# How to parameterize the solution set of a matrix?

## Question:

How to parameterize the solution set of a matrix?

## System of Equations

A system of linear equations can have no solution, a unique solution or infinitely many solutions with different number of free variables.

To solve a system of equations, we perform Gaussian elimination to eliminate variables from the equations in such a manner that the matrix of the system is a staircase type matrix or the form is a row echelon form.

A system is consistent if the augmented matrix has a non- pivot column in the last column.

A row echelon form with at least-one non-pivot column in the matrix, will have infinitely many solutions, if the system is consistent.

The number of non-pivot columns gives the number of free parameters in the (infinite) solution.

## Answer and Explanation:

When solving a system of equations {eq}\displaystyle A\vec{x}=\vec{b}\iff \left[\begin{array}{cccc} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{array}\right] \left[\begin{array}{c} x_1\\ x_2\\ x_3\\ x_4 \end{array}\right] =\left[\begin{array}{c} b_1\\ b_2\\ b_3\\ b_4 \end{array}\right] {/eq}

whose augmented matrix is consistent and in a row echelon form has multiple non-pivot columns,

like {eq}\displaystyle\left[\begin{array}{cccc|c} \boxed{b_{11}}&*&*&*&*\\ 0& 0&\boxed{b_{23}}&*&*\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{array}\right], {/eq}

we obtain infinitely many solutions with the variables corresponding to the non-pivot columns being the free variables, denoted with different names.

For example, in the matrix above, the second and fourth columns are non-pivot columns, therefore the second and fourth variables will be denoted with different letters, like

{eq}\displaystyle x_2=s, x_4=t, s, t\in\mathbb{R}. {/eq}

For example, the system with the augmented matrix {eq}\displaystyle \begin{align} \left[A|\vec{b}\right]=\left[\begin{array}{cccc|c} 1&2&1&-1&-3\\ 1&2&-1&1&-1\\ 0&0&1&-1&-1 \end{array}\right] \end{align} {/eq} has a row echelon form given as {eq}\displaystyle \left[\begin{array}{cccc|c} 1&2&1&-1&-3\\ 1&2&-1&1&-1\\ 0&0&1&-1&-1 \end{array}\right]\overset{-R_1+ R_2}{\iff}\left[\begin{array}{cccc|c} 1&2&1&-1&-3\\ 0&0&-2&2&2\\ 0&0&1&-1&-1 \end{array}\right]\overset{\frac{1}{2}R_2+R_3}{\iff}\left[\begin{array}{cccc|c} \boxed{1}&2&1&-1&-3\\ 0&0&\boxed{-2}&-2&2\\ 0&0&0&0&0 \end{array}\right] {/eq}.

So, the second and fourth columns of the matrix are a non-pivot columns, therefore, the free variables are {eq}\displaystyle \begin{align}&x_2=t, x_4=s, t, s\in\mathbb{R}\\ &\text{and obtain the solution in terms of the two parameters}\\ &\begin{cases} x_1+2x_2+x_3-x_4=-3\\ x_2=t\\ -2x_3-2x_4=2\\ x_4=s \end{cases}\iff \begin{cases} x_1=-2-2t +2s\\ x_2=t\\ x_3=-1-s\\ x_4=s \end{cases}, t, s\in\mathbb{R} \end{align} {/eq}

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from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 8