## monroe17 Group Title If A= x 3 y 3 (matrix) determine the values of x and y for which A^2=A one year ago one year ago

1. satellite73 Group Title

did you square it?

2. monroe17 Group Title

like 3^2?

3. satellite73 Group Title

i mean square the matrix

4. satellite73 Group Title

i got $\left[ \begin{array}{ c c } x^2+3y & 3x+9 \\ xy+3y & 3y+9 \end{array} \right]$

5. satellite73 Group Title

assuming the matrix is $\left[ \begin{array}{ c c } x & 3 \\ y & 3 \end{array} \right]$

6. satellite73 Group Title

that means $3x+9=3$ so $x=-2$ and also $$y=-2$$

7. monroe17 Group Title

once you square the matrix, how do you determine the x and y values though?

8. monroe17 Group Title

oh nevermind. how'd you get 3x+9=3

9. satellite73 Group Title

you have $\left[ \begin{array}{ c c } x^2+3y & 3x+9 \\ x^2+3y & 3y+9 \end{array} \right]=\left[ \begin{array}{ c c } x & 3 \\ y& 3 \end{array} \right]$

10. satellite73 Group Title

assuming my squaring is correct

11. satellite73 Group Title

so that means each entry has to be the same for the matrices to be equal

12. satellite73 Group Title

therefore $$3x+9=3\implies x=-2$$ and $$3y+3=3\implies y=-2$$ and so the matrix must be $\left[ \begin{array}{ c c } -2 & 3 \\ -2 & 3 \end{array} \right]$ by substituting back

13. satellite73 Group Title

sorry i meant $$3y+9=3\implies x=-2$$

14. monroe17 Group Title

ohhh okay!

15. satellite73 Group Title

you can check that this is right by squaring and seeing that you get the same matrix back

16. satellite73 Group Title

also you might want to check that i squared the matrix correctly, but i think it is right

17. satellite73 Group Title

gotta run

18. monroe17 Group Title

okay:) So, the steps to a problem like this is to first follow the formula (A^2=A) for the matrix. Then just solve..

19. monroe17 Group Title

when it said A^2=A I first thought it mean like x^2 and 3^2.. i haven't learned this yet lol