Let A=\left[\begin{array}{cc}\alpha & -1 \ 6 & \beta]\end{array}, \alpha>0, such that {\det}(A)=0 and \alpha+\beta=1. If I denotes 2 \times 2 identity matrix, then the matrix, (I+A)^8 is [2 Apr 2025 Shift 1]

Correct Answer is : [766−2551530−509]

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

\left[\begin{array}{cc}257 & -64 \ 514 & -127]\end{array}

\left[\begin{array}{cc}4 & -1 \ 6 & -1]\end{array}

\left[\begin{array}{cc}1025 & -511 \ 2024 & -1024]\end{array}

Matrices and Determinants Part 4

Section: Mathematics

Question : 30 of 58

Marks: +1, -0

Let A=[

α	−1
6	β

],α>0, such that det(A)=0 and α+β=1. If I denotes 2×2 identity matrix, then the matrix, (I+A)8 is

[2 Apr 2025 Shift 1]

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766	−255
1530	−509

257	−64
514	−127

1025	−511
2024	−1024

3	−1
6	−2