Business Maths

3rd Tutorial

AX = C 

A–1AX = A–1C 

IX = A–1C 

X = A–1C

It should be noted that the order in the multiplication above is important and is not at all arbitrary. Recall that, for matrices, multiplication is not commutative. That is, AB is almost never equal to BA. So multiplying the matrix equation "on the left" (to get A–1AX) is not at all the same thing as multiplying "on the right" (to get AXA–1). And you can not say that the product AXA–1 equals A–1AX, because you can't switch around the order in the multiplication. Instead, you have to multiply A–1 on the left, putting it right next to the A in the original matrix equation. And since you have to do the same thing to both sides of an equation when you're solving, you must multiply "on the left" on the right-hand side of the equation as well, resulting in A–1C. You cannot be casual with your placement of the matrices; you must be precise, correct, and consistent. This is the only way to successfully cancel off A and solve the matrix equation.

The matrix inverse of a square matrix  may be taken in Mathematica using the function Inverse[m].

For a  matrix

| (2) |

the matrix inverse is

| | | (3) |

| | | (4) |

For a  matrix

| (5) |

the matrix inverse is

| (6) |

A general  matrix can be inverted using methods Gaussian elimination,

Find the inverse of the following matrix.

First, I write down the entries the matrix A, but I write them in a double-wide matrix:

In the other half of the double-wide, I write the identity matrix:

Now I'll do matrix row operations to convert the left-hand side of the double-wide into the identity. (As always with row operations, there is no one "right" way to do this. What follows are just the steps that happened to occur to me. Your calculations could easily look quite different.)

Now that the left-hand side of the double-wide contains the identity, the right-hand side contains the inverse. That is, the inverse matrix is the following:

Note...