Let A = {X = (x, y, z) T : PX = 0 and
x² + y² + z² = 1} where
P = [ 1, 2, 1; - 2, 3, - 4; 1, 9, - 1 ],
then the set A :

Solution

Maths sir · 4 steps

Find the line of solutions of PX = 0, then see where the unit sphere cuts it.

1. Check the determinant. If |P| ≠ 0, the only solution of PX = 0 is X = 0. That point fails x² + y² + z² = 1, so first find out whether |P| is zero.

   |P| = 1(-3+36) - 2(2+4) + 1(-18-3) = 33 - 12 - 21 = 0

2. A line of solutions. Rows 1 and 2 are not proportional, and row 3 equals 3R₁ + R₂. So the rank is 2, and the solutions form one line through the origin. Its direction is perpendicular to (1, 2, 1) and (-2, 3, -4), and their cross product gives (-11, 2, 7).
3. Apply the sphere. Put X = t(-11, 2, 7) into the sphere equation: t²(121 + 4 + 49) = 1, so 174t² = 1 and t = ±1/√174.
4. So. A line through the centre of the unit sphere meets it at exactly two opposite points, ±1/√174(-11, 2, 7). The set A has exactly two elements, option (c).

The idea to keep. When |P| = 0 with rank 2, PX = 0 gives a line through the origin. A unit-length condition then picks out exactly two points, ± the unit vector along that line.

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