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
Steve M
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

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