1
$\begingroup$

Find the value of $k$ for which the system of equations

$x-3y-z=0 $

$3x-5y-z=0$

$-x+ky+2z=k^2-4$

has:

(i) no real solutions

(ii) infinitely many solutions

(iii) a unique solution

I rewritten the system of equations in the augmented matrix form and simplified to $z= \frac{k^2-4}{-k+5}$. Now I think that the answers are as follows:

(i) $k=5$, because the denominator will be zero

(ii) I believe there is no such a possibility because for infinitely many solutions it should be rewritten as $ az=b $ and both $a$ and $b$ would have to be zero - but it's not possible if written as $z(-k+5)=k^2-4$.

(iii) Any $k \ne 5 $.

Please correct me if I'm wrong in any of the parts of the exercise, I am especially not sure about (ii).

$\endgroup$
  • 2
    $\begingroup$ I would suggest just taking the augmented matrix to row-echelon form and studying its rank. $\endgroup$ – ÍgjøgnumMeg Mar 17 '16 at 19:43
  • 1
    $\begingroup$ All seems fine to me, except that z should be twice the value you have given above. There is a unique solution for k not 5 and no solution for k = 5. $\endgroup$ – almagest Mar 17 '16 at 20:01
2
$\begingroup$

Let $A$ denote the coefficient matrix of the system and $b$ denote the vector of constants. The augmented matrix $(A\lvert b)$ is as follows;

$$ (A\lvert b) = \left(\begin{array}{rrr|r} 1 & -3 & -1 & 0 \\ 3 & -5 & -1 & 0 \\ -1 & k & 2 & k^2 - 4 \end{array}\right). $$

In order to check the number of solutions of the system, we take $(A \lvert b)$ to row-echelon form and study its rank. The row-echelon form of $(A \lvert b)$, which I will henceforth denote by $P$ is;

$$ P = \left(\begin{array}{rrr|r} 1 & -3 & -1 & 0 \\ 0 & 4 & 2 & 0 \\ 0 & 0 & \frac{1}{2}(5-k) & k^2 - 4 \end{array}\right). $$

(Somebody please correct me if any errors are found).

Now we consider the cases.

$(i)$ No solutions; if we can find a $k$ such that $\frac{1}{2}(5-k) = 0$ while $k^2 – 4 \neq 0$ then we have that $\text{rank}(A\lvert b) > \text{rank}(A)$ and, as such, that there are no solutions. Indeed, if $k = 5$ then we have that

$$0x + 0y + 0z = 21$$

which is clearly nonsense.

$(ii)$ Exactly one solution; in this case we have only to find any $k$ such that $\text{rank}(A\lvert b) = \text{rank}(A) = 3$. In other words, a value of $k$ such that we can reduce the matrix further to reduced row-echelon form.

$(iii)$ Infinitely many solutions; in this case, we look for a $k$ such that $\frac{1}{2}(5-k) = k^2 - 4 = 0$. However since $k^2 - 4 = (k-2)(k+2)$, there is clearly no way for both to be zero simultaneously. Hence, there exists no $k$ such that there are infinitely many solutions to the system.

Hopefully this helps.

$\endgroup$
  • $\begingroup$ For the third case you have two solution for $k$ $\endgroup$ – N74 Mar 17 '16 at 22:56
  • $\begingroup$ There are two solutions for $k^2 - 4 = 0$ but no way for $k$ to make both $k^2 - 4 = 0$ AND $\frac{1}{2}(5 - k) = 0$ simultaneously. $\endgroup$ – ÍgjøgnumMeg Mar 18 '16 at 0:11
1
$\begingroup$

Let's work on the first and second equations:

$$\begin{cases} x-3y-z & = 0\\ 3x-5y-z & = 0 \end{cases} \Rightarrow \begin{cases} x = & y\\ z = & -2y \end{cases}. $$

Substitute this solution in the third equation:

$$-y+ky+2(-2y)=k^2-4 \Rightarrow y(k-5) = k^2-4.$$

If $k \neq 5$, then you have a unique solution:

$$\begin{cases} x = & \frac{k^2-4}{k-5} \\ y = & \frac{k^2-4}{k-5} \\ z = & -2 \frac{k^2-4}{k-5} \end{cases} $$

On the other hand, if $k = 5$, then the equation $y(k-5) = k^2-4$ becomes:

$$y(5-5) = 5^2-4 \Rightarrow 0 = 21,$$

which, obviously, is impossible, and hence the whole system is impossible.

Summarizing:

  • One solution for $k \neq 5$
  • No solution for $k = 5$
$\endgroup$

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.