Question #0ed08

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Answer 1 · 2016-05-05T09:14:22

$(x, y) = (-4, 1)$

Performing the multiplication, first, we have

$A^2=((3,2),(1,1))^2 = ((3,2),(1,1))((3,2),(1,1))=((11,8),(4,3))$

$xA = x((3,2),(1,1)) = ((3x,2x),(x,x))$

$yI = y((1,0),(0,1)) = ((y,0),(0,y))$

Then, summing, we have

$A^2+xA+yI = ((11,8),(4,3))+((3x,2x),(x,x))+((y,0),(0,y))$

$=((3x+y+11,2x+8),(x+4,x+y+3)) = ((0,0),(0,0))$

Equating entries with corresponding indices, we get the following system of equations from the bottom row:

${(x+4 = 0), (x+y+3 = 0):}$

From the first equation, we have $x = -4$

Substituting this into the second equation, we get

$-4+y+3=0$

$=> y = 1$

If we substitute these values in for $x$ and $y$ we find that they fulfill the given equation, and thus we have $(x, y) = (-4, 1)$