MATRICES: How you determine x and y so that ... ?
#((0,x),(x,y))²# = #((x,x),(x,x+y))#
1 Answer
We have four possible solution
# x=0, y=0 #
# x=0, y=1 #
# x=1, y=0 #
# x=1, y=1 #
Explanation:
We seek
# ( (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
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Thus, we have four possible solution
# x=0, y=0 #
# x=0, y=1 #
# x=1, y=0 #
# x=1, y=1 #