JEE Main 2022 · 29 June · evening · Single correct · +4 / −1
Let A = ( 2, - 1; 0, 2 ). If B = I - ⁵C₁(adj A) + ⁵C₂(adj A)² - ..... - ⁵C₅(adj A)⁵, then the sum of all elements of the matrix B is
- (a) -5
- (b) -6
- (c) -7
- (d) -8
Answer and solution
Answer (c) -7
Solution
Maths sir · 4 steps
That alternating sum is a binomial expansion in disguise: B is just (I - adj A)⁵.
- Spot the binomial. I commutes with every matrix, so the binomial theorem works for I and adj A. The alternating signs and the coefficients ⁵Cᵣ give B = (I - adj A)⁵.
- Find adj A. For a 2 × 2 matrix, swap the diagonal entries and change the sign of the off-diagonal ones. This gives adj A = (2, 1; 0, 2), so I - adj A = (-1, -1; 0, -1) = -(I + N), where N = (0, 1; 0, 0).
- Higher powers vanish. N² = O, so every term with N² or a higher power vanishes when you expand (I + N)⁵. Only I + 5N survives.
B = (-1)⁵ (I + N)⁵ = -(I + 5N) = (-1, -5; 0, -1)
- So. The sum of all elements is -1 - 5 + 0 - 1 = -7, which is option (c).
The idea to keep. When you see ⁿCᵣ coefficients on powers of a matrix, collapse the sum into (I ± M)ⁿ. Then split off a nilpotent part (N² = O) so the power has only two terms.
Answer from the NTA final answer key. Solution by PrepWiser.