Matrices & Determinants
Matrices and Determinants
Grade Class 12
Question:
Let α and β be the distinct roots of the equation x<sup>2</sup> + x - 1 = 0. Consider the set T = {1, α, β}. For a 3 × 3 matrix M = (a<sub>ij</sub>)<sub>3×3</sub>, define R<sub>i</sub> = a<sub>i1</sub> + a<sub>i2</sub> + a<sub>i3</sub> and C<sub>j</sub> = a<sub>1j</sub> + a<sub>2j</sub> + a<sub>3j</sub> for i = 1, 2, 3 and j = 1, 2, 3. Match each entry in List-I to the correct entry in List-II.<br><b>List-I</b><br>(P) The number of matrices M = (a<sub>ij</sub>)<sub>3×3</sub> with all entries in T such that R<sub>i</sub> = C<sub>j</sub> = 0 for all i, j, is<br>(Q) The number of symmetric matrices M = (a<sub>ij</sub>)<sub>3×3</sub> with all entries in T such that C<sub>j</sub> = 0 for all j, is<br>(R) Let M = (a<sub>ij</sub>)<sub>3×3</sub> be a skew symmetric matrix such that a<sub>ij</sub> ∈ T for i > j. Then the number of elements in the set { (x, y, z) ∈ R<sup>3</sup> : M(x, y, z)<sup>T</sup> = 0 } is<br>(S) Let M = (a<sub>ij</sub>)<sub>3×3</sub> be a matrix with all entries in T such that R<sub>i</sub> = 0 for all i. Then the absolute value of the determinant of M is<br><b>List-II</b><br>(1) 1<br>(2) 12<br>(3) infinite<br>(4) 6<br>(5) 0
(A) (P) → (4) (Q) → (2) (R) → (5) (S) → (1)
(B) (P) → (2) (Q) → (4) (R) → (1) (S) → (5)
(C) (P) → (2) (Q) → (4) (R) → (3) (S) → (5)
(D) (P) → (1) (Q) → (5) (R) → (3) (S) → (4)
Step-by-Step Solution
Key Concept: Use properties of roots of quadratic equations (sum = -1, product = -1) and properties of matrix rows/columns sums to determine constraints on matrix entries.
The roots of x^2 + x - 1 = 0 are \alpha, \beta = (-1 \pm \sqrt{5})/2. Note \alpha + \beta = -1 and \alpha\beta = -1. For (P), R_i = C_j = 0 implies row and column sums are zero. For (Q), symmetry and column sum constraints. For (R), skew-symmetric matrix with diagonal 0, M(x,y,z)^T = 0 has non-trivial solutions if det(M)=0. For (S), if row sums are 0, then (1,1,1)^T is an eigenvector with eigenvalue 0, so det(M)=0.
Correct Answer: 3