Download Inverses

Dr. Neal, WKU
MATH 307
Inverses
Let A be an n × n matrix. We say that A is invertible (or non-singular) if there is another
n × n matrix B such that AB = In = BA . If no such matrix B exists, then matrix A is
called non-invertible or singular.
If such an n × n matrix B does exist, then B is called “ A inverse” and is then
€
−1 €
−1
A
denoted
by
the
symbol
. Thus the defining property
€
€ of A is
€
€
€
€
€
€
€
AA−1 €
= In = A−1 A . €
€
Theorem 1.4. Let A be invertible. Then A−1 is invertible and (A−1 )−1 = A .
€
Proof. Because A−1 A = In = AA−1, the matrix A has the inverse property when acting on
−1
the matrix
of A−1 . That is, €A−1 is invertible and A = (A−1 )−1 .
€ A . So A is the inverse
€
€
€
 a€11 a12 
Let A = 

 a21 a22 
€
 a11

 a21
Inverse
€ of a 2 × 2 Matrix
€
€
€ 1 a
22 −a12 
be a 2 × 2 matrix with det( A ) ≠ 0. Then A−1 =

.
det(A)  −a21 a11 
€
€
Because
det(A) = a11a22 €
− a12 a21, it can easily be shown that
€
a12 
1  a22 −a12  1 0
1  a22 −a12   a11
×

=
=

 ×
a€
a11  0 1 det(A) −a21 a11   a21
22  det(A) −a21
a12 
;
a22 
thus, A−1 is as stated.
€
 4 −2
€ Example 1. Let A = 
 . Then det(A) = 4(−6) − (−2)(3) = −18 . Thus,
 3 −6
A−1 = −
€
€
€
1  −6 2  1/3 −1/9 

=

18  −3 4 € 1/6 −2 /9
and
 4 −2 1/3 −1/9  1 0

 ×
=
.
 3 −6 1/6 −2 /9 0 1
∗ Your calculator can easily compute the inverse of an n × n matrix,
or tell you if the€matrix is singular. ∗
Simply enter the matrix, say as [A] on a TI-84, or as ma on a TI-89. Then
€
TI-84: Enter [A]−1 then MATH 1.
€TI-89: Enter ma^(−1)
Use your TI to compute A−1 in the above example and to show A × A−1 = I2 .
€
€
€
€
Dr. Neal, WKU
Some More Theorems about Inverses and Determinants
We shall assume the following basic facts:
(i) det (In ) = 1. Because the n × n identity matrix In is diagonal, its determinant is the
product of the main diagonal which gives 1.
(ii) Any two n × n matrices can be multiplied together to give another n × n matrix.
€
(iii) For n × n matrices A and B , det (A B) = det ( A) × det (B) .
€
(iv) Matrix
multiplication is associative.
€
€
Theorem 1.5. Let A and B be n × n matrices.
−1
(a) If A is invertible, then A is unique.
(b) If A and B are invertible,
then A B is invertible and (AB)−1 = B −1 A−1 .
€
1
−1
−1
(c) If A exists, then det (A) ≠ 0 and det (A ) =
.
det (A)
−1
exists and A A −1 = In = A −1 A . We must show
−1
that A is the only matrix with this property. So suppose there is another n × n matrix
C such that AC = In = C A .
Proof. (a) Because A is invertible, A
−1
Then, C = C In = C(A A−1 ) = (C A)A−1 = In A−1 = A−1 . Thus, C = A , which means
−1
€
that A is unique.
€
−1
−1
−1 −1
(b) €Assume both A and B exist. We claim that B€ A behaves as the inverse of
−1 −1
AB . To demonstrate this, we multiply AB on both sides by B A to get In :
−1 −1
AB × B A
= A(B B−1 )A−1 = (A In )A−1 = A A −1 = In
and
€
€
B−1A −1 × AB = B−1 (A−1 A) B = (B−1 In ) B = B−1 B = In .
−1 −1
Thus, B A
−1 −1
is the inverse of AB . Therefore, A B is invertible and (AB)−1 = B A .
€
€
(c) We know that det (A) × det( A −1 ) = det(A A−1 ) = det (In ) = 1. Thus, neither det (A)
nor det (A −1 ) can be 0 because their product is non-zero. So, det (A) ≠ 0. Also because
det (A) × det( A −1 ) = 1, we have det (A −1 ) = 1 / det ( A) .
€
Dr. Neal, WKU
To find the inverse of an n × n matrix “by hand”, we actually have to solve n
separate systems of n equations and n unknowns. We will illustrate with a symbolic
3 × 3 matrix which requires that we solve 3 systems of 3 equations, 3 unknowns.
For a 3 × 3 matrix A , we need another 3 × 3 matrix B such that AB = I3 . Each
€
€
column of B (if it exists) has 3 unknowns x1 , y1 , z1 and we need 3 equations to solve for
€
€
them. We obtain the first system from the inner products of the three rows of A with
the first column of B which yields the first column of I3 :
€
€
€
 a11 a12 a13   x1 − − € €
 1 €0 0
a11 x1 + a12 y1 + a13 z1 = 1
€

