# Thread: Finding x and y where squared matrix = unsquared matrix

1. ## Finding x and y where squared matrix = unsquared matrix

Hi, trying to figure out this question, made a guess at how to do it but ended up with the wrong answer .

A =
[x -4]
[y 2]

I need to find the x and y where A^2 = A.

I tried figuring out A^2 and got:

[x²-4y -4x-8]
[xy+2y -4y+4]

I set that equal to the original A and took the resulting equations:
x
² - 4y - 4x - 8 = x - 4
xy + 2y - 4y + 4 = y + 2

And rearranged/substituted and ended with:
x = 1
y = -2

Unfortunately it seems that wasn't the correct answer. So at this point I'm not really sure how to solve this question, can anyone help with this?

Thanks!

2. Hi cb220, try not to make it so complex, ignore the elements in the 1st column giving

-4x-8=-4
-4y+4=2

I get x=-1, y=-0.5

3. Ahh, cool I didn't know you could just use some columns like that. That is easier! Thanks for the help.

4. Not a bad idea to check the solution against the more complicated columns for consistency.

5. Originally Posted by cb220
Ahh, cool I didn't know you could just use some columns like that. That is easier! Thanks for the help.
You understand, don't you, that saying two matrices are equal means that indvidual corresponding entries are equal.
$\displaystyle A^2= A$
$\displaystyle \begin{bmatrix}x^2-4y & -4x-8 \\ xy+2y & -4y+4\end{bmatrix}= \begin{bmatrix}x & -4 \\ y & 2\end{bmatrix}$
means that $\displaystyle x^2- 4y= x$, $\displaystyle -4x- 8= -4$, $\displaystyle xy+ 2y= y$, and $\displaystyle -4y+ 4= 2$.

You, for some reason, wrote the rows as if they were single entries rather than two separte columns.