#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 62 maths textbook solution

Answer: Hence proved, $A^{n}=\operatorname{diag}\left(a^{n} \quad b^{n} \quad c^{n}\right)$ for all positive integer n.

Hint: We use the principle of mathematical induction.

Given:$A=\operatorname{diag}\left(\begin{array}{lll}a & b & c\end{array}\right)$

Prove: $A^{n}=\operatorname{diag}\left(a^{n} \quad b^{n} \quad c^{n}\right)$ for all positive integer n             …(i)

Solution: step 1:

put n=1 in eqn (i)

$A^{1}=\operatorname{diag}\left(a^{1} \quad b^{1} \quad c^{1}\right) \ \ [ x^1 = x ]$

$A=\operatorname{diag}\left(\begin{array}{lll}a & b & c\end{array}\right)$

So, $A^n$ is true for n=1.

Step 2: let $A^n$ be true for n=k, so

$A^{k}=\operatorname{diag}\left(a^{k} \quad b^{k} \quad c^{k}\right)$

Step 3: now, we have to show that

$A^{k+1}=\operatorname{diag}\left(a^{k+1} \quad b^{k+1} \quad c^{k+1}\right)$

Now,

$\\ \begin{array}{l} A^{k+1}=A^{k} A \\\\ =d i a g\left(a^{k} \quad b^{k} \quad c^{k}\right) d i a g(a \quad b \quad c) \\\\ A^{k+1}=\left[\begin{array}{ccc} a^{k} & 0 & 0 \\ 0 & b^{k} & 0 \\ 0 & 0 & c^{k} \end{array}\right]\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right] \\\\\\ =\left[\begin{array}{ccc} a^{k} \times a+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+b^{k} \times b+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+c^{k} \times c \end{array}\right] \\\\ =\left[\begin{array}{ccc} a^{k+1} & 0 & 0 \\ 0 & b^{k+1} & 0 \\ 0 & 0 & c^{k+1} \end{array}\right] \\\\\\ A^{k+1}=\operatorname{diag}\left(a^{k+1} \quad b^{k+1} \quad c^{k+1}\right) \end{array}$

So, $A^n$ is true for n=k+1 whenever $A^n$ is true for n=k

Hence, by principle of mathematical induction $A^n$ is true for all positive integers.