27.2. APPLICATIONS 507

Example 27.2.3 Find the inverse of the matrix

A =



12 0 1

2

− 16

13 − 1

2

− 56

23 − 1

2

First find its determinant. This determinant is 1

6 . The inverse is therefore equal to

6



∣∣∣∣∣ 1/3 −1/22/3 −1/2

∣∣∣∣∣ −

∣∣∣∣∣ −1/6 −1/2−5/6 −1/2

∣∣∣∣∣∣∣∣∣∣ −1/6 1/3

−5/6 2/3

∣∣∣∣∣−

∣∣∣∣∣ 0 1/22/3 −1/2

∣∣∣∣∣∣∣∣∣∣ 1/2 1/2−5/6 −1/2

∣∣∣∣∣ −

∣∣∣∣∣ 1/2 0−5/6 2/3

∣∣∣∣∣∣∣∣∣∣ 0 1/21/3 −1/2

∣∣∣∣∣ −

∣∣∣∣∣ 1/2 1/2−1/6 −1/2

∣∣∣∣∣∣∣∣∣∣ 1/2 0−1/6 1/3

∣∣∣∣∣



T

.

Expanding all the 2×2 determinants this yields

6

 1/6 1/3 1/61/3 1/6 −1/3−1/6 1/6 1/6

T

=

 1 2 −12 1 11 −2 1

Always check your work. 1 2 −1

2 1 11 −2 1

 1/2 0 1/2−1/6 1/3 −1/2−5/6 2/3 −1/2

=

 1 0 00 1 00 0 1

and so we got it right. If the result of multiplying these matrices had been something otherthan the identity matrix, you would know there was an error. When this happens, youneed to search for the mistake if you are interested in getting the right answer. A commonmistake is to forget to take the transpose of the cofactor matrix.

Proof of Theorem 27.2.1: From the definition of the determinant in terms of expansionalong a column, and letting (air) = A, if det(A) ̸= 0,

n

∑i=1

air cof(A)ir det(A)−1 = det(A)det(A)−1 = 1.

Now considern

∑i=1

air cof(A)ik det(A)−1

when k ̸= r. Replace the kth column with the rth column to obtain a matrix Bk whosedeterminant equals zero by Theorem 27.1.21. However, expanding this matrix Bk along thekth column yields

0 = det(Bk)det(A)−1 =n

∑i=1

air cof(A)ik det(A)−1