If A & B are two non singular matrices of order 3 × 3 such that A<sup>T</sup> + B = I & BA<sup>T</sup> = -B, then which is/are always true (where X<sup>T</sup> denotes transpose of X and I denotes unit matrix)-
Step-by-Step Solution
Key Concept: Given A^T + B = I, so B = I - A^T. Substitute this into BA^T = -B to get (I - A^T)A^T = -(I - A^T), which simplifies to A^T - (A^T)^2 = -I + A^T, implying (A^T)^2 = I. Thus |A^T|^2 = |I| = 1, so |A|^2 = 1, meaning |A| = 1 or -1. Also, from BA^T = -B, B(A^T + I) = 0. Since A^T + B = I, A^T = I - B. Substituting this, B(I - B + I) = 0, so B(2I - B) = 0, which means B^2 = 2B. Since B is non-singular, B = 2I. Then |B| = |2I| = 2^3 = 8.
Given A<sup>T</sup> + B = I ⇒ B = I - A<sup>T</sup>. Also BA<sup>T</sup> = -B ⇒ B(A<sup>T</sup> + I) = 0. Substituting B = I - A<sup>T</sup>, we get (I - A<sup>T</sup>)(A<sup>T</sup> + I) = 0 ⇒ I - (A<sup>T</sup>)<sup>2</sup> = 0 ⇒ (A<sup>T</sup>)<sup>2</sup> = I. Thus |A|<sup>2</sup> = 1 ⇒ |A| = ±1. From BA<sup>T</sup> = -B, since B is non-singular, we can multiply by B<sup>-1</sup> to get A<sup>T</sup> = -I. But A<sup>T</sup> + B = I ⇒ -I + B = I ⇒ B = 2I. Thus |B| = |2I| = 2<sup>3</sup> = 8. So |B|=8 and |A| can be 1 or -1.
Correct Answer: 2,3