For what value of k does this system equations have infinite solutions? For what value of $k$ the system
$
\left\{ 
\begin{array}{c}
kx+ay=5 \\ 
ax+ky=k \\ 
\end{array}
\right. 
$
has infinite solutions?
Honestly, i don't know how to start this problem, i saw that this have something to do with matrices and the determinants, but this is meant to be done without that.
The possible answers are: 
$A) $ $-5$ or $5$
$B) \sqrt 5$ or $-\sqrt 5$ 
$C) -25$ or $25$
$D) 0$
$E)$ Can't be determinated
 A: Since both the equations have infinite solutions, they must coincide with each other
This gives:
$\frac{k}{a}$ =  $\frac{a}{k}$ =  $\frac{5}{k}$
Considering the equation $\frac{a}{k}$ =  $\frac{5}{k}$ and solving it, $a$ turns out to be $5$.
Considering the equation $\frac{k}{a}$ =  $\frac{a}{k}$ and knowing that $a = 5$ and solving it,  $k^2$ turns out to be 25.
This means $k$ can be $5$ or $-5$.
Hence option $A$ is right.
A: Write 
$$D= \left(%
\begin{array}{cc}
  k &a \\
  a & k \\
\end{array}%
\right) = k^2-a^2$$
Then system has infinite or $0$ solution iff $D=0$. So when $k=a$ or $k=-a$.
Now if it has infite solutions then $$D_x= \left(%
\begin{array}{cc}
  5 &a \\
  k & k \\
\end{array}%
\right) = 5k-ak =0$$
and $$D_y= \left(%
\begin{array}{cc}
  k &5 \\
  a & k \\
\end{array}%
\right) = k^2-5a=0$$  
From 1.st one we get $k=0$ or $a=5$.
So if $k=0$ then $a=0$. In this case there is no solutions.
if $k\ne 0$ then $a=5$ so $k=\pm 5$.
A: Let consider the augmented matrix
$$\begin{bmatrix}
  k &a&5 \\
  a & k&k \\
\end{bmatrix}\to \begin{bmatrix}
  ak &a^2&5a \\
  ak & k^2&k^2 \\
\end{bmatrix}\to \begin{bmatrix}
  ak &a^2&5a \\
  0 & k^2-a^2&k^2-5a \\
\end{bmatrix}$$
Then to have infinitely many solution we need


*

*$k^2-a^2=0$

*$k^2-5a=0$


that is $a^2-5a=0\implies a(a-5)=0$ then check for 


*

*$a=0\implies k=0$ and

*$a=5\implies k=\pm 5$

