For what value(s) of $m$ does this system not have any solutions?

Question

I got this system:

$$2x-my=-2$$

$$mx-2y=m+4$$

For what values of $m$ does the system not have any solutions? I have figured out $m = 2$ doesn't yield any solutions, but I don't really know how to get that value?

Answer 1

When the determinant is nonzero, then the matrix is invertible, which means that there is exactly one solution to your system.

When the determinant is zero, there may be no solutions, or an infinity of them. You need to check which of these it is, for each case that makes the determinant zero separately.

So you have solved the exercise correctly in the comments!

Answer 2

From first equation, put $y = \frac{2x+2}{m}$ in second equation

You will get :

$$mx- 2\Big(\frac{2x+2}{m}\Big)=m+4$$

On simplifying:

$$m^2x-4x-4=m^2+4m \implies x = \frac{m^2+4m+4}{m^2-4}$$

For these equations to have solution, x must be defined ;

$$m^2-4 \neq 0 \implies m \neq 2$$

Since for $m= -2$ equations are identical i.e. infinite solutions

Therefore, for no solution $m= 2$

Answer 3

Multiply the first equation by $m$ and the second by $2$.

$$2mx-m^2y=-2m$$

$$2mx-4y=2m+8$$

Subtract them:

$$(4-m^2)y=-4m-8$$

Now, if $m^2 \neq 4$, this yields a formula for $y$: $y = \frac{-4m-8}{4-m^2}$ and you can plug it in one of the original equations to see that it also yields a formula for $x$.

That leaves the two cases $m = 2$ and $m = -2$. For $m = 2$, this gives

$$0y=-16$$ which is not possible. Luckily, for $m = -2$, it gives

$$0y=0$$; that's why it does work in this case. In fact, because $m = -2$ makes your original equations linearly dependent, you'll get infinite solutions $(x, y)$ in this case.

Answer 4

from the first equation we get $$x=-1+\frac{m}{2}$$ plugging this in the second equation we get $$m\left(-1+\frac{m}{2}y\right)-2y=m+4$$ or $$y\left(\frac{m^2}{2}-2\right)=2m+4$$ if now $$m^2=4$$ this meansd $$m=\pm 2$$ then the left hand side is $0$ (for $m=2$) and the right hand side isn't zero thus no solution exists.