A First Course in Linear Algebra, 2017a

3.2. Applications of the Determinant 133
This tells us that our calculation for A −1 is correct. It is left to the reader to verify that AA −1 = I.
♠
The verification step is very important, as it is a simple way to check your work! If you multiply A −1 A
and AA −1 and these products are not both equal to I, be sure to go back and double check each step. One
common error is to forget to take the transpose of the cofactor matrix, so be sure to complete this step.
We will now prove Theorem 3.40.
Proof. (of Theorem 3.40) Recall that the (i, j)-entry of adj(A) is equal to cof(A) ji . Thus the (i, j)-entry of
B = A · adj(A) is :
n
n
B ij = ∑ a ik adj(A) kj = ∑ a ik cof(A) jk
k=1
k=1
By the cofactor expansion theorem, we see that this expression for B ij is equal to the determinant of the
matrix obtained from A by replacing its jth row by a i1 ,a i2 ,...a in — i.e., its ith row.
If i = j then this matrix is A itself and therefore B ii = detA. If on the other hand i ≠ j, then this matrix
has its ith row equal to its jth row, and therefore B ij = 0 in his case. Thus we obtain:
A adj(A)=det(A)I
Similarly we can verify that:
adj(A)A = det(A)I
And this proves the first part of the theorem.
Further if A is invertible, then by Theorem 3.24 we have:
1 = det(I)=det ( AA −1) = det(A)det ( A −1)
and thus det(A) ≠ 0. Equivalently, if det(A)=0, then A is not invertible.
Finally if det(A) ≠ 0, then the above formula shows that A is invertible and that:
A −1 = 1
det(A) adj(A)
This completes the proof.
♠
This method for finding the inverse of A is useful in many contexts. In particular, it is useful with
complicated matrices where the entries are functions, rather than numbers.
Consider the following example.
Example 3.43: Inverse for Non-Constant Matrix
Suppose
⎡
A(t)= ⎣
Show that A(t) −1 exists and then find it.
e t 0 0
0 cost sint
0 −sint cost
⎤
⎦
Solution. First note det(A(t)) = e t (cos 2 t + sin 2 t)=e t ≠ 0soA(t) −1 exists.