# How to show whether a $3\times 4$ matrix has no solution, a unique solution or infinitely many solutions?

The system is :

$$\begin{matrix} 1 & -4 & 6 & a & | & 0 \\ -2 & 5 & -4 & -1 & | & b \\ 1 & -10 & 22 & 8 & | & c \end{matrix}$$

After Gaussian elimination, I found that

$$\begin{array}{cccc|cc} 1 & -4 & 6 & a & & 0 \\ 0 & 1 & -\tfrac{8}{3} & - \left( 2a- \tfrac{1}{3} \right) & & - \tfrac{1}{3}b \\ 0 & 0 & 0 & 10-5a & & c-2b \end{array}$$

Is it correct and I can continue to determine whether there is no solution, a unique solution or infinitely many solutions?

Here are the operations:

$$\begin{matrix} 1 & -4 & 6 & a & | & 0 \\ -2 & 5 & -4 & -1 & | & b \\ 1 & -10 & 22 & 8 & | & c \end{matrix}$$

$R_2+2R_1\rightarrow R_2$ $$\begin{matrix} 1 & -4 & 6 & a & | & 0 \\ 0 & -3 & 8 & 2a-1 & | & b \\ 1 & -10 & 22 & 8 & | & c \end{matrix}$$

$R_3-R_1\rightarrow R_3$ $$\begin{matrix} 1 & -4 & 6 & a & | & 0 \\ 0 & -3 & 8 & 2a-1 & | & b \\ 0 & -6 & 16 & 8-a & | & c \end{matrix}$$

$R_3-2R_2\rightarrow R_3$ $$\begin{matrix} 1 & -4 & 6 & a & | & 0 \\ 0 & -3 & 8 & 2a-1 & | & b \\ 0 & 0 & 0 & 10-5a & | & c-2b \end{matrix}$$

$-\frac 13(R_2)\rightarrow R_2$ $$\begin{matrix} 1 & -4 & 6 & a & | & 0 \\ 0 & 1 & -\frac 83 & -\tfrac{2a-1}{3} & | & -\frac 13b \\ 0 & 0 & 0 & 10-5a & | & c-2b \end{matrix}$$

• Welcome at MSE. Could you please describe the row operations you used to arrive at this result? – Lutz Lehmann Mar 1 '14 at 9:30
• @CarlJon: In addition to what LutzL asked, the solution is not correct, so posting the row operations used would be helpful to spot the issue. Regards – Amzoti Mar 1 '14 at 14:12
• I have posted the processes, thanks for your help. – CarlJon Mar 3 '14 at 11:58
• Probably a typo: your $-\frac{2a-1}3$ in the row operations has become $-(2a - \frac 13)$ in your solution. – Peter Phipps Mar 3 '14 at 12:11
• You're right, I want to express - 2a-1 over 3. – CarlJon Mar 3 '14 at 12:31

There's no solution if there's a row of the form $0 0 0 0 | Q$ where $Q$ is not zero. Which row could possibly look like that? What would have to happen (to $a$, $b$, and $c$) for the row to look like that?

If there is a solution, then the variable represented by the 3rd column is arbitrary, so there are infinitely many solutions.

Constraints on $a$

In general, we have existence of a solution when the data vector is in the column space of the target matrix $\mathbf{A}$. That is, if the data vector can be written as a linear combination of the fundamental columns of $\mathbf{A}$.

Additionally, if the nullspace $\mathcal{N} \left( \mathbf{A}^{*} \right)$ is trivial, then the solution is also unique.

Look for the fundamental columns. Start with the reduced row echelon form: \begin{align} \mathbf{A} &\mapsto \mathbf{E}_{\mathbf{A}} \\ % \left[ \begin{array}{rrrr} 1 & -4 & 6 & a \\ -2 & 5 & -4 & -1 \\ 1 & -10 & 22 & 8 \\ \end{array} \right] % &\mapsto % \left[ \begin{array}{rrrr} \color{blue}{1} & 0 & -\frac{14}{3} & 0 \\ 0 & \color{blue}{1} & -\frac{8}{3} & 0 \\ 0 & 0 & 0 & \color{blue}{1} \\ \end{array} \right] % \end{align} The fundamental columns are marked by blue pivots. Because the 3rd column is linearly dependent, we can ignore it and study a simpler, equivalent form: $$\hat{\mathbf{A}} = % \left[ \begin{array}{rrrr} 1 & -4 & a \\ -2 & 5 & -1 \\ 1 & -10 & 8 \\ \end{array} \right]$$ The two matrices have the same range space: $$\mathcal{R} \left( \mathbf{A} \right) = \mathcal{R} \left( \hat{ \mathbf{A} } \right)$$ The determinant will determine when there is a nullspace. $$\det \hat{\mathbf{A}} = 15 \left( a - 2 \right)$$ When $a=2$ there are no longer three linearly independent columns. If a solution exists, it will not be unique.

Constraints on $b$, $c$

If $b=c=0$, we are out of the range space and probing the null space. When $a\ne2$ the nullspace is $$\mathcal{N} \left( \mathbf{A}^{*} \right) = \text{span } % \left\{ \, \left[ \begin{array}{c} 14 \\ 8 \\ 3 \\ 0 \end{array} \right] \, \right\}$$ When $a=2$ the dimension of the nullspace increases $$\mathcal{N} \left( \mathbf{A}^{*} \right) = \text{span } % \left\{ \, \left[ \begin{array}{c} 14 \\ 8 \\ 3 \\ 0 \end{array} \right] ,\, \left[ \begin{array}{c} 2 \\ 1 \\ 0 \\ 1 \end{array} \right] \, \right\}$$

When $a\ne 2$ and either $b\ne0$ or $c\ne0$ we are guaranteed a unique solution because $\mathcal{R} \left( \mathbf{A} \right) = \mathbb{C}^{3}$.