Let A=[2, a, 0; 1, 3, 1; 0, 5, b]. If A³=4 A²-A-21 I, where I is the identity matrix of order 3 × 3, then 2 a+3 b is equal to

Solution

Maths sir · 5 steps

Read the given equation as A's characteristic equation, then match trace, minors and determinant.

1. Use Cayley–Hamilton. Every 3 × 3 matrix satisfies its characteristic equation λ³ - (trA)λ² + Sλ - |A| = 0. Here S is the sum of the three principal 2 × 2 minors. The given relation is A³ - 4A² + A + 21I = O, so match it term by term.

   trA = 4, S = 1, |A| = -21

2. Trace gives b. trA = 2 + 3 + b = 4, so b = -1.
3. Minors give a. The principal minors are 3b - 5 = -8, 2b - 0 = -2 and 6 - a. Their sum is -4 - a = 1, so a = -5.
4. Check determinant. |A| = 2(3b - 5) - a(b - 0) = 2(-8) - (-5)(-1) × (-1). That works out to -16 + a = -21, which matches. So a = -5 and b = -1 are consistent.
5. So. 2a + 3b = -10 - 3 = -13, which is option (c).

The idea to keep. A cubic in A equal to zero is a hint to use Cayley–Hamilton: the coefficients give the trace, the sum of principal minors and the determinant.

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