If A is an invertible matrix. Let B be the inverse of A. Then by the properties in this section, we have
detA detB = detAB = detI = 1.
In particular, we just proved the ⇒ direction of the following property.
To prove the ⇐ direction, we need to use the assumption detA ≠ 0 to construct an inverse of A. We illustrate the idea by a 4 by 4 matrix.
A = [  a_{11}  a_{12}  a_{13}  a_{14}  ]. 
a_{21}  a_{22}  a_{23}  a_{24}  
a_{31}  a_{32}  a_{33}  a_{34}  
a_{41}  a_{42}  a_{43}  a_{44} 
Recall the cofactor expansion along the first row
a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} + a_{14}C_{14} = det[  a_{11}  a_{12}  a_{13}  a_{14}  ]. 
a_{21}  a_{22}  a_{23}  a_{24}  
a_{31}  a_{32}  a_{33}  a_{34}  
a_{41}  a_{42}  a_{43}  a_{44} 
If we replace the terms a_{11}, a_{12}, a_{13}, a_{14} in the formula by a_{21}, a_{22}, a_{23}, a_{24}, then we have
a_{21}C_{11} + a_{22}C_{12} + a_{23}C_{13} + a_{24}C_{14} = det[  a_{21}  a_{22}  a_{23}  a_{24}  ]. 
a_{21}  a_{22}  a_{23}  a_{24}  
a_{31}  a_{32}  a_{33}  a_{34}  
a_{41}  a_{42}  a_{43}  a_{44} 
The determinant vanishes by the following result.
A has a zero row (column) ⇒ detA = 0.
A row (column) in A is a scalar multiple of another row (column) ⇒ detA = 0.
Proof If A has a zero row, then by cofactor expansion along the zero row, we get detA = 0. If [row i] is r multiple of [row j], then apply (r)[row i] + [row j] to A to get B. By this property, we have detA = detB. Moreover, the jth row of B is zero, so that detB = 0.
The argument we had clearly generalizes to the following result.
a_{i1}C_{j1} + a_{i2}C_{j2} + a_{i3}C_{j3} + a_{i4}C_{j4} = {  detA  i = j 
0  i ≠ j 
In other words, if we define the adjugate
adjA = [  C_{11}  C_{21}  C_{31}  C_{41}  ], 
C_{12}  C_{22}  C_{32}  C_{42}  
C_{13}  C_{23}  C_{33}  C_{43}  
C_{14}  C_{24}  C_{34}  C_{44} 
then we have
A adjA = (detA)I.
Since A is a square matrix, by this property, we conclude
We had the 2 by 2 case of this result before.
Example In an earlier example, we used row operations to find the inverse of
A = [  3  1  1  ] 
1  1  1  
2  2  1 
Now we use our new formula. First,
detA = det[  3  4  4  ] 


1  0  0  
2  4  1 
Then
C_{11} = (1)^{1+1}det[  1  1  ] = 3, C_{21} = (1)^{2+1}det[  1  1  ] = 3, C_{31} = (1)^{3+1}det[  1  1  ] = 0, 
2  1  2  1  1  1  
C_{12} = (1)^{1+2}det[  1  1  ] = 1, C_{22} = (1)^{2+2}det[  3  1  ] = 5, C_{32} = (1)^{3+2}det[  3  1  ] = 4, 
2  1  2  1  1  1  
C_{13} = (1)^{1+3}det[  1  1  ] = 4, C_{23} = (1)^{2+3}det[  3  1  ] = 4, C_{33} = (1)^{3+3}det[  3  1  ] = 4, 
2  2  2  2  1  1 
and
A^{1} = (1/12)[  3  3  0  ] = [  1/4  1/4  0  ]. 
1  5  4  1/12  5/12  1/3  
4  4  4  1/3  1/3  1/3 
One find the result to be the same as the earlier computation. However, the computation here is much more complicated. It is even more complicated if you try the inverse of matrices of bigger size. Therefore row operation is still the practical way of computing the inverse.