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: \[ \begin{bmatrix} 2 & 1 \\4 & 2 \end{bmatrix} \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} 1 \\1 \end{bmatrix} \]