Question

Let α and β be the distinct roots of the equation \(x^2 + x - 1 = 0\). Consider the set T = {1, α, β}. For a 3 x 3 matrix \(M = (a_{ij})_{3 \times 3}) \), define \(R_i\) = \(a_{i 1 }+ a_{i 2} + a_{i3}\) and \(C_j\) = \(a_{1j }+ a_{2j }+ a_{3j}\) for i = 1, 2, 3 and j = 1, 2, 3 

Match each entry in List-I to the correct entry in List-II.

	List-I		List-II
(P)	The number of matrices M = \((a_{ij})_{3 \times 3}\) with all entries in T such that \(R_i\) = \(C_j\) = 0 for all i, j is	(1)	1
(Q)	The number of symmetric matrices M = \((a_{ij})_{3 \times 3}\) with all entries in T such that \(C_j\) = 0 for all j is	(2)	12
(R)	Let M = \((a_{ij})_{3 \times 3}\) be a skew symmetric matrix such that \(a_{ij}\) ∈ T for i > j. Then the number of elements in the set \(\left\{\left(\begin{array}{l} x \\ y \\ z \end{array}\right): x, y \cdot z \in R, M\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} a_{12} \\ 0 \\ -a_{23} \end{array}\right)\right\}\) is	(3)	Infinite
(S)	Let M = \((a_{ij})_{3 \times 3}\) be a matrix with all entries in T such that \(R_i\) = 0 for all i. Then the absolute value of the determinant of M is	(4)	6
		(5)	0

The correct option is 

(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)

Answer 1

Correct option is (C) (P) → (2) (Q) → (4) (R) → (3) (S) → (5) 

\(x^2+x-1=0 \rightarrow\) roots are \(\alpha\) and \(\beta\)

\(\alpha+\beta=-1 \quad \alpha \beta=-1\)

Set \(T=\{1, \alpha, \beta\} \quad M=\left(a_{i j}\right)_{3 \times 3}\)

\(R_i=a_{i 1}+a_{i 2}+a_{i 3} \quad C_j=a_{1 j}+a_{2 j}+a_{3 j}\)

(P) \(R_i=C_j=0\) for all \(i, j\)

\(\alpha+\beta=-1 \quad T=\{1, \alpha, \beta\}\)

Number of matrices

(Q) Number of symmetric matrices = ?

\(C_j=0\) \(\forall\) j

Number of symmetric matrices

\(=\lfloor\underline{3} \times 1=6\left[\begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha \end{array}\right]\)

(R) \(M \rightarrow\) skew symmetric of \(3 \times 3\)

\(|M|=0 \quad a_{i j} \in T \text { for } i>j\)

\(M\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} a_{12} \\ 0 \\ -a_{23} \end{array}\right) \)

\( {\left[\begin{array}{ccc} 0 & -a_{21} & -a_{31} \\ a_{21} & 0 & -a_{32} \\ a_{31} & a_{32} & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} a_{12} \\ 0 \\ -a_{23} \end{array}\right]}\)

As x, y, z ∈ R and \(a_{12}\) & \(a_{23}\) ∈ R & |M| = 0

\(\therefore\) System has infinite solutions.

(S) \(R_i=0 \forall i\)

\(M =\left[\begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha \end{array}\right]\)

\(C_1 \rightarrow C_1 + C_2 + C_3 \) \(|M| =\left[\begin{array}{ccc} 1+\alpha +\beta & \alpha & \beta \\ 1+\alpha +\beta & \beta & 1 \\ 1+\alpha +\beta & 1 & \alpha \end{array}\right] = 0\)

\((P) \rightarrow (2) (Q) \rightarrow (4) (R) \rightarrow (3) (S) \rightarrow (5)\)