# When do ab = c matrix have a solution?

## Question:

When do {eq}ab = c {/eq} matrix have a solution?

## Consistent Systems of Equations

A system of equations is consistent if the last column in the augmented matrix is no a pivot column.

A matrix with less rows than columns, will have a number of pivot columns equal or less than the number of rows, therefore the matrix will have at least one non-pivot column.

If an augmented matrix is consistent and has non-pivot columns, then there are infinitely many solutions with a number of free parameters, given by the number of non-pivot columns.

The system of equations given as {eq}\displaystyle A\vec{x}=\vec{b}\ \boxed{\text{ has a solution if the augmented matrix } \left[A|\vec{b}\right] \text{ is consistent}} {/eq}

We can have a unique solution if the system is consistent (the last column of the augmented matrix is a non-pivot column) and all the other columns are pivot columns,

or we can have infinitely many solutions if the system is consistent and there is at least one non-pivot column, except the last column.

For example, the system with the augmented matrix given as {eq}\displaystyle \left[ \begin{array}{ccc|c} 1&2&1&2 \\ 1&2&1&1 \\ 1&1&0&0 \end{array} \right] {/eq} has a solution, obtain as follows.

{eq}\displaystyle \begin{align} &\left[ \begin{array}{cccc} 1&2&1&1 \\ 1&2&1&1 \\ 1&1&0&0 \end{array} \right] \overset{-R_1+R_2}{\implies }\left[ \begin{array}{ccc|c} 1&2&1&1 \\ 0&0&0&0\\ 1&1&0&0 \end{array} \right] \\\\ &\overset{R_2\text{ interchanged with } R_3}{\implies }\left[ \begin{array}{ccc|c} 1&2&1&1 \\ 1&1&0&0 \\ 0&0&0&0\\ \end{array} \right] \overset{-R_1+R_2}{\implies }\left[ \begin{array}{ccc|c} \boxed{1}&2&1&1 \\ 0&\boxed{-1}&-1&-1 \\ 0&0&0&0 \\ \end{array} \right] , &\left[\text{ where the boxed numbers are the pivot positions }\right]\\ \implies &\text{the system is consistent because the last column is not a pivot columns, so there is a solution}\\ &\text{and because there is a non-pivot column, there are infinitely many solutions} \\ &x_3=s, x_2=-1+s \text{ and } x_1=1-2(-1+s)-s=3-3s\implies \text{ the solutions are } \vec{x}=\left[ \begin{array}{c} 3 \\ -1 \\ 0\\ \end{array} \right] +s\left[ \begin{array}{c} -3\\ 1 \\ 1\\ \end{array} \right], s\in\mathbb{R}. \end{align} {/eq} 