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.
Best Answer
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.