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**Hint:**We will tackle both the statements one by one. By the use of PMI, we will try to establish the given result. If we are successful, then both are true otherwise whichever option is applicable.

**Complete step-by-step answer:**

Let us consider the Statement I for now that is “If $A = \left( {\begin{array}{*{20}{c}}

a&0 \\

0&b

\end{array}} \right)$ ……(1),

then ${A^n} = \left( {\begin{array}{*{20}{c}}

{{a^n}}&0 \\

0&{{b^n}}

\end{array}} \right)$ for all $n \in {\rm N}$”.

We will use the Principle of Mathematical Induction.

Principle of Mathematical Induction is done by two steps which are as follows:

Step 1. Show it is true for the first case, usually n = 1.

Step 2. Show that if n = k is true then n = k + 1 is also true.

Applying this on ${A^n} = \left( {\begin{array}{*{20}{c}}

{{a^n}}&0 \\

0&{{b^n}}

\end{array}} \right)$:

Step 1: Putting n = 1 in ${A^n} = \left( {\begin{array}{*{20}{c}}

{{a^n}}&0 \\

0&{{b^n}}

\end{array}} \right)$, we get : $A = \left( {\begin{array}{*{20}{c}}

a&0 \\

0&b

\end{array}} \right)$ which is obviously true.

Step 2: Let it be true for n = k.

So, ${A^k} = \left( {\begin{array}{*{20}{c}}

{{a^k}}&0 \\

0&{{b^k}}

\end{array}} \right)$ ……..(2)

Now let n = k + 1

So, we need to prove that ${A^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{{a^{k + 1}}}&0 \\

0&{{b^{k + !}}}

\end{array}} \right)$.

We can write ${A^{k + 1}} = {A^k}.A$.

${A^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{{a^k}}&0 \\

0&{{b^k}}

\end{array}} \right).\left( {\begin{array}{*{20}{c}}

a&0 \\

0&b

\end{array}} \right)$ (Using (1) and (2))

$\therefore {A^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{{a^k}.a + 0}&0 \\

{a.0 + {b^k}.0}&{0 + {b^k}.b}

\end{array}} \right)$ (By Multiplication of matrices)

Hence, ${A^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{{a^{k + 1}}}&0 \\

0&{{b^{k + 1}}}

\end{array}} \right)$

Hence, proved.

Therefore, by principle of mathematical induction ${A^n} = \left( {\begin{array}{*{20}{c}}

{{a^n}}&0 \\

0&{{b^n}}

\end{array}} \right)$ is true for all $n \in {\rm N}$.

Now, we will start the same way with Statement II: If $P = \left( {\begin{array}{*{20}{c}}

3&{ - 4} \\

1&{ - 1}

\end{array}} \right)$, ……(3)

then ${P^n} = \left( {\begin{array}{*{20}{c}}

{1 + 2n}&{ - 4n} \\

n&{1 - 2n}

\end{array}} \right)$ for all $n \in {\rm N}$.

Step 1: Putting n = 1 in ${P^n} = \left( {\begin{array}{*{20}{c}}

{1 + 2n}&{ - 4n} \\

n&{1 - 2n}

\end{array}} \right)$, we get : $P = \left( {\begin{array}{*{20}{c}}

3&{ - 4} \\

1&{ - 1}

\end{array}} \right)$ which is obviously true.

Step 2: Let it be true for n = k.

So, ${P^k} = \left( {\begin{array}{*{20}{c}}

{1 + 2k}&{ - 4k} \\

k&{1 - 2k}

\end{array}} \right)$ ……..(3)

Now let n = k + 1

So, we need to prove that ${P^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{1 + 2(k + 1)}&{ - 4(k + 1)} \\

{k + 1}&{1 - 2(k + 1)}

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

{3 + 2k}&{ - 4k - 4} \\

{k + 1}&{ - 1 - 2k}

\end{array}} \right)$.

We can write ${P^{k + 1}} = {P^k}.P$.

${P^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{1 + 2k}&{ - 4k} \\

k&{1 - 2k}

\end{array}} \right).\left( {\begin{array}{*{20}{c}}

3&{ - 4} \\

1&{ - 1}

\end{array}} \right)$ (Using (3) and (4))

$\therefore {P^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{3 + 6k - 4k}&{ - 4 - 8k + 4k} \\

{3k + 1 - 2k}&{ - 4k - 1 + 2k}

\end{array}} \right) = \left( {\begin{array}{*{20}{c}}

{3 + 2k}&{ - 4 - 4k} \\

{k + 1}&{ - 2k - 1}

\end{array}} \right)$ (By Multiplication of matrices)

Hence, ${P^{k + 1}} = \left( {\begin{array}{*{20}{c}}

{3 + 2k}&{ - 4k - 4} \\

{k + 1}&{ - 1 - 2k}

\end{array}} \right)$

Hence, proved.

Therefore, by principle of mathematical induction ${P^n} = \left( {\begin{array}{*{20}{c}}

{1 + 2n}&{ - 4n} \\

n&{1 - 2n}

\end{array}} \right)$ is true for all $n \in {\rm N}$.

**So, the correct answer is “Option D”.**

**Note:**The students might make the mistake of starting to solve such question by find the square, cube and so on of matrices, but that will give us a vague proof because, we have not actually shown that it is happening for all n, we just represented that it is happening for few values of n.

Also take care while multiplying the matrices because that is a confusing task in itself.

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