Question #0ed08

1 Answer
May 5, 2016

Answer:

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

Explanation:

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)#