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If \mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right], prove that \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{ll}1 & \mathrm{n} \\ 0 & 1\end{array}\right] for all \mathrm{n} \in \mathrm{N}.
CBSE | Class 12 | Exercise 5C | Maths | Matrices | RS Aggarwal
If \mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right], prove that \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{ll}1 & \mathrm{n} \\ 0 & 1\end{array}\right] for all \mathrm{n} \in \mathrm{N}.

Solution:

We have \boldsymbol{A}=\left(\begin{array}{ll}\mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1}\end{array}\right).
We need to show: A^{n}=\left(\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right) ; \forall n \in N.

(i) Let’s compute first A^{2}
A^{2}=A A=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right)
Therefore the result is true for \boldsymbol{n}=\mathbf{2}.

(ii) Now suppose the result is true for \boldsymbol{n}=\boldsymbol{k}-\mathbf{1}.
Therefore we have \boldsymbol{A}^{\boldsymbol{k}-\mathