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Statement

If A=1011 and...

Question

$If A = (\begin{matrix} 1 & 0 1 & 1 \end{matrix}] and I = (\begin{matrix} 1 & 0 0 & 1 \end{matrix}], then which of the following holds for all n \in N ?$

A

A^{n} = nA - (n - 1) I

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B

A^{n} = 2^{n - 1} A - (n - 1) I

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C

A^{n} = nA + (n - 1) I

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D

A^{n} = 2^{n - 1} A + (n - 1) I

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Solution

The correct option is A $A^{n} = nA - (n - 1) I$
$We shall prove by principle of mathematical induction that first option is correct . Let P (n) be the statement, P (n) : A^{n} = nA - (n - 1) I Consider P (1) LHS = A and RHS = A - (1 - 1) I = A Thus P (n) is true for n = 1 . Assume that P (k) is true for some k \in N . Then A^{k} = kA - (k - 1) I . . . (1) Now, we will show that P (k + 1) is true whenever P (k) is true . Consider A^{k + 1} = A^{k} . A = (kA - (k - 1) I) . A (Using (1)] = {kA}^{2} - (k - 1) A A^{2} = A . A = (\begin{matrix} 1 & 0 1 & 1 \end{matrix}] (\begin{matrix} 1 & 0 1 & 1 \end{matrix}] = (\begin{matrix} 1 & 0 2 & 1 \end{matrix}] we get A^{2} = 2 A - I So, A^{k + 1} = k (2 A - I) - (k - 1) A = 2 kA - kI - (k - 1) A = (2 k - k + 1) A - kI = (k + 1) A - ((k + 1) - 1) I Thus, P (k + 1) is true whenever P (k) is true . Hence, by principle of mathematical induction, P (n) is true for all natural number . Hence, the first option is correct .$ $Then, clearly third option is wrong . Now, consider the second option . A^{n} = 2^{n - 1} A - (n - 1) I For n = 2, LHS = A^{2} = (\begin{matrix} 1 & 0 2 & 1 \end{matrix}] and RHS = 2 A - I = (\begin{matrix} 1 & 0 2 & 1 \end{matrix}] \Rightarrow P (n) is true for n = 2 .$
$For n = 3, LHS = A^{3} = (\begin{matrix} 1 & 0 3 & 1 \end{matrix}] and RHS = 4 A - 2 I = (\begin{matrix} 2 & 0 4 & 2 \end{matrix}] Thus, for n = 3, LHS \neq RHS$

$Similarly, we can show that the fourth option is not true for n = 2 .$

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