# Solve different solutions for linear system

I am currently in a linear algebra class and I have to answer the following questions, for what values of $a$ and $b$ does the system below have:
a) No solution
b) Only one solution
c) Infinitely many solutions

\begin{cases} x_{2} + 3x_{3} = 1 \\ x_{1} + 2x_{2} + 6x_{3} = 1 \\ x_{2} - 6x_{3} = -1 \\ 2x_{1} + 2x_{2} + ax_{3} = b \end{cases}

To attempt this I tried to reduce the augmented matrix to echelon form and ended up with this

\begin{bmatrix} 1 & 2 & 6 & 1 \\ 0 & 2 & 3 & 1 \\ 0 & 0 & -9 & -2 \\ 0 & 0 & a & b + 1 \end{bmatrix}

From my understanding, for this particular system to have infinitely many solutions, both the third and the forth row has to be only zeros. Because since $a$ is in the third column, this is the only way to achieve $x_{3}$ being a free variable. The problem is that I don't know how to reduce this system in such a way and wondered if someone could help! I think if I understand how to arrange the matrix I will be able to answer the other questions. Thank you!

• The question asks for the number of solutions, so why don't you calculate determinant of the matrix? Commented Feb 14, 2018 at 15:18

The given system has augmented matrix $$\begin{bmatrix}0&1&3&|&1\\1&2&6&|&1\\0&1&-6&|&-1\\2&2&a&|&b\end{bmatrix}\xrightarrow{(1)}\begin{bmatrix}0&1&3&|&1\\1&2&6&|&1\\0&1&-6&|&-1\\0&-2&a-12&|&b-2\end{bmatrix}\xrightarrow{(2)}\begin{bmatrix}0&1&3&|&1\\1&0&0&|&-1\\0&0&-9&|&-2\\0&0&a-6&|&b\end{bmatrix}$$ where the steps are \begin{align}&(1)\qquad -2R_2+R_4\mapsto R_4\\&(2)\qquad \begin{cases}-2R_1+R_2\mapsto R_2\\-R_1+R_3\mapsto R_3\\2R_1+R_4\mapsto R_4\end{cases}\end{align} The system is inconsistent (no solutions) when there is a nonzero row in the coefficient matrix with a zero entry in the augmented vector on the right hand side. The system has one solution when the system is consistent and the matrix has rank $3$. Otherwise, there are infinitely many solutions.

One can perform a final step via $$(3)\qquad\begin{cases}-3R_3+R_1\mapsto R_1\\(6-a)R_3+R_4\mapsto R_4\end{cases}$$ $$\begin{bmatrix}0&1&3&|&1\\1&0&0&|&-1\\0&0&1&|&\frac{2}{9}\\0&0&a-6&|&b\end{bmatrix}\xrightarrow{(3)}\begin{bmatrix}0&1&0&|&\frac{1}{3}\\1&0&0&|&-1\\0&0&1&|&\frac{2}{9}\\0&0&0&|&b+\frac{2}{9}(6-a)\end{bmatrix}$$

• thanks, why do you have the leftmost 1 in the second row? we're taught to have it on the top. I guess it doesn't matter, just curious Commented Feb 14, 2018 at 19:00
• I just do row swaps at the end.
– Dave
Commented Feb 14, 2018 at 19:22

you have to write augmented matrix. then you can to perform row operations to convert this into echelon form

https://www.wikihow.com/Reduce-a-Matrix-to-Row-Echelon-Form

• I know how to do it, but I don't know how to do it with this particular problem since I dont know what to do with column 3. Commented Feb 14, 2018 at 15:11

Let's start from here:

\begin{bmatrix} 1 & 2 & 6 & 1 \\ 0 & 2 & 3 & 1 \\ 0 & 0 & -9 & -2 \\ 0 & 0 & a & b + 1 \end{bmatrix}

For equations 1 to 3 we calculate determinant:

$$\begin{vmatrix} 1 & 2 & 6 \\ 0 & 2 & 3 \\ 0 & 0 & -9 \\ \end{vmatrix}=18>0$$

This means the first $3$ equations have a unique solution.

Our system is overdetermined, so if the fourth equation is identical to the third one, the whole system has 1 unique solution and otherwise it has no solution.

If $a=-9$ AND $b+1=-2$ then we have 1 solution. Otherwise, we have no solution.

• We haven't touched on determinants (yet) in my course. But it's interesting what you are saying about the relation between the third and forth equation, thank you Commented Feb 14, 2018 at 19:03

There's a general result for this:

Let $A\,\mathbf x =\mathbf b$ be a (possibly non-homogeneous) linear system of size $m\times n$ over the field $K$. We'll denote $\;[A\mid \mathbf b]$ the augmented matrix. This linear system has

• no solution if $\;\DeclareMathOperator{\rank}{rank} \rank A<\rank A\mid\mathbf b$;

• solutions if $\;\rank A=\rank A\mid\mathbf b$.

Furthermore, if there are solutions, it is a single solution if $A$ has maximal rank, i.e. if $\rank A=\min(m,n)$. It has an infinity of solutions if $\rank A<\min(m,n)$.

In all cases, the set of solutions is an affine subspace of $K^n$ with dimension $\operatorname{codim} A$.