# When does a singular system have infinitely many solutions?

## Question:

When does a singular system have infinitely many solutions?

## Consistent Systems of Equations

To solve a system of equations, we use Gaussian elimination on the augmented matrix of the system to determine if the system is consistent and solve it.

Gaussian elimination is a technique to solve a system of linear equations by eliminating the unknowns from the equations in such a manner that the matrix of the system is an upper triangular matrix.

A system is consistent if the augmented matrix has no pivot column in the last column.

If a system is consistent and the matrix had non-pivot columns, then the number of the non-pivot column gives the number of free parameters to use to solve the system.

A square matrix {eq}\displaystyle A_{n\times n} {/eq} is singular if the determinant of the matrix is zero,

so solving a system of equations with a singular matrix, like {eq}\displaystyle A\vec{x}=\vec{b} {/eq} and {eq}\displaystyle \det(A)=0, {/eq}

implies that the reduced row echelon for of the matrix, {eq}\displaystyle A {/eq} will have at least one non-pivot column

{eq}\displaystyle \boxed{\text{and if the system is consistent, then there are infinitely many solutions}}. {/eq}

For the system to be consistent, when there is at least one non-pivot column, we need to have the last column of the augmented matrix a non-pivot column,

like {eq}\displaystyle \left[A\bigg\vert \vec{b}\right] =\left[\begin{array}{ccc|c} 1&0&*&*\\ 0&1&*&*\\ 0&0&0&0 \end{array} \right] \text{ or } \left[\begin{array}{ccc|c} 1&*&*&*\\ 0&0&0&0\\ 0&0& 0&0 \end{array} \right] {/eq}

For example, the system {eq}\displaystyle A\vec{x}=\vec{b}, {/eq} where {eq}\displaystyle A=\left[\begin{array}{ccc} 1&1&0\\ 2&1&1\\ 1&1&0 \end{array} \right] \text{ and } \vec{b}=\left[\begin{array}{c} 1\\ 2\\ 1 \end{array} \right] {/eq}

has the reduced echelon form of the augmented matrix given by {eq}\displaystyle \left[A\bigg\vert \vec{b}\right]=\left[\begin{array}{ccc|c} 1&1&0&1\\ 2&1&1&2\\ 1&1&0&1 \end{array} \right]\overset{-2\cdot R_1+R_2}{\iff} \left[\begin{array}{ccc|c} 1&1&0&1\\ 0&-1&1&0\\ 1&1&0&1 \end{array} \right] \overset{- R_1+R_3}{\iff} \left[\begin{array}{ccc|c} 1&1&0&1\\ 0&-1&1&0\\ 0&0&0&0 \end{array} \right]\\\\ \text{which is a consistent system with one non-pivot column, so, one free parameter, therefore there are infinitely many solutions}. {/eq}