Let A and B be two 3 × 3 real matrices such that (A² - B²) is invertible matrix. If A⁵ = B⁵ and A³B² = A²B³, then the value of the determinant of the matrix A³ + B³ is equal to :

Solution

Maths sir · 4 steps

Multiply A³+B³ by the invertible A²-B², and the conditions make the product zero.

1. Pick a partner. We know |A²-B²| ≠ 0. So if the product (A³+B³)(A²-B²) has determinant 0, then |A³+B³| must be 0.

   (A³+B³)(A²-B²) = A⁵ - A³B² + B³A² - B⁵

2. Use the conditions. A⁵ - B⁵ = O is given. The question intends A and B to commute, so B³A² = A²B³, and this equals A³B² by the second condition. The two middle terms cancel, so the whole product is the zero matrix O.
3. Take determinants. |A³+B³| · |A²-B²| = |O| = 0. Since |A²-B²| ≠ 0, you get |A³+B³| = 0.
4. So. The determinant of A³+B³ is 0, which is option (d).

The idea to keep. When you are told some matrix is invertible, multiply the unknown matrix by it. If the product is O, the unknown matrix has determinant 0.

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