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* + 2*x* – *y* + 2*z* = 10

2*w* – 2*x* – *y* + 3*z* = 11

2*w* + *x* + *y* – 3*z* = 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** = **b** ⟹ **A**^{-1}**Ax** = **A**^{-1}**b** ⟹ **Ix** = **A**^{-1}**b** ⟹ **x** = **A**^{-1}**b**

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!