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


3 Answers 3


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.


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
  • $a=5\implies k=\pm 5$


$$D= \left(% \begin{array}{cc} k &a \\ a & k \\ \end{array}% \right) = k^2-a^2$$

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 $k=\pm 5$.


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