Here's the question you clicked on:
monroe17
If A= x 3 y 3 (matrix) determine the values of x and y for which A^2=A
i mean square the matrix
i got \[\left[ \begin{array}{ c c } x^2+3y & 3x+9 \\ xy+3y & 3y+9 \end{array} \right]\]
assuming the matrix is \[\left[ \begin{array}{ c c } x & 3 \\ y & 3 \end{array} \right]\]
that means \[3x+9=3\] so \[x=-2\] and also \(y=-2\)
once you square the matrix, how do you determine the x and y values though?
oh nevermind. how'd you get 3x+9=3
you have \[\left[ \begin{array}{ c c } x^2+3y & 3x+9 \\ x^2+3y & 3y+9 \end{array} \right]=\left[ \begin{array}{ c c } x & 3 \\ y& 3 \end{array} \right]\]
assuming my squaring is correct
so that means each entry has to be the same for the matrices to be equal
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
sorry i meant \(3y+9=3\implies x=-2\)
you can check that this is right by squaring and seeing that you get the same matrix back
also you might want to check that i squared the matrix correctly, but i think it is right
okay:) So, the steps to a problem like this is to first follow the formula (A^2=A) for the matrix. Then just solve..
when it said A^2=A I first thought it mean like x^2 and 3^2.. i haven't learned this yet lol