Question

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))`

Answer 1

`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`