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 × 3 matrix M=(a_ i j )_ 3 × 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

Correct answer: C. (P) arrow(2) (Q) arrow(4) (R) arrow(3) (S) arrow(5) aligned & α, β are roots of x^2+x-1=0 & α+β=-1 1+α+β=0 & M= [ array lll a_ 11 & a_ 12 & a_ 13 a_ 21 & a_ 22 & a_ 23 a_ 31 & a_ 32 & a_ 33 array ] aligned (P) M=[ array lll 1 & α & β α & β & 1 β & 1 & α array ] 3!× 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 3!× 2=12 ways (Q) M=[ array lll x & a & b a & y & c b & c & z array ] Consider one such arrangement with a=α, b=β, c=1 M= [ array lll 1 & α & β α & β & 1 β & 1 & α array ] a, b, c can be arranged in 3 ! ways and corresponding entries can be arranged in 1 way. (R) [ array ccc 0 & a & b -a & 0 & c -b & -c & 0 array ] [ array l x y z array ]= [ array c a 0 -c array ] aligned & a y+b z=a \& -a x+c z=0 \& -b x-c y=-c aligned It is observed that D=D_ x =D_ y =D_ z =0 infinite solution (S) [ array lll 1 & α & β β & α & 1 α & 1 & β array ] α β-1-α β^2+α^2+β^2-α^2 β=0 ( since α β=α+β=-1 )

JEE Advanced 2024Paper 1

Exam: JEE Advanced
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Matrices

JEE Advanced 2024 · Paper 1 · Mathematics · Matrices

Let α and β be the distinct roots of the equation x^2+x-1=0 ....

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. $Let α and β be the distinct roots of the equation x^2+x-1=0 . Consider the set T=\ 1, α, β\ . For a 3 × 3 matrix M=(a_$ The correct option is

Options

A $(\mathrm{P}) \rightarrow(4) \quad(\mathrm{Q}) \rightarrow(2) \quad(\mathrm{R}) \rightarrow(5) \quad(\mathrm{S}) \rightarrow(1)$
B $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(5)$
C $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)$
D $(\mathrm{P}) \rightarrow(1) \quad(\mathrm{Q}) \rightarrow(5) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(4)$

Correct answer

C. $(\mathrm{P}) \rightarrow(2) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(3) \quad(\mathrm{S}) \rightarrow(5)$

Step-by-step solution

\begin{aligned} & \alpha, \beta \text{ are roots of } \mathrm{x}^2+\mathrm{x}-1=0 \\ & \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)

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)

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 M=

\left[

\begin{array}{lll}1 & \alpha & \beta \\ \alpha & \beta & 1 \\ \beta & 1 & \alpha\end{array}

\right]

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]

\begin{aligned}& 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)

\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$ )

Practice Matrices on Quantrex Academy →

More from Matrices