#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 57 math

Answer: Hence proved, $A^{n}=\left[\begin{array}{cc}a^{n} & b\left(\frac{a^{n}-1}{a-1}\right) \\ 0 & 1\end{array}\right]$ for all positive integers n.

Hint: We use the principle of mathematical induction.

Given:

$A=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]$

Prove:

$A^{n}=\left[\begin{array}{cc}a^{n} & b\left(\frac{a^{n}-1}{a-1}\right) \\ 0 & 1\end{array}\right]$ for every positive integer n                  …(i)

Solution:

step 1:

Put n=1 in eqn (i)

$\\A^{1}=\left[\begin{array}{cc}a^{1} & b\left(\frac{a^{1}-1}{a-1}\right) \\ 0 & 1\end{array}\right]\\\\ A=\left[\begin{array}{cc}a & b\left(\frac{a-1}{a-1}\right) \\ 0 & 1\end{array}\right] \quad \therefore\left\{a^{1}=a\right\} \\\\ A=\left[\begin{array}{ll}a & b \\ 0 & 1\end{array}\right]$

So,  $A^n$ is true for n=1

Step 2: let $A^n$ be true for n=k, so

$A^{k}=\left[\begin{array}{cc} a^{k} & b\left(\frac{a^{k}-1}{a-1}\right) \\ 0 & 1 \end{array}\right] ... (ii)$

Step 3: we have to show that

$\\A^{k+1}=A^{k} \times A\\\\ =\left[\begin{array}{cc} a^{k} & b\left(\frac{a^{k}-1}{a-1}\right) \\ 0 & 1 \end{array}\right] \times\left[\begin{array}{ll} a & b \\ 0 & 1 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} a^{k+1}+0 & a^{k} b+b\left(\frac{a^{k}-1}{a-1}\right) \\ 0+0 & 0+1 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} a^{k+1} & \frac{a^{k+1} b-a^{k} b+a^{k} b-b}{a-1} \\ 0 & 1 \end{array}\right]\\\\\\ A^{k+1}=\left[\begin{array}{cc} a^{k+1} & \frac{b\left(a^{k+1}-1\right)}{a-1} \\ 0 & 1 \end{array}\right]$

So,

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

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