System of equations. Find when it has one solution, multiple solutions or no solutions

Question

I have a linear algebra class and I'm struggling to understand the solution to the following question:

I have a system of equations given by:

$$x_1 + x_2 + ax_3 = 2\\ 2x_1 + x_2 + (2a + 1)x_3 = 5\\ 3x_1 + (a-1)x_2 + 2x_3 = b + 2$$

As you can see there is also an a and b in the equations. I am to find a and b such that the system has a single solution. Has no solution and has multiple solution.

So my way to find it was first to perform Gaussian elimination.

$$ M = \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 2 & 1 & 2a + 1 & 5\\ 3 & a-1 & 2 & b + 2 \end{array}\right) \xrightarrow{R3 := R3 - 3R1} \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 2 & 1 & 2a + 1 & 5\\ 0 & a-4 & 2 - 3a & b - 4 \end{array}\right)\\ \xrightarrow{R2 := R2 - 2R1} \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 0 & -1 & 1 & 1\\ 0 & a-4 & 2 - 3a & b - 4 \end{array}\right) $$

And I know that when I have the echolon form there should be n pivots for the equations to have a single solution. So $a - 4$ has to be 0. Which means $a = 4$ I choose $b = -6$ because this is easy to work with.

So a unique solution is easy but I can't find how I can fill in a or b to get no or multiple solutions.

No solutions would require a row to all 0 before the | and a constant after it which I can never get and mutiple solutions requires a row which is all 0 which I also cannot get...

I have no clue how to proceed now so any help is appreciated!

Thank you for reading.

Answer 1

Determinant of the system is:

$$\triangle =\begin{vmatrix} 1 & 1 & a \\ 2 & 1 & 2a+1 \\ 3 & a-1 & 2 \\ \end{vmatrix}=2+2a$$

The system has unique solution for $\triangle\ne0$ or $a\ne-1$.

For $a=-1$ the system becomes:

$$x_1 + x_2 - x_3 = 2\tag{1}$$ $$2x_1 + x_2 - x_3 = 5\tag{2}$$ $$3x_1 -2 x_2 + 2x_3 = b + 2\tag{3}$$

Subtract (1) from (2) and you get:

$$x_1=3$$

Equations (2) and (3) now become:

$$x_2-x_3=-1$$

$$-2x_2+2x_3=b-7$$

or:

$$-2x_2+2x_3=2$$

$$-2x_2+2x_3=b-7$$

You get infinitely many solutions for $b-7=2$ or for $b=9$. For example, if $x_2=z$ then $x_3=z+1$ (for any $z\in R$) and we already know that $x_1=3$.

For $b\ne9$ there are no solutions.

EDIT: If you insist on the echelon form, here is the right order of steps:

$$M = \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 2 & 1 & 2a + 1 & 5\\ 3 & a-1 & 2 & b + 2 \end{array}\right) \xrightarrow{R3 := R3 - 3R1} \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 2 & 1 & 2a + 1 & 5\\ 0 & a-4 & 2 - 3a & b - 4 \end{array}\right)\\ \xrightarrow{R2 := R2 - 2R1} \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 0 & -1 & 1 & 1\\ 0 & a-4 & 2 - 3a & b - 4 \end{array}\right) \xrightarrow{R3 := (a-4)R2 + R3} \left(\begin{array}{ccc|c} 1 & 1 & a & 2\\ 0 & -1 & 1 & 1\\ 0 & 0 & -2 - 2a & a+b - 8 \end{array}\right) $$

And now, when you are done with the last row, you are ready to make some conclusions (not before it):

Obviously for $-2-2a\ne0$, or $a\ne-1$ you will have a unique solution. For $a=-1$ the last row has all zeros so the system either has infinitely many or no solutions at all. It has to be discussed separately and I already did that in the original version of my answer.

The condition $a=4$ or $a\ne4$ should never be mentioned. It has noting to do with the final solution.

Answer 2

If $a=4$, then yes, you have one unique solution no matter what $b$ is. However, you aren't done yet, and that isn't the only possibility for a unique solution!

You examined all possibilities when $a=4$ (there's only one), now you must continue and analyse all other possibilities. From here on, you can assume that $a\neq 4$, so you can divide the final row $R_3$ by $a-4$ and get $$\begin{pmatrix}0 & 1 & \frac{2-3a}{a-4} & | & \frac{b-4}{a-4}\end{pmatrix}$$ as your final row.

Then, continue creating the echelon form.