# 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:

$\left[{\begin{array}{cccc}{1}&{1}&{1}&{1}\\{1}&{2}&{{-}{1}}&{2}\\{2}&{{-}{2}}&{{-}{1}}&{3}\\{2}&{1}&{1}&{{-}{3}}\end{array}}\right]\left[{\begin{array}{c}{w}\\{x}\\{y}\\{z}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{5}\\{10}\\{11}\\{0}\end{array}}\right]$

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

${\mathbf{A}}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{59}\left[{\begin{array}{cccc}{1}&{4}&{12}&{15}\\{4}&{16}&{{-}{1}{1}}&{1}\\{39}&{{-}{2}{1}}&{{-}{4}}&{{-}{5}}\\{15}&{1}&{3}&{{-}{11}}\end{array}}\right]$

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:

${\mathbf{x}}\hspace{0.33em}{=}\hspace{0.33em}{\mathbf{A}}^{{-}{1}}{\mathbf{b}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{59}\left[{\begin{array}{cccc}{1}&{4}&{12}&{15}\\{4}&{16}&{{-}{1}{1}}&{1}\\{39}&{{-}{2}{1}}&{{-}{4}}&{{-}{5}}\\{15}&{1}&{3}&{{-}{11}}\end{array}}\right]\hspace{0.33em}\left[{\begin{array}{c}{5}\\{10}\\{11}\\{0}\end{array}}\right]\hspace{0.33em}{=}\hspace{0.33em}\left[{\begin{array}{c}{3}\\{1}\\{{-}{1}}\\{2}\end{array}}\right]$

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