Lecture 8: Properties of Inverse Matrices
Not all matrices have inverse matrices! The following 2x2 matrix does not have
an inverse because one cannot find the first column of the inverse Ab1 = e1.
1 1 
1 
A=
and e1 =  

2 2
0
 1 1   b11  1 
 2 2  b  =  0 

  21   
Or , 1b11 + 1b21 = 1 and
2b11 + 2b21 = 0.
Then b11 = −b21 , and the first equation gives
1(−b21 ) + 1b21 = 1
0 = 1, a contradiction !
If A does have an inverse, then it can be used to solve Ax = d where x = A-1d.
The proof of this is very simple and makes use of the associative property of a matrix
product and the definition of an inverse matrix:
A(A-1d) = (A A-1)d = I d = d.
Some other properties are given in Proposition 4.
Proposition 4. Let A, A1 and A2 be nxn matrices.
1. If A-1 = B, then A(col k of B) = ek.
2. If A has an inverse matrix, then there is only one inverse matrix.
3. If A1 and A2 have inverses, then A1A2 has an inverse and (A1A2)-1= A2-1 A1-1.
4. If A has an inverse, then x = A-1d is the solution of Ax = d and this is the only
solution.
5. The following are equivalent:
(i).
A has an inverse.
(ii).
det(A) is not zero.
(iii).
Ax = 0 implies x = 0.
Let B1 and B2 be inverses of A so that AB1 = I and AB2 = I.
Proof of 2.
Subtract these two equations to get
AB1 – AB2 = I – I = 0
A(B1 – B2) = 0
A-1(A(B1 – B2)) = A-1 0
(A-1A)(B1 – B2) = 0
(I)(B1 – B2) = 0
(B1 – B2) = 0
So, B1 = B2.
Let A1 and A2 have inverses.
Proof of 3.
We must show (A1A2)-1(A1A2) = I and (A1A2)(A1A2)-1 = I.
(A1A2)-1(A1A2) = (A2-1A1-1)(A1A2)
= A2-1(A1-1(A1A2) )
= A2-1((A1-1A1)A2)
= A2-1((I)A2)
= A2-1A2
=I
Proof of 5. This will have to wait for a more detailed course on matrix algebra.
Example of Property 3 in Proposition 4. Find the inverse matrix of
 4 0 0  4 0 0  1 0 0
A =  −4 2 0  =  0 2 0   −2 1 0  = A1 A2
 0 0 1   0 0 1   0 0 1 
A1 is a diagonal matrix with an inverse
1/ 4 0 0 
A =  0 1/ 2 0  .
 0
0 1 
−1
1
A2 is an elementary matrix with an inverse
1 0 0
A =  2 1 0  .
 0 0 1 
 1 0 0  1/ 4 0 0   1/ 4 0 0 
−1
−1 −1
So, A = A2 A1 =  2 1 0   0 1/ 2 0  =  2 / 4 1/ 2 0  .
 0 0 1   0
0 1   0
0 1 
−1
2
Application of the Two-loop Circuit.
1   i1   0 
 1 −1
R
0 − R3  i2  =  E1  .
 1
 0 − R2 − R3   i3   E2 
Let R1 = 1, R2 = 2, R3 = 3, E1 = 10 and E2 = 20.
Ax = d
1 −1 1   x1   0 
1 0 −3  x  = 10  .

 2  
 0 −2 −3  x3   20 
−1
 x1  1 −1 1   0 
 x  = 1 0 −3 10 
 2 
  
 x3  0 −2 −3  20 
 6 /11 5 /11 −3 /11  0 
=  −3 /11 3 /11 −4 /11 10 
 2 /11 −2 /11 −1/11  20 
 −10 /11
=  −50 /11 .
 −40 /11
Block Gauss Elimination and Inverse Matrices. Often an nxn matrix A is broken into
blocks of matrices. This is done either to reflect substructures of a model or to
accommodate memory hierarchy of a computer. Consider the following 2x2 block
representation where A11 is n1xn1, A22 is n2xn2 and n = n1 + n2.
A
A =  11
 A21
A12 
and
A22 
 A11
A
 21
A12   x1   d1 
=
A22   x2   d 2 
If A11 has an inverse, then we can use a block elementary matrix to zero the 21-block.
0   A11 A12   x1   I1
 I1
=
 − A A−1 I   A
A22   x2   − A21 A11−1
 21 11
2   21
 A11 A12   x1   d1 
where
=

ˆ   x2   dˆ 
0
A

22 
 2
Aˆ = A − A A−1 A
22
22
21 11
0   d1 
I 2   d 2 
12
−1
21 11 1
dˆ2 = d 2 − A A d
If both A11and Aˆ 22 = A22 − A21 A11−1 A12 have inverses,
the block backward substitution step can be executed
−1
x = Aˆ dˆ
2
22
2
−1
11
x1 = A (d1 − A12 x2 ).
Examples of this can be found in the Matlab demo blockg_el.m
Homework.
1. In the proof of part 3 in Proposition 4 prove the other equality.
2. In part 4 of Proposition 4 prove there is only one solution.
3. In part 5 in Proposition prove (i) implies (iii).
4. Use the inverse matrix to solve
 1 3 5   x1   3 
0 2 1  x  = 4 .

 2  
 2 2 0   x3   5 
5. Use property 3 to find the inverse of
1 0 0  1 0 0   1 0 0 
0 2 0 = 0 2 0  0 1 0 .

 


 6 0 3  0 0 3   2 0 1 