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Section 2.2
Matrix Inverse
In its most basic form a matrix A has an inverse if there is a matrix B such that
AB  BA  I
and if this matrix B exists at all then we label it
B  A −1
Theorem 4 hints at a future method to determine if a matrix has an inverse or not.
There is a function called the determinant that you apply it to a square matrix and the
function spits out a number. If the collection of matrices are matrices with only real
numbers then the output number is a real numbers and if we allow complex numbers
into our matrix then the output number is a complex number.
det : Collection of Square Matrices → Either  or ℂ
For a 2x2 matrix
a b
det
c d
5 3
det
2 4
 ad − cb
 5∗4−2∗3
 20 − 6
 14
det
2  3i i
2i
4
 2  3i4 − 2i ∗ i
 8  12i − 2i 2
 10  12i
Important Fact 1
1
A has an inverse or is called invertible if detA ≠ 0
Important Fact 2
If A is a n x n matrix and A is invertible then for each b in the range of linear
transformation T there is a unique solution to
Ax  b
A −1 Ax  A −1 b
x  A −1 b
If in the back of your mind you are thinking doesn’t this mean that the transformation T
is one - to - one they you would be correct.
Example 3, page 104
This is a very interesting example for the Civil Engineers and the Material Scientists in
terms of the "stiffness matrix" and the deflection vector.
Example 2, page 109
A
3 2
8 5
detA  35 − 82
 15 − 16
 −1
Since the determinant is not zero matrix A does have an inverse.
2
1
detA
A −1 
 −1 ∗

5
−2
−8
3
5
−2
−8
3
−5
2
8
−3
and check
A −1 A 
−5
2
3 2
8
−3
8 5
1 0

0 1
 I2
Example 6, page 109
Use the inverse to solve
7x 1  3x 2  −9
− 6x 1 − 3x 2  4
A
7
3
−6 −3
, b
−9
4
Find the inverse
A −1 
1
1
−2 − 73
Get the solution
3
x  A −1 b


1
−9
1
−2 −
−5
26
3
7
3
4

x1
x2
Example 12, page 109
Use matrix algebra to show that if A is invertible and D satisfies
AD  I
then
D  A −1
Proof
AD  I
A −1 AD  A −1 I
A −1 AD  A −1
ID  A −1
D  A −1
Example 18, page 110
Solve the matrix equation AB  BC for A assuming that A, B,and C are all invertible.
Solution
AB  BC
ABB −1  BCB −1
A  BCB −1
4