The Fundamental Question

The fundamental problem in the first part of a standard linear algebra course is solving systems of equations. For example, solving for $x$ and $y$ in the equations $$x + y = 6$$ $$2x - y = 0$$

Using matrix multiplication, we can write an equivalent problem using matrices and vectors. $$\begin{bmatrix} 1 & 1 \\2 & -1 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 6 \\0 \end{bmatrix}$$ The matrix is the coefficient matrix. It stores the coefficients of the variables in the equations: $$A = \begin{bmatrix}1 & 1\\2 & -1\end{bmatrix}$$ The vector on the left is the variable vector. It tells the order of the variables that the coefficients in the columns of $A$ correspond to. $$\overrightarrow{v} = \begin{bmatrix}x\\y\end{bmatrix}$$ The vector on the right is the set of values that each equation equals. Usually it is denoted with $\overrightarrow{b}.$ $$\overrightarrow{b} = \begin{bmatrix}6\\0\end{bmatrix}$$

The equation can the be written $$A\overrightarrow{v} = \overrightarrow{b}$$ The question is, what is $\overrightarrow{v}$?

Let's solve the $1$-dimensional case because it is the easiest but also gives a little insight into what's to come. The fundamental question is, given $a$ and $b,$ how do you solve $av = b$ for $v?$

There are three cases.

Case 1: If $a \neq 0$ then we can divide by $a.$ The solution is $v = b/a.$

Case 2: If $a = 0$ and $b = 0$ then the equation is $0v = 0$ and any number is a solution. In particular, there are infinitely many solutions.

Case 3: If $a = 0$ and $b \neq 0$ then there are no solutions. For example, if $a = 0$ and $b = 2,$ there is no number $v$ that solves $0v = 2.$

All three of these cases will come up in higher dimensions but they are easy to recognize. The example $$\begin{bmatrix} 1 & 1 \\2 & -1 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 6 \\0 \end{bmatrix}$$ is like Case 1 since there is only one solution.

This is like Case 2 which has infinitely many solutions: $$\begin{bmatrix} 2 & 3 \\4 & 6 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 5 \\10 \end{bmatrix}$$

This is like Case 3 which has no solutions: $$\begin{bmatrix} 2 & 1 \\4 & 2 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 1 \\1 \end{bmatrix}$$