# Solve a linear system with more variables than equations

Suppose that, after a series of elementary row operations the augmented matrix of a linear system with variables $x_1$, $x_2$, $x_3$, $x_4$ is transformed into reduced row echelon form as follows:

$$\left(\begin{array}{cccc|c}1 & 0 & 0 & 1 & 0\\0 & 1 & 0 & 2 & 1 \\0 & 0 & 1 & 3 & 0 \end{array}\right)$$.

Can I solve the linear system as below?

Let $t$ be an arbitrary real number. Then solving each linear equation corresponding to the augmented matrix for leading variable and setting $x_4=t$, we get $x_1=-t, x_2=1-2t$, and $x_3=-3t$. Thus the general solution of the linear system is

\begin{align} x_1=-t\\ x_2=1-2t\\ x_3=-3t\\ \end{align}

where t is an arbitrary real number.

• I don't get your question, did you not just solve it? Normally, when you have more variables than equations you have infinitely many solutions because you have a degree of freedom. – user88595 Dec 11 '13 at 15:50
• You haven't solved the problem, you have reduced it from 4 variables to 1 variable. – ja72 Dec 11 '13 at 15:52
• You solved the system in the sense that you produced a parametric solution to the system. However this system is underdetermined, it has more variables than equations. This means it will have infinitely many solutions. Some people might say, in that sense, it can be solved because the solution is not unique. – Wintermute Dec 11 '13 at 16:03

## 2 Answers

you have the answer already. you have 3 equations and 4 variables and infinitely many solutions.

What you have done is a decomposition. Given an underconstrained system $A x =b$ you can split the variables to dependent and independent components such that

$$x = T x_{dep} + U x_{ind}$$

in your example

$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} + \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \begin{pmatrix} x_4 \end{pmatrix}$$

Now you reduce the system with

\begin{aligned} A T x_{dep} + A U x_{ind} & = b \\ A T x_{dep} & = b - A U x_{ind} \\ x_{dep} &= (A T)^{-1} \left(b - A U x_{ind}\right) \end{aligned}

The final values are

$$x = T (A T)^{-1} b + \left({\bf 1_{4×4}} - T (A T)^{-1} A\right) U x_{dep} \\ x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{pmatrix}0\\1\\0\end{pmatrix} + \begin{bmatrix} 0&0&0&-1 \\0&0&0&-2 \\ 0&0&0&-3 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \begin{pmatrix} x_4 \end{pmatrix}$$ $$x = \begin{pmatrix} -x_4 \\ 1-2 x_4 \\ -3 x_4 \\ x_4 \end{pmatrix}$$