Gujarati
3 and 4 .Determinants and Matrices
normal

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 $M=\left(a_{\ell}\right) 3 \times 3_3$, define $R_l=a_{l 1}+a_{l 2}+a_\beta$ and $C_j=a_{1 j}+a_{2 l}+a_{3 j}$ 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 $M=\left(a_{i j}\right)_3 \times 3$ 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 $M=\left(a_{i j}\right) 3 \times 3$ with all entries in $T$ such that $C_j=0$ for all $j$ is ($2$) ($2$)
($R$) Let $M=\left(a_{i j}\right) 3 \times 3$ be a skew symmetric matrix such that $a_{i j} \in T$ 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 is ($3$) Infinite
($S$) Let $M=\left(a_{i j}\right)_3 \times 3$ 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

$( P ) \rightarrow(4)( Q ) \rightarrow(2)( R ) \rightarrow(5)( S ) \rightarrow(1)$

B

$(P) \rightarrow (2) (Q) \rightarrow (4) (R) \rightarrow (1) (S) \rightarrow (5)$

C

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

D

$(P) \rightarrow (1) (Q) \rightarrow (5) (R) \rightarrow (3) (S) \rightarrow (4)$

(IIT-2024)

Solution

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

$\therefore \alpha+\beta=-1 \Rightarrow 1+\alpha+\beta=0$

$M=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$

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

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.

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

$a y+b z=a$

$-a x+c z=0$

$-b x-c y=-c$

It is observed that $D=D_x=D_y=D_z=0$

$\therefore$ infinite solution

$(S)$ $\left[\begin{array}{lll}1 & \alpha & \beta \\ \beta & \alpha & 1 \\ \alpha & 1 & \beta\end{array}\right]$

$\Rightarrow \alpha \beta-1-\alpha \beta^2+\alpha^2+\beta^2-\alpha^2 \beta=0 \quad$ (since $\alpha \beta=\alpha+\beta=-1$ )

Standard 12
Mathematics

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