied by the adjoint matrix of 𝐴 is equal
to the adjoint of 𝐴 multiplied by 𝐴 is equal to the determinant of 𝐴 multiplied
by 𝐼, where 𝐼 is the identity matrix of order 𝑛.
We’ll begin by noting that if the
matrix is singular, that is, the determinant of 𝐴 is equal to zero, then the
inverse of 𝐴 does not exist. And we’ll come back to this case a
little later on. If, on the other hand, an 𝑛-by-𝑛
matrix is nonsingular, that means its inverse does exist, then to show whether the
given statement is true or false, we can use the theorem that for a nonsingular
𝑛-by-𝑛 matrix 𝐴, 𝐴 inverse — that’s 𝐴 negative one — is the adjoint of 𝐴
divided by the determinant of 𝐴. And recall that the adjoint of
matrix 𝐴 is the transpose of the matrix of its cofactors.
So let’s assume that the inverse of
our matrix exists. This then means that the
determinant is nonzero. To show whether this statement is
true or false, let’s take our definition of the inverse of 𝐴 and multiply through
on the left by the matrix 𝐴. Now we know that for any
nonsingular 𝑛-by-𝑛 matrix, 𝐴 multiplied by its inverse is the identity
matrix. So now on the left-hand side, we
have the 𝑛-by-𝑛 identity matrix. And now since the determinant of
the matrix 𝐴 is a scalar, we can multiply through by this. And this gives us the determinant
of 𝐴 multiplied by the identity matrix is 𝐴 multiplied by its adjoint. And this shows that our first
statement 𝐴 adjoint 𝐴 is indeed equal to the determinant of 𝐴 multiplied by 𝐼
for a nonsingular matrix 𝐴.
Now, if we begin again with our
formula for the inverse of 𝐴 and multiply through this time on the right by our
matrix 𝐴, we again have the identity matrix on the left since 𝐴 inverse 𝐴 is the
identity matrix. And again multiplying through by
the determinant of 𝐴, which we can do since it’s a scalar, we have the determinant
of 𝐴 multiplied by the identity matrix is equal to the adjoint of 𝐴 multiplied by
𝐴. And so we can say that our second
expression is also true for a nonsingular matrix. The whole of the given expression
is therefore true for a nonsingular matrix 𝐴.
We still don’t know whether the
given statement is true or false if our matrix 𝐴 is singular. That’s when the determinant is
equal to zero. So writing out our statement again,
using slightly different notation for the determinant just for clarity, and now
suppose we take the determinant of each of our expressions, we can use the theorem
that for 𝑛-by-𝑛 matrices 𝐴 and 𝐵, the determinant of the product 𝐴𝐵 is equal
to the determinant of matrix 𝐴 multiplied by determinant of matrix 𝐵. Our first expression is then the
determinant of 𝐴 multiplied by the determinant of the adjoint of 𝐴. And similarly, for our second
expression, and on the right we leave our expression as it is.
Now, if 𝐴 is a singular matrix,
then the inverse doesn’t exist and the determinant is equal to zero. In our first two expressions then,
we’re multiplying by zero. And anything multiplied by zero is
equal to zero. On our right-hand side, we’re
multiplying the identity matrix by zero, which gives us the zero matrix. And the determinant of the zero
matrix is zero. And so the left-hand, the middle
term, and the right-hand term are all equal to zero. This means then that for a singular
matrix where the determinant is equal to zero, then the given statement is also
true. Hence, 𝐴 multiplied by adjoint 𝐴
is equal to adjoint 𝐴 multiplied by 𝐴 is equal to the determinant of 𝐴 multiplied
by 𝐼 is true for any square matrix 𝐴 of order 𝑛.
