MAT 2401 Discovery Lab 2.3 Names___________________________________
Objectives
 To introduce inverse matrices.
 To explore the idea of solving matrix equations by inverses.
 To explore how to find inverse matrices by formula and row operations.
 Use SageMath to facilitate the calculations for large system.
Instructions
 Do not look up any references including the textbook and internet.
 Do not use a calculator. You are supposed to learn SageMath.
 Use correct notations and do not skip steps.
 Two persons per group. Do not communicate with other groups.
Matrix Equations
In §2.1, we see that a system of linear equations such as
 x  2 y  3z  9

 4
 x  3 y
 2 x  5 y  5 z  17

can be encoded into the matrix equation
AX  b
where
 1 2 3
 x
9




A   1 3 0 , X   y  , b   4 .
 2 5 5
 z 
17 
If we can solve the matrix equation for X , we have the solutions
system of linear equations.
(1)
(2)
 x, y, z 
for the
Idea
To solve the matrix equation (2), we will attempt to find a matrix B such that BA  I .
We can then solve for X because
AX  b
B  AX   Bb
 BA X  Bb
IX  Bb
X  Bb
(3)
B belongs to a type of special matrices called inverse matrices which are described
below.
1
Inverse Matrix
Let A be a square matrix. An inverse for A is a square matrix, denoted by A1 , of the
same size as A such that
AA1  A1 A  I .
If such an inverse exists, A is called invertible or non-singular.
 2 1
 1 1
1. Let A  
. Verify that B  

 is an inverse of A by computing AB
1 1
 1 2 
and BA .
 2 1  1 1
AB  


1 1  1 2 
and
 1 1  2 1
BA  


 1 2  1 1
Formula
a b 
For 2  2 matrices A  
 , there is a convenient formula to compute its inverse:
c d 
1  d b 
A1 
.
(4)
ad  bc  c a 
 2 1
2. Use the inverse formula (4) to compute A1 for A  
.
1 1
A1 
You see that your answer is the same as the matrix B in problem 1.
2
2 x  y  5
3. Use an matrix inverse to solve the linear system 
.
 x y 3



The system is represented by AX  b , where A  



5 
b    . (You know you have computed A1 already, right?)
3



 x
 , X   y  , and
 



From (3), We know the solution is given by
X  A1b

So, x 
,y
.
3
Row Operations
For bigger matrices, there are no convenient formula to compute the inverses. It turns out
that we can use elementary row operations to find the inverse matrices.
 2 1
4. Let A  
 . We set up an augmented matrix:
1 1
2 1 1 0
1 1 0 1 


Note that the augmented matrix is of the form
the augmented matrix becomes
I
A
I  . Perform row operations so that
1 0 * *
*  
.
0 1 * *
2 1 1 0

1 1 0 1  


The row operations effectively multiplied the left side of the augmented matrix, which is
A , by A1 . And the end result is A1 A  I . As a result, the right side of the augmented
matrix, which is I , is also multiplied by A1 . So the end result is A1I  A1 .
A
row
I  
  I
operations
A1 
This solution method can be extended to bigger matrices. Let us try it out with the
following example.
4
 1 2 3
5. Let A   1 3 0  . Find A1 by using row operations.


 2 5 5 
(a) Set up the augmented matrix  A I  . Use row operations to transform it into
 I A1  . I am expecting you to be able to do this by hand. To save time, let us use
SageMath. We introduced the rref() commend in Lab 1.3. Below is a recap of the
syntax.
A=matrix([[...matrix.entries...],[...matrix.entries...],[...matrix.entries...]]);
show(A);
show(A.rref());
A
I 
A1 
5
 x  2 y  3z  9

(b) Use (a) to solve the system  x  3 y
 4 .
 2 x  5 y  5 z  17





The system is represented by AX  b , where A  




9
and b   4  . From (3), We know the solution is given by
 
17 




x
 , X   y ,

 

 z 



X  A1b

So, x 
, y
, z
.
Practice Problems (For you to do at home. Do not turn in!)
2 x  4 y  z  2
 2 y  2 z  12


1.   x  y  z  0
2.  3x  y  3z  2
 x  4y
 x  2 y  3z  13
1


Answers
1. x  3, y  1, z  4
2. x  0, y  5, z  1
6
Properties of Inverses
A 
(ii)  A 
1 1
(i)
k 1
(iii)
 cA
(iv)
A 
(v)
 AB 
T
A
  A1  , where k is a positive integer.
1
k

1 1
A , where c is a non-zero constant.
c
1
  A1 
1
 B 1 A1
T
Many of the properties above are easy to conceive. The only possible exception is the last
1
property  AB   B 1 A1 .
6. Show that B 1 A1 is the inverse of AB by computing
B
1
A1   AB  . Do not skip details.
 AB   B 1 A1 
and
 AB   B 1 A1   A  BB 1  A1
 A  I  A1
 AA1
I
B
1
A1   AB  
7