 



→ a21 x1 + a22 y1 + a23 z1 = 0
 a21 a22 a23 ×  y1 − −  =  0 1 0€

 



a31 x1 + a32 y1 + a33 z1 = 0
 a31 a32 a33   z1 − − 
 0 0 1
€
A
×
B
=
I3
€
€
We obtain the second system from the inner products of the three rows of A with
the second column of B which yields the second column of I3 :
 a11

 a21

 a31
a12
a22
a32
a13   −
 
a23 ×  −
 
a33   −
A
×
1
x1 − 


y1 −  =  0


z1 − 
0
B
=
0 0

1 0

0 1
I3
€→
a11 x1 + a12 y1 + a13 z1 = 0
€
a21 x1 + a22 y1 + a23 z1 = 1
a31 x1 + a32 y1 + a33 z1 = 0
€
€
We obtain the third system from the inner products of the three rows of A with the
third column of B which yields the third column of I3 :
 a11

 a21

 a31
a12
a22
a32
a13   −
 
a23 ×  −
 
a33   −
A
×
1
− x1


− y1 =  0


− z1 
0
B
=
0
1
0
0

0
€ 
1
→
a11 x1 + a12 y1 + a13 z1 = 0
€
a21 x1 + a22 y1 + a23 z1 = 0
a31 x1 + a32 y1 + a33 z1 = 1
I3
€
€
• Each of these systems has a unique solution if and only if det( A ) ≠ 0 . •
The process is the same for n × n matrices, which allows us to state:
€
€
€
Lemma 1. Let A be an n × n matrix. If det( A ) ≠ 0, then there exists another n × n matrix
€
B such that AB = In .
€
€
But is B€actually the inverse of A ?
We still must show that BA = I n also.
€
Dr. Neal, WKU
However, suppose AB = In . Then 1 = det (In ) = det ( AB) = det (A) × det( B) ; thus,
det (B) ≠ 0 . Applying Lemma 1 to matrix B , there exists another n × n matrix C such
that BC = In . Then,
B = BI n = B( AB) = ( BA)B
€
€
Multiplying on the right by C , we obtain In = BC = (BA )BC = (BA) In = BA .
So if AB = In , then we also must have BA = I n . We now can state:
€
Theorem 1.6. Let A be an n × n matrix. If det( A ) ≠ 0, then A−1 exists.
But €
we also €
proved in Theorem€ 5(c) that if€ A−1 exists then det( A ) ≠ 0. So we
actually have:
−1
Theorem 1.7. Let A be an n × n matrix. Then
€ A exists if and only
€ if det( A ) ≠ 0.
€ 2

Example 2. Let A =  3

0
€
€
 2 1 3
1 3



−1 6 and B =  3 −1 6  .



 1 −2 3 
2 3
€
 5/7

Use your calculator to show that det (A) = −21 and A −1 =  3 / 7

 −2 / 7
−1 / 7
−2 / 7
4 / 21
−3 / 7

1/ 7 .

5 / 21
Use your calculator to show that det (B) = 0 and trying to compute its inverse
results in a singular matrix message.
Solving Matrix Equations with Inverses
We can use inverses to solve matrix algebra equations and also to solve a system of n
equations and n unknowns.
Example 3. (a) Suppose C is invertible and BC = D . Solve for B .
−1
€ (b) Suppose P A P = D . Solve for A .
−1
Solution. (a) Multiply
each side of
€
€ BC = D on the right by C :
−1
BIn = DC −1 →€ B = DC .
€
BCC −1 = DC −1 →
−1
−1
(b) Multiply each side of P €AP = D on the left by P €
and on€the right by P :
PP −1 APP −1 = PDP −1 → In A In = P D P −1 → A = PDP −1 .
€
Dr. Neal, WKU
Theorem 1.8. Let A be an invertible n × n matrix. Every system of equations AX = B has
−1
a unique solution given by X = A B .
−1
Proof. Because A exists,€we know that det (A) ≠ 0 . So AX = B has a unique solution
−1
−1
−1
for X . To obtain the solution, multiply on the left by A : A AX = A B →
−1
In X = A −1B → X = A B .
 2 −1 2 
 3


 
Example 4. Consider the system x + y − 2z = 6 . Let A =  1 1 −2 and B =  6 .


 
 −3 2 −1
 3
−3x + 2y − z = 3
2x − y + 2z = 3
(a) Enter these matrices
into your calculator and show that A
€
−1
 3
 
× B =  7 .
 
 2
(b) Adjust A to A B , then compute rref (A B) to obtain the same solution.
€