Let , such that det(A) = 0 and . If I denotes identity matrix, then the matrix is: [2025]

Question

Let $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], α > 0$ , such that det(A) = 0 and $α + β = 1 .$ If $I$ denotes $2 \times 2$ identity matrix, then the matrix ${(I + A)}^{8}$ is: [2025]

Smart Achievers · Accepted Answer

(4)

We have, $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], | A | = 0$

$\Rightarrow α β + 6 = 0 \Rightarrow α β = - 6$

Also, $α + β = 1$ [Given]

$\Rightarrow α = 3 and β = - 2$ ( $∵ α > 0$ )

Now, $A = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] \Rightarrow A^{2} = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}] = [\begin{matrix} 3 & - 1 \\ 6 & - 2 \end{matrix}]$

$\Rightarrow A^{2} = A$

So, $A = A^{2} = A^{3} = A^{4} = A^{5} = A^{6} = A^{7}$

$∴ {(I + A)}^{8} = I + {}^{8}C_{1} A^{7} + {}^{8}C_{2} A^{6} + . . . + {}^{8}C_{8} A^{8}$

$= I + A ({}^{8}C_{1} + {}^{8}C_{2} + . . . + {}^{8}C_{8}) = I + A (2^{8} - 1)$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + [\begin{matrix} 765 & - 255 \\ 1530 & - 510 \end{matrix}] = [\begin{matrix} 766 & - 255 \\ 1530 & - 509 \end{matrix}]$ .

Solution

Q.
Let $A = [\begin{matrix} α & - 1 \\ 6 & β \end{matrix}], α > 0$ , such that det(A) = 0 and $α + β = 1 .$ If $I$ denotes $2 \times 2$ identity matrix, then the matrix ${(I + A)}^{8}$ is: [2025]

1 $[\begin{matrix} 257 & - 64 \\ 514 & - 127 \end{matrix}]$

2 $[\begin{matrix} 1025 & - 511 \\ 2024 & - 1024 \end{matrix}]$

3 $[\begin{matrix} 4 & - 1 \\ 6 & - 1 \end{matrix}]$

4 $[\begin{matrix} 766 & - 255 \\ 1530 & - 509 \end{matrix}]$

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Solution

Q. Let A=[α–16β], 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: [2025]

1 [257–64514–127]

2 [1025–5112024–1024]

3 [4–16–1]