A2=100101010100101010=100110101A3=100101011100110101=100201110A3−A2=0001−1101−1and A−I=0001−1101−1⇒ A3−A2=A−I (1)Now, det⁡An−I=det⁡(A−I)I+A+A2+…+An−1=det.(A−I)×det.I+A+A2+…+An−1=0From (1), A4 - A3 = A2 - A (2)Again from (1), A5 - A4 = A3 - A2 = A - IThus, if n is even, An - An-1 = A2 - A (3)If n is odd, An - An-1 = A - I (4)Consider that n is even.∴ An−An−1=A2−A [from(3)]and An−1−An−2=A−I [from(4)]Adding these, we getAn−An−2=A2−I⇒ An=An−2+A2−I =An−4+A2−I+A2−I =An−6+A2−I+2A2−I ⋯ … … ⋯ ⋯ … =A2+n−22A2−I∴ An=n2A2−n−22I∴ A200=100A2−99I =100100110101−99100010001=1001001010001

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If $A = [\begin{matrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix}]$ , then

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$A^{3} - A^{2} = A - I$

$\det . (A^{100} - I) = 0$

$A^{200} = [\begin{array}{ccc} 1 & 0 & 0 \\ 100 & 1 & 0 \\ 100 & 0 & 1 \end{array}]$

$A^{100} = [\begin{array}{ccc} 1 & 1 & 0 \\ 50 & 1 & 0 \\ 50 & 0 & 1 \end{array}]$

answer is A, B, C.

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Detailed Solution

$\begin{array}{l} A^{2} = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}] [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}] \\ A^{3} = [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}] [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 2 & 0 & 1 \\ 1 & 1 & 0 \end{array}] \\ A^{3} - A^{2} = [\begin{array}{ccc} 0 & 0 & 0 \\ 1 & - 1 & 1 \\ 0 & 1 & - 1 \end{array}] \end{array}$

and $A - I = [\begin{array}{ccc} 0 & 0 & 0 \\ 1 & - 1 & 1 \\ 0 & 1 & - 1 \end{array}]$

$\Rightarrow A^{3} - A^{2} = A - I$ (1)

$\begin{matrix} Now, \det (A^{n} - I) = \det ((A - I) (I + A + A^{2} + \dots + A^{n - 1})) \\ = \det . (A - I) \times \det . (I + A + A^{2} + \dots + A^{n - 1}) \\ = 0 \end{matrix}$

From (1), A⁴ - A³ = A² - A (2)

Again from (1), A⁵ - A⁴ = A³ - A² = A - I

Thus, if n is even, Aⁿ- A^n-1= A² - A (3)

If n is odd, Aⁿ - A^n-1 = A - I (4)

Consider that n is even.

$∴ A^{n} - A^{n - 1} = A^{2} - A$ [from(3)]

and $A^{n - 1} - A^{n - 2} = A - I$ [from(4)]

Adding these, we get

$A^{n} - A^{n - 2} = A^{2} - I$

$\begin{array}{l} \Rightarrow A^{n} = A^{n - 2} + A^{2} - I \\ = (A^{n - 4} + A^{2} - I) + A^{2} - I \\ = (A^{n - 6} + A^{2} - I) + 2 (A^{2} - I) \\ \dots \dots \dots \\ \dots \dots \dots \\ = (A^{2}) + \frac{n - 2}{2} (A^{2} - I) \end{array}$

$\begin{array}{l} ∴ A^{n} = (\frac{n}{2}) A^{2} - (\frac{n - 2}{2}) I \\ ∴ A^{200} = 100 A^{2} - 99 I \end{array}$

$= 100 [\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix}] - 99 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 100 & 1 & 0 \\ 100 & 0 & 1 \end{array}]$