# When can a singular system have a solution?

## Question:

When can a singular system have a solution?

## Singular Systems of Equations

A singular system of equations is a system for which the matrix of the system is a singular matrix or the determinant is zero.

To solve a system, we perform a Gaussian elimination to eliminate 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.

A consistent matrix with non-pivot columns has infinitely many solutions.

A system is singular if the square matrix of the system, {eq}\displaystyle A {/eq} is singular or the determinant of the matrix is zero.

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

we obtain the reduced row echelon form of the matrix, {eq}\displaystyle A {/eq} that has at least one non-pivot column,

{eq}\displaystyle \boxed{\text{ and if the last column of the augmented matrix is not a pivot column, then the system is consistent, so we have a solution for the system}}. {/eq}

For example, the singular 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\\ 3 \end{array} \right] {/eq}

has the reduced echelon form of the augmented matrix given as {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&3 \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&3 \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&3 \end{array} \right]\\\\ \text{which is an inconsistent system because the last column is a pivot column, so there is no solution}. {/eq}

But, the singular system with the right hand side vector {eq}\displaystyle \vec{b}=\left[\begin{array}{c} 1\\ 2\\ 1 \end{array} \right] {/eq} has the reduced row echelon given as {eq}\displaystyle \left[A\bigg\vert \vec{b}\right]= \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 there is a solution with one free parameter, so infinitely many}. {/eq}