System of Equations, Part 4

Please read the previous posts on this topic if you do not understand this one.

To illustrate the power of matrices, consider the following system of equations:

w + x + y + z = 5
w + 2xy + 2z = 10
2w – 2xy + 3z = 11
2w + x + y – 3z = 0

This would take a while using the substitution or elimination methods. I will solve this using the matrix method. This system, in matrix form, is:


This system is in the form Ax = b and as shown in my last post, the solution to this is found by pre-multiplying both side by A-1. Now admittedly, finding A-1 manually for a 4 × 4 matrix would take some time. However, taking advantage of the internet (or a modern calculator), I find that


Here I factored out the 1/59 out of each element in the matrix to make it look nicer.

So from my last post, you know that the answer is found by pre-multiplying the b matrix:

Ax = bA-1Ax = A-1bIx = A-1bx = A-1b

So the solution is:


So w = 3, x = 1, y = -1, and z = 2. Isn’t that fantastic!