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JEE Advance - Mathematics (2024 - Paper 1 Online - No. 14)

Let $\alpha$ and $\beta$ be the distinct roots of the equation $x^2+x-1=0$. Consider the set $T=\{1, \alpha, \beta\}$. For a $3 \times 3$ matrix $M=\left(a_{i j}\right)_{3 \times 3}$, define $R_i=a_{i 1}+a_{i 2}+a_{i 3}$ and $C_j=a_{1 j}+a_{2 j}+a_{3 j}$ 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})_{3x3} $ 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})_{3x3} $ with all entries in $ T $ such that $ C_j = 0 $ for all $ j $, is (2) 12
(R) Let $ M = (a_{ij})_{3x3} $ be a skew symmetric matrix such that $ a_{ij} \in T $ for $ i > j $.

Then the number of elements in the set

$ \left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} : x, y, z \in \mathbb{R}, M \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_{12} \\ 0 \\ a_{13} \end{pmatrix} \right\} $ is		 (3) infinite
(S) Let $ M = (a_{ij})_{3x3} $ 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

The correct option is
(P) $\rightarrow$ (4) $\quad$ (Q) $\rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad$ (S) $\rightarrow$ (1)
$(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)$
$(\mathrm{P}) \rightarrow(2) \quad$ (Q) $\rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad$ (S) $\rightarrow$ (5)
(P) $\rightarrow$ (1) $\quad$ (Q) $\rightarrow$ (5) $\quad$ (R) $\rightarrow$ (3) $\quad$ (S) $\rightarrow$ (4)

Explanation

$$\alpha, \beta$$ are roots of $$x^2+x-1=0$$

$$\begin{aligned} & \therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0 \\ & M=\left[\begin{array}{lll} \mathrm{a}_{11} & \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{21} & \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right] \end{aligned}$$

(P) $$\quad M=\left[\begin{array}{lll}1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha\end{array}\right] \Rightarrow 3 ! \times 2=12$$

For one arrangement of row 1 we can arrange other two rows exactly in two ways and row 1 can be arranged in 3! ways

$$\therefore 3!\times 2=12 \text { ways }$$

(Q) $$\quad M=\left[\begin{array}{lll}x & a & b \\ a & y & c \\ b & c & z\end{array}\right] \Rightarrow$$ Consider one such arrangement with $$a=\alpha, b=\beta, c=1$$

a, b, c can be arranged in 3! ways and corresponding entries can be arranged in 1 way.

(R)

$$\begin{aligned} & {\left[\begin{array}{ccc} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} a \\ 0 \\ -c \end{array}\right]} \\ & a y+b z=a \\ & -a x+c z=0 \\ & -b x-c y=-c \end{aligned}$$

It is observed that $$\mathrm{D}=\mathrm{D}_{\mathrm{x}}=\mathrm{D}_{\mathrm{y}}=\mathrm{D}_{\mathrm{z}}=0$$

$$\therefore$$ infinite solution

(S)

$$\begin{aligned} & {\left[\begin{array}{lll} 1 & \alpha & \beta \\ \beta & \alpha & 1 \\ \alpha & 1 & \beta \end{array}\right] } \\ & \Rightarrow \alpha \beta-1-\alpha \beta^2+\alpha^2+\beta^2-\alpha^2 \beta=0 \quad(\text { since } \alpha \beta=\alpha+\beta=-1) \end{aligned}$$

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