# The number of solutions to a system of equations

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).

• I would suggest just taking the augmented matrix to row-echelon form and studying its rank. – ÍgjøgnumMeg Mar 17 '16 at 19:43
• 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. – almagest Mar 17 '16 at 20:01

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.

• For the third case you have two solution for $k$ – N74 Mar 17 '16 at 22:56
• 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. – ÍgjøgnumMeg Mar 18 '16 at 0:11

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 $k = 5$