10.6 Solving linear equations

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This module explains how to use matrices to solve linear equations.

At this point, you may be left with a pretty negative feeling about matrices. The initial few ideas—adding matrices, subtracting them, multiplying a matrix by a constant, and matrix equality—seem almost too obvious to be worth talking about. On the other hand, multiplying matrices and taking determinants seem to be strange, arbitrary sequences of steps with little or no purpose.

A great deal of it comes together in solving linear equations. We have seen, in the chapter on simultaneous equations, how to solve two equations with two unknowns. But suppose we have three equations with three unknowns? Or four, or five? Such situations are more common than you might suppose in the real world. And even if you are allowed to use a calculator, it is not at all obvious how to solve such a problem in a reasonable amount of time.

Surprisingly, the things we have learned about matrix multiplication, about the identity matrix, about inverse matrices, and about matrix equality, give us a very fast way to solve such problems on a calculator!

Consider the following example, three equations with three unknowns:

$x+2y-z=11$
$2x-y+3z=7$
$7x-3y-2z=2$

Define a 3×3 matrix [A] which is the coefficients of all the variables on the left side of the equal signs:

$\left[A\right]=$ $\left[\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 7& -3& -2\end{array}\right]$

Define a 3×1 matrix [B] which is the numbers on the right side of the equal signs:

$\left[B\right]=$ $\left[\begin{array}{c}\text{11}\\ 7\\ 2\end{array}\right]$

Punch these matrices into your calculator, and then ask the calculator for [A-1][B]: that is, the inverse of matrix [A], multiplied by matrix [B].

The calculator responds with a 3×1 matrix which is all three answers . In this case, $x=3$ , $y=5$ , and $z=2$ .

The whole process takes no longer than it takes to punch a few matrices into the calculator. And it works just as quickly for 4 equations with 4 unknowns, or 5, etc.

Huh? why the heck did that work?

Solving linear equations in this way is fast and easy. But with just a little work—and with the formalisms that we have developed so far about matrices—we can also show why this method works.

Step 1: in which we replace three linear equations with one matrix equation

First of all, consider the following matrix equation:

$\left[\begin{array}{c}x+2y-z\\ 2x-y+3z\\ 7x-3y-2z\end{array}\right]$ $=$ $\left[\begin{array}{c}\text{11}\\ 7\\ 2\end{array}\right]$

The matrix on the left may look like a 3×3 matrix, but it is actually a 3×1 matrix. The top element is $x+2y-z$ (all one big number), and so on.

Remember what it means for two matrices to be equal to each other. They have to have the same dimensions ( ). And all the elements have to be equal to each other . So for this matrix equation to be true, all three of the following equations must be satisfied:

$x+2y–z=11$
$2x–y+3z=7$
$7x–3y–2z=2$

Look familiar? Hey, this is the three equations we started with! The point is that this one matrix equation is equivalent to those three linear equations . We can replace the original three equations with one matrix equation, and then set out to solve that.

Step 2: in which we replace a simple matrix equation with a more complicated one

Do the following matrix multiplication. (You will need to do this by hand—since it has variables, your calculator can’t do it for you.)

$\left[\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 7& -3& -2\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

If you did it correctly, you should have wound up with the following 3×1 matrix:

$\left[\begin{array}{c}x+2y-z\\ 2x-y+3z\\ 7x-3y-2z\end{array}\right]$

Once again, we pause to say…hey, that looks familiar! Yes, it’s the matrix that we used in Step 1. So we can now rewrite the matrix equation from Step 1 in this way:

$\left[\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 7& -3& -2\end{array}\right]$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ $=$ $\left[\begin{array}{c}\text{11}\\ 7\\ 2\end{array}\right]$

Stop for a moment and make sure you’re following all this. I have shown, in two separate steps, that this matrix equation is equivalent to the three linear equations that we started with.

But this matrix equation has a nice property that the previous one did not. The first matrix (which we called [A] a long time ago) and the third one ([B]) contain only numbers. If we refer to the middle matrix as [X] then we can write our equation more concisely:

$\left[A\right]\left[X\right]=\left[B\right]$ , where $\left[A\right]=$ $\left[\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 7& -3& -2\end{array}\right]$ , $\left[X\right]=$ $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ , and $\left[B\right]=$ $\left[\begin{array}{c}\text{11}\\ 7\\ 2\end{array}\right]$

Most importantly, $\left[X\right]$ contains the three variables we want to solve for! If we can solve this equation for $\left[X\right]$ we will have found our three variables $x$ , $y$ , and $z$ .

Step 3: in which we solve a matrix equation

We have rewritten our original equations as $\left[A\right]\left[X\right]=\left[B\right]$ , and redefined our original goal as “solve this matrix equation for $\left[X\right]$ .” If these were numbers, we would divide both sides by $\left[A\right]$ . But these are matrices, and we have never defined a division operation for matrices. Fortunately, we can do something just as good, which is multiplying both sides by $\left[A{\right]}^{–1}$ . (Just as, with numbers, you can replace “dividing by 3” with “multiplying by $\frac{1}{3}$ .”)

Solving a Matrix Equation
$\left[A\right]\left[X\right]=\left[B\right]$ The problem.
$\left[A{\right]}^{–1}\left[A\right]\left[X\right]=\left[A{\right]}^{–1}\left[B\right]$ Multiply both sides by $\left[A{\right]}^{–1}$ , on the left. (Remember order matters! If we multiplied by $\left[A{\right]}^{–1}$ on the right, that would be doing something different.)
$\left[I\right]\left[X\right]=\left[A{\right]}^{–1}\left[B\right]$ $\left[A{\right]}^{–1}\left[A\right]=\left[I\right]$ by the definition of an inverse matrix.
$\left[X\right]=\left[A{\right]}^{–1}\left[B\right]$ $\left[I\right]$ times anything is itself, by definition of the identity matrix.

So we’re done! $\left[X\right]$ , which contains exactly the variables we are looking for, has been shown to be $\left[A{\right]}^{–1}\left[B\right]$ . This is why we can punch that formula into our calculator and find the answers instantly.

Let’s try one more example

$5x–3y–2z=4$
$x+y–7z=7$
$10x–6y–4z=10$

We don’t have to derive the formula again—we can just use it. Enter the following into your calculator:

$\left[A\right]=$ $\left[\begin{array}{ccc}5& -3& -2\\ 1& 1& -7\\ \text{10}& -6& -4\end{array}\right]$ $\left[B\right]=$ $\left[\begin{array}{c}4\\ 7\\ \text{10}\end{array}\right]$

Then ask the calculator for $\left[A{\right]}^{–1}\left[B\right]$ .

The result?

What happened? To understand this error, try the following:

Hit ENTER to get out of the error, and then hit <MATRX> 1<MATRX>1 ) ENTER

Aha! Matrix $\left[A\right]$ has a determinant of 0. A matrix with 0 determinant has no inverse. So the operation you asked the calculator for, $\left[A{\right]}^{–1}\left[B\right]$ , is impossible.

What does this tell us about our original equations? They have no solution. To see why this is so, double the first equation and compare it with the third—it should become apparent that both equations cannot be true at the same time.

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