Let A=[ array cc α & -1 6 & β array ], α 0 , such that det(A)=0 and α+β=1 . If I denotes 2 × 2 identity matrix, then the matrix (1+A)^8 is:

Correct answer: D. [ array cc 766 & -255 1530 & -509 array ] aligned & |A|=0 & α β+6=0 & α β=-6 & α+β=1 & α=3, β=-2 & ~A=[ array ll 3 & -1 6 & -2 array ] & A^2=[ array ll 3 & -1 6 & -2 array ][ array ll 3 & -1 6 & -2 array ]=[ array ll 3 & -1 6 & -2 array ] & A^2=A & A=A^2=A^3=A^4=A^5 & (I+A)^8 & =I+ ^8 C_1 ~A^7+ ^8 C_2 ~A^6+ . .+ ^8 C_8 ~A^8 & =I+A( ^8 C_1+ ^8 C_2+ . .+ ^8 C_8) & =I+A(2^8-1) & =[ array ll 1 & 0 0 & 1 array ]+[ array cc 765 & -255 1530 & -510 array ] aligned =[ array cc 766 & -255 1530 & -509 array ]

JEE Main 20252 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2025 · 2 April · Morning Shift · Mathematics · Matrices

Let A=[ array cc α & -1 6 & β array ], α 0 , such that det(A)=0 and...

Let

A=\left[

\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}

\right], \alpha \gt 0

, such that

\operatorname{det}(A)=0

and

\alpha+\beta=1

. If I denotes

2 \times 2

identity matrix, then the matrix

(1+\mathrm{A})^8

is:

Options

A $\left[$ $\begin{array}{ll}4 & -1 \\ 6 & -1\end{array}$ $\right]$
B $\left[$ $\begin{array}{cc}257 & -64 \\ 514 & -127\end{array}$ $\right]$
C $\left[$ $\begin{array}{cc}1025 & -511 \\ 2024 & -1024\end{array}$ $\right]$
D $\left[$ $\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}$ $\right]$

Correct answer

D. $\left[$ $\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}$ $\right]$

Step-by-step solution

\begin{aligned} & |\mathrm{A}|=0 \\ & \alpha \beta+6=0 \\ & \alpha \beta=-6 \\ & \alpha+\beta=1 \\ & \Rightarrow \alpha=3, \beta=-2 \\ & \mathrm{~A}=\left[\begin{array}{ll}3 & -1 \\ 6 & -2\end{array}\right] \\ & \mathrm{A}^2=\left[\begin{array}{ll}3 & -1 \\ 6 & -2\end{array}\right]\left[\begin{array}{ll}3 & -1 \\ 6 & -2\end{array}\right]=\left[\begin{array}{ll}3 & -1 \\ 6 & -2\end{array}\right] \\ & \therefore \mathrm{A}^2=\mathrm{A} \\ & \mathrm{A}=\mathrm{A}^2=\mathrm{A}^3=\mathrm{A}^4=\mathrm{A}^5 \\ & (\mathrm{I}+\mathrm{A})^8 \\ & =\mathrm{I}+{ }^8 \mathrm{C}_1 \mathrm{~A}^7+{ }^8 \mathrm{C}_2 \mathrm{~A}^6+\ldots . .+{ }^8 \mathrm{C}_8 \mathrm{~A}^8 \\ & =\mathrm{I}+\mathrm{A}\left({ }^8 \mathrm{C}_1+{ }^8 \mathrm{C}_2+\ldots . .+{ }^8 \mathrm{C}_8\right) \\ & =\mathrm{I}+\mathrm{A}\left(2^8-1\right) \\ & =\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}765 & -255 \\ 1530 & -510\end{array}\right]\end{aligned}

=\left[

\begin{array}{cc}766 & -255 \\ 1530 & -509\end{array}

$\right]

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