Transcript Chapter 2 Section 4

Chapter 2 Section 4
The Inverse Matrix
Problem
• Find X in the a matrix equation:
AX=B
(recall that A is the matrix of the coefficients, X is the matrix of the
variables, and B is the matrix of the constants).
• There is no division when dealing with
matrices!
• We will use the inverse matrix!
Definition
• Given a matrix A , then A – 1 is the inverse
matrix of matrix A if and only if:
A  A–1 = A–1  A = I
(where I is the identity matrix)
Restriction / Caution
• Restriction:
– A matrix has to be a square matrix in order to
have an inverse!!!
• Caution:
– Not all square matrices have inverses!!!
How to Find an Inverse Matrix
• We will let the calculator do all the work for
us.
• Enter the matrix using the matrix editor, and
then exit to the home screen.
• Call up the matrix that you want the inverse
for, then hit the [ x-1 ] key.
• The resulting matrix (if there is one) is the
inverse matrix
Example of an Inverse Matrix
A=
4
8
7
A– 1 =
–2
–3
–2
3
5
4
4 –2
8 –3
7 –2
Then the inverse of A is:
3
5
4
–1
=
–2 2 –1
3 –5 4
5 –6 4
Example of a Square Matrix that Does
Not have an Inverse
A=
1
0
1
3
1
5
2
4
10
does not have an inverse!
The following message will appear on your calculator if you
try to take the inverse of the above matrix: [A] – 1 .
ERR: SINGULAR MAT
1:Quit
2:Goto
WARNING
• The inverse matrix will solve a matrix
equation ONLY when there is a unique
solution to the system of equations.
• If there is a possibility that there is either an
infinite number of solutions or no solution,
then use the Gauss-Jordan elimination
method (rref) as described in sections 1 and
2.
Exercise 21 (page 98)
Use the fact that:
9
0
2
– 20 – 9 – 5
4
0
1
–4 –2 –1
0
5
0
1
–1
Solve:
9x + 2z = 1
– 20x – 9y – 5z + 5w = 0
4x + z = 0
– 4x – 2y – z + w = – 1
Using a matrix equation!
=
1
0
–4
0
0
1
0
2
–2
0
0 –5
9
0
1 –9
Exercise 21 continued (a)
9x + 2z = 1
– 20x – 9y – 5z + 5w = 0
4x + z = 0
– 4x – 2y – z + w = – 1
9
– 20
4
–4
0
2
–9 –5
0
1
–2 –1
0
5
0
1
x
y
z
w
=
1
0
0
–1
Exercise 21 continued (b)
x
y
z
w
x
y
z
w
=
9
– 20
4
–4
0
–9
0
–2
=
1
0
–4
0
0
1
0
2
2
–5
1
–1
0
5
0
1
–2 0
0 –5
9
0
1 –9
–1
1
0
0
–1
1
0
0
–1
Exercise 21 continued (c)
x
y
z
w
=
1
5
–4
9
Thus x = 1, y = 5, z = – 4, and w = 9
Exercise 31 (page 98)
Using a matrix equation, solve:
2x – 4z + 7z = 11
x + 3y – 5z = – 9
3x – y + 3z = 7
2
1
3
–4
3
–1
A
7
–5
3
x
y
z
•
X
=
=
11
–9
7
B
Exercise 31continue (a)
So
A·X=B
X = A– 1 · B
=
=
–1
2 –4 7
1 3 –5
3 –1 3
– 0.8
5.6
5
·
=
So x = – 4/5, y = 28/5, and z = 5
11
–9
7
– 4/5
28/5
5
A Repeat of the WARNING
• If the system of equations has the either an
infinite number of solutions or no solutions,
then there will not be an inverse matrix ( i.e.
The ERR: SINGULAR MAT error in the
calculator). Unfortunately the error message
will not tell you which of the two situations
caused the error message; for that you will
have to use the Gauss-Jordan elimination
method to make the determination.