# Consider the following system of linear equations..

Story cut short, I have an exam in a weeks time and this is a question off a previous exam paper - I'm unsure as to how I should go about it as there are 4 variables with only three linear equations..

Any help will be appreciated:

• Maybe a quick search using the search engine would have done the trick... math.stackexchange.com/questions/539798/… – Dmoreno Aug 15 '14 at 11:46
• There are 3 unknowns. $a$ is a parameter. – user5402 Aug 15 '14 at 11:47
• You apply elementary row operations to the augmented matrix to get $\begin{bmatrix} 1 & 0 & -4 & -2(-1+a) \\ 0 & 1 & 2 & -1+a \\ 0 & 0 & -3-2a+a^2 & -2-a+a^2 \end{bmatrix}$. I leave the rest to you. For your reference: (i) $a\neq 3$ (ii) $a=3$ (iii) $a=-1$. – Radz Aug 15 '14 at 12:02
• Dear @Jeff, Also add $a\neq -1$ for (i). – Radz Aug 15 '14 at 12:10

No, actual there are only 3 variables. a is a Parameter. You shall solve this system dependent on a, that means your solution will depend on the parameter a. After that, you should say, which values of a are suitable for the three possibilities (i), (ii), (iii).

Hint:

• Write your system in matrix form: $$A \mathbf{x} = b, \quad \mathbf{x}=x_i, \quad A = \left( \begin{array}{ccc} 1 & 2 & 0 \\ 1 & 3 & 2 \\ 1 & 3 & a^2-2a-1 \end{array} \right), \quad b = \left( \begin{array}{c} 0 \\ a-1 \\ a^2-3\end{array}\right).$$
• Remember Rouché theorem.

Bigger hint:

Note that $\det{A} = 0 \iff a = -1$ or $a=3$, hence for $a\neq \{-1,3\}$ $A$ is invertible and there exists unique solution given by $\mathbf{x} = A^{ -1} b$. On the other hand, for $a = -1$, notice that the 2nd and 3rd rows of the augmented matrix $[A|b]$ coincide, making the 2nd and the 3rd equations redundant. Maybe this indicates us something... Of course, for $a = 3$ the 2nd and 3rd rows of $A$ coincide while the 2nd and the 3rd rows of $b$ do not. Again, this tries to tell us something...

This can be approached by simple inspection. No matrices necessary.

Note that when the coefficient of $x_3$ in the third equation equals two, i.e. $$a^2 -2a -1 = 2 \iff a^2-2a-3 = 0 \iff (a-3)(a+1) =0 \iff (a = 3 \text{ or } a = -1$$

We have $x_1 + 3x_2 + 2x_3$ for both the second and third equation.

• At $a = -1$, the last equation matches the second equation: $x_1 + 3x_2 + 2x_3 = -2$. This means that the system, when $a = -1$ is underdetermined, which means the system has infinitely many solutions.

• At $a = 3$, however, the system is inconsistent, meaning no solution exists. That is, it cannot both be true that $x_1 + 3x_2 +2x_3 = 2$ and $x_1 + 3x_2 + 2x_3 = 6$.

At all other values for $a$, the system should have a unique solution.