1
$\begingroup$

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:

Question Image

$\endgroup$
4
  • 1
    $\begingroup$ Maybe a quick search using the search engine would have done the trick... math.stackexchange.com/questions/539798/… $\endgroup$ – Dmoreno Aug 15 '14 at 11:46
  • 1
    $\begingroup$ There are 3 unknowns. $a$ is a parameter. $\endgroup$ – user5402 Aug 15 '14 at 11:47
  • 1
    $\begingroup$ 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) $a=-1$. $\endgroup$ – Radz Aug 15 '14 at 12:02
  • $\begingroup$ Dear @Jeff, Also add $a\neq -1$ for (i). $\endgroup$ – Radz Aug 15 '14 at 12:10
2
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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...

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.