Let A=[a_ i j ] be a matrix of order 3 × 3 , with a_ i j =( 2 )^ i+j . If the sum of all the elements in the third row of A^2 is α+β 2 , α, β ∈ Z , then α+β is equal to :

Correct answer: B. 224 aligned & A=[ array lll ( 2 )^2 & ( 2 )^3 & ( 2 )^4 ( 2 )^3 & ( 2 )^4 & ( 2 )^5 ( 2 )^4 & ( 2 )^5 & ( 2 )^6 array ] & A=[ array ccc 2 & 2 2 & 4 2 2 & 4 & 4 2 4 & 4 2 & 8 array ] & A^2=2^2[ array ccc 1 & 2 & 2 2 & 2 & 2 2 2 & 2 2 & 4 array ][ array ccc 1 & 2 & 2 2 & 2 & 2 2 2 & 2 2 & 4 array ] aligned =4[ array ccc - & - & - - & - & - (2+4+8) & (2 2 +4 2 +8 2 ) & (4+8+16) array ] Sum of elements of 3^ nd row =4(14+14 2 +28) aligned & =4(42+14 2 ) & =168+56 2 & α+β 2 & α+β=168+56=224 aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Matrices

Let A=[a_ i j ] be a matrix of order 3 × 3 , with a_ i j =( 2 )^...

Let

A=\left[a_{i j}\right] be a matrix of order

3 \times 3

, with

a_{i j}=(\sqrt{2})^{i+j}

. If the sum of all the elements in the third row of

A^2

is

\alpha+\beta \sqrt{2}, \alpha, \beta \in \mathbf{Z}

, then

\alpha+\beta is equal to :

Options

A280
B224
C210
D168

Correct answer

B. 224

Step-by-step solution

\begin{aligned} & A=\left[\begin{array}{lll}(\sqrt{2})^2 & (\sqrt{2})^3 & (\sqrt{2})^4 \\ (\sqrt{2})^3 & (\sqrt{2})^4 & (\sqrt{2})^5 \\ (\sqrt{2})^4 & (\sqrt{2})^5 & (\sqrt{2})^6\end{array}\right] \\ & A=\left[\begin{array}{ccc}2 & 2 \sqrt{2} & 4 \\ 2 \sqrt{2} & 4 & 4 \sqrt{2} \\ 4 & 4 \sqrt{2} & 8\end{array}\right] \\ & A^2=2^2\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\left[\begin{array}{ccc}1 & \sqrt{2} & 2 \\ \sqrt{2} & 2 & 2 \sqrt{2} \\ 2 & 2 \sqrt{2} & 4\end{array}\right]\end{aligned}

=4\left[

\begin{array}{ccc}- & - & - \\ - & - & - \\ (2+4+8) & (2 \sqrt{2}+4 \sqrt{2}+8 \sqrt{2}) & (4+8+16)\end{array}

\right]

Sum of elements of

3^{\text{nd }}

row

=4(14+14 \sqrt{2}+28)

\begin{aligned} & =4(42+14 \sqrt{2}) \\ & =168+56 \sqrt{2} \\ & \alpha+\beta \sqrt{2} \\ \therefore & \alpha+\beta=168+56=224 \end{aligned}

Practice Matrices on Quantrex Academy →

More from Matrices