MATRICES: How you determine x and y so that ... ?

Question

$((0,x),(x,y))²$ = $((x,x),(x,x+y))$

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Answer 1 · 2017-10-10T16:42:00

We have four possible solution

$x=0, y=0$
$x=0, y=1$
$x=1, y=0$
$x=1, y=1$

We seek $x$ and $y$ such that:

$( (0,x),(x,y) )^2 = ( (x,x),(x,x+y) )$

And so:

$( (0,x),(x,y) ) ( (0,x),(x,y) ) = ( (x,x),(x,x+y) )$

So then multiplying the two matrices together we have:

$( (0+x^2,0+xy),(0+xy,x^2+y^2) ) = ( (x,x),(x,x+y) )$

$:. ( (x^2,xy),(xy,x^2+y^2) ) = ( (x,x),(x,x+y) )$

If we equate elements we get four equations:

$[A]: x^2 = x$
$[B]: xy = x$
$[C]: xy = x$
$[D]: x^2+y^2 = x+y$

From Equation [A] we have:

$x^2=x => x^2-x= 0$
$:. x(x-1)= 0$
$:. x=0,1$

From Equation [B] (identical to [C]) we get:

$xy = x => xy - x =0$
$:. x(y-1) = 0$
$:. x=0,y=1$

From Equation [D] we get:

$x^2+y^2 = x+y$

When $x=0=> y^2=y => y(y-1) = 0 => y=0,1$
When $x=1=> 1+y^2=1+y => y^2=y => y=0,1$
When $y=1=> x^2+1=x+1 => x^2=x => x=0,1$

Thus, we have four possible solution

$x=0, y=0$
$x=0, y=1$
$x=1, y=0$
$x=1, y=1$