## Saturday, November 24, 2012

For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the
first column of B to get
(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9

For those having difficulty finding the inverse of a matrix, here's part of a chapter from my book on this topic.

Multiplicative Inverses of Matrices

Recall that the multiplicative inverse of any real number x is the number when multiplied by x equals 1. In this case the multiplicative inverse of x is 1/x. Suppose we have matrix A of the form

If we multiply A by what is known as the identity matrix, we still get A. We will call the identity matrix I, note the illustration below of AI = A.

If A is an n by n matrix, then there exists another matrix A-1, called A inverse, so that AA-1 = I and A-1A = I. So we basically are looking for a matrix that when multiplied by the original matrix, equals the identity matrix and vice versa. The identity matrix is always a matrix with 1's along the diagonal from upper left to lower right and 0's everywhere else.

Example: Show that B is the multiplicative inverse of A where

For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the first column of B to get

(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9
= 15/27 + 12/27
= 1

Next we multiply the first row of A with the second column of B to get
(-3)(4/27) + (4)(1/9) = -12/27 + 4/9
= -12/27 + 12/27
= 0

Now proceed to multiply the second row of A with the first row of B to get
(3)(-5/27) + (5)(1/9) = -15/27 + 5/9
= -15/27 + 15/27
= 0

Finally, multiply the second row of A by the second column of B to get
(3)(4/27) + (5)(1/9) = 12/27 + 5/9
= 12/27 + 15/27
= 1

Therefore B is the multiplicative inverse of A. Notice the illustration of this below.