Solving Linear Equations using Matrices: Inverse and Invertibility - Prof. A. M. Sathaye, Study notes of Mathematics

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• Lecture continuation of 2.
  • Ma 162 Spring - Ma 162 Spring
    • February 11-

Matrices for solving equations.

Now we describe the main use of matrices for solving systems of linear equations. In this lecture, we would mainly consider systems where the number of equations equals the number of variables. A linear system of n equations in n variables can be descibes by a single matrix equation of the form AX = B. For example: The equations 2 x − 3 y = 1 , x − 2 y = 5 can be written as [ 2 − 3 1 − 2

] [ x y

]

[ 1 5

] .

The main idea of solution.

The solution, as before can be written as IX = C which reduces to X = C. Thus, the solution to the above system is x = 4 , y = 2 , z = −3 and can be written as  

 

 

x y z

  =

 

 .

Both the solutions above can be described by the following simple philosophy. Let the original equations be AX = B where we assume that A is a square n × n matrix, X is the column of n variables and B denotes the right hand sides. We find an n × n matrix M such that AM = MA = In.

Inverse Defined.

Multiplying both sides of the equation AX = B by M on the left, we get MAX = MB which becomes

IX = MB and yields the solution X = MB.

Thus, it would be good to have a mechanism for finding such a matrix M when possible. We define the inverse of a square matrix A to be a square matrix M such that MA = AM = I. The matrix M can be shown to be uniquely defined by A , when it exists and is called the inverse of A. The matrix A is said to be invertible (or non singular) if its inverse exists and it is said to be non invertible or singular otherwise.

Inverse Calculation continued.

It is easy to check that M

[ 1 5

]

[ − 13 − 9

] is the old solution. As a side note, we observe that in this case the inverse M is the same as A , or AA = A^2 = I. Such matrices are said to be unipotent. Important notation. When the inverse of A exists, it is denoted by the convenient notation A −^1. Do not ever write (^) A^1 in place of A −^1 ; it is both illegal and meaningless.

An observation.

We solved the equation AX = B above as X = MB = A −^1 B for a specific 2 × 2 matrix A. Note that B did not enter the calculation until the product MB. Thus we observe that if A is invertible, then the equation AX = B has a unique solution X = A −^1 B. This should be compared with the statement: If a,b are numbers and if a 6 = 0 then the equation ax = b has a uniques solution x = ba.

What happens if ∆ = 0. Let P =

[ 1 2 2 4

] and consider the

equations PX = Q where Q =

[ u v

] .

We invite you to check that when v = 2 u this system has infinitely many solution, but it has no solution when v 6 = 2 u.

Example of inverse.

We now illustrate the procedure on our 3 × 3 matrix in the second example. Start with:

[ A | I ] =

  

  .

When we do row transformations R 2 − R 1 , R 3 − 2 R 1 and R 3 − 32 R 2 , we get    

   

This is REF.

Inverse continued.

Now we go on to make RREF. The operations

2 R 3 , R 2 − 3 R 3 , R 1 + R 3 ,

R 2 , R 1 + R 2

produce the RREF:   

  

Thus the desired inverse is:

A −^1 =

  

  

Testing Invertibility.

Thus, it would be useful to know if we are likely to find an inverse before doing the full work. Luckily, we already have such a tool. Let A be an n × n matrix. We can convert [ A | I ] to REF. It is easy to see that we have exactly n pivots. The inverse exists if and only if all the pivots are on the left hand side of the searator bar. It is instructive to see that for the matrix [ 1 2 1 0 2 4 0 1

]

The operation R 2 − 2 R 1 gives REF with p.p. ( 1 , 3 ). Since ≥ 2 we have no inverse!