MATRICES: Determine a in this matrix so that the matrix is nilpotent with p = 3. ?

#((a,1,3),(5,2,6),(-2,-1,-3))^3 = ((0,0,0),(0,0,0),(0,0,0))#

1 Answer
Oct 10, 2017

#a=1#

Explanation:

Let:

# bb(A) = ( (a,1,3), (5,2,6), (-2,-1,-3) )# so that #bb(A)^3=bb(0) #

Let us compute #bb(A)^2#:

# bb(A)^2 = ( (a,1,3), (5,2,6), (-2,-1,-3) ) ( (a,1,3), (5,2,6), (-2,-1,-3) ) #

# \ \ \ \ \ = ( (a^2+5-6,a+2-3,3a+6-9), (5a+10-12,5+4-6,15+12-18), (-2a-5+6,-2-2+3,-6-6+9) ) #

# \ \ \ \ \ = ( (a^2-1,a-1,3a-3), (5a-1,3,9), (1-2a,-1,-3) ) #

And now we can compute #bb(A)^3#:

# bb(A)^3 = ( (a^2-1,a-1,3a-3), (5a-1,3,9), (1-2a,-1,-3) ) ( (a,1,3), (5,2,6), (-2,-1,-3) ) #

# \ \ \ \ \ = ( (a^3-2a+1,a^2-a,3a^2-3a), (5a^2-2a-3,5a-5,15a-15), (a+1-2a^2,2-2a,6-6a) ) #

Equating # bb(A)^3# with the null matrix, we have:

# ( (a^3-2a+1,a^2-a,3a^2-3a), (5a^2-2a-3,5a-5,15a-15), (a+1-2a^2,2-2a,6-6a) ) = ( (0,0,0), (0,0,0), (0,0,0) ) #

Randomly equating row #3# and column #3# we have:

# 6-6a = 0 => a = 1 #

And we can quickly verify that #a=1# simultaneously satisfies the remaining #8# elements .

Hence #a=1#