Singularity and inverses

Let A be an M*M(M rows by M columns) matrix, an inverse of A is also an M*M matrix with the special property that A*A-1=In Where In is an N*N identity matrix.

Before we even start finding inverses, it should be noted that NOT all matrices have inverses. Some matrices do not have inverses at all and other matrices might have more than one inverse. There are such thing as left inverse and right inverse. Typically letters of the alphabet will be used to identify a matrix and inverse for any matrix is usually shown as the letter of the matrix with a negative 1 exponent.

For example if we had a matrix T= , then T inverse would be written as T-1

For two by two matrices, there is an easy way to find the inverse and here it is.

Let A= Where Q,R,S and T are all real numbers, then A-1=1/(Q*T-SR)*

We change the sign of the numbers that sit on the bottom left corner and upper right corner, and switch the other two. This method of finding inverses only applies to two by two matrices. It does not work for any N by N matrix. Lets use this to find the inverses of the following matrices.

Ex1. A= A-1=1/(6*1-3*5)* = 1/(-9)* =

Ex2.B= B-1=1/(3*4-2*(-1))* = 1/(14)* = =

Ex3. C=C-1=1/(1*4-2*2)* = 1/(0)* = no inverse. This matrix does not have an inverse. Remember when your Algebra professor told you 1/0 was undefined? Whenever you end up with 1/0, then you can stop right there and correctly conclude that there is not inverse for that particular matrix.

Finding the inverse of a 3 by 3 matrix is considerably more challenging. It is better to use a calculator for anything bigger than 3 by 3 because it is time consuming.

Find the inverse of the following matrix?

Ex3.

Let A=First we find the determinant of the matrixDet(A)=1(1*0-1*0)-3(1*0-6*0)-2(