#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 56 maths

Answer: Hence proved, $A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$ for all possible integers n.

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

We use the principle of mathematical induction to prove. $A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$

Given:

$A^{n}=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]$

Prove:

$A^{n}=\left[\begin{array}{cc}1 & n \\ 0 & 1\end{array}\right]$ for all possible integers n                         …(i)

Solution:

Step 1: put n=1 in eqn (i)

$A^{1}=\left[\begin{array}{cc}1 & 1 \\ 0 & 1\end{array}\right]$

So, $A^n$ is true for n=1

Step 2 :let,$A^n$  be true for n=k, then

$A^{k}=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right] ... (ii)$

Step 3 : we have to show that $A^{k+1}=\left[\begin{array}{cc}1 & k+1 \\ 0 & 1\end{array}\right]$

So,

$\\A^{k+1}=A^{k} \times A=\left[\begin{array}{ll}1 & k \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \\\\\\\ A^{k+1}=\left[\begin{array}{cc}1+0 & 1+k \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}1 & 1+k \\ 0 & 1\end{array}\right]$

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

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