The number of elements in the set { A = ( a, b; 0, d ):a,b,d ∈ { - 1,0,1} and (I - A)³ = I - A³ }, where I is 2 × 2 identity matrix, is :

Solution

Maths sir · 3 steps

Expand (I - A)³ like an ordinary cube; the condition collapses to A² = A.

1. Expand the cube. I commutes with every matrix, so the binomial expansion works here. The condition becomes I - 3A + 3A² - A³ = I - A³, which leaves 3A² - 3A = O.

   A² = A ⇒ (a², ab + bd; 0, d²) = (a, b; 0, d)

2. Compare entries. a² = a gives a ∈ {0, 1}, and d² = d gives d ∈ {0, 1}. So -1 is ruled out for both. The top-right entry gives b(a + d) = b, that is b(a + d - 1) = 0.

   (a, d)	a + d - 1	Allowed b	Count
   (0, 0)	-1	b = 0 only	1
   (1, 1)	1	b = 0 only	1
   (1, 0)	0	any of -1, 0, 1	3
   (0, 1)	0	any of -1, 0, 1	3

3. So. The total is 1 + 1 + 3 + 3 = 8 matrices.

The common slip: 12. Letting b take all three values in every case ignores b(a + d - 1) = 0 and gives 4 × 3 = 12, not 8.

The idea to keep. Any polynomial identity in A and I expands like ordinary algebra, because I commutes with A. Reduce the condition first, then compare entries.

Answer from the NTA final answer key. Solution by PrepWiser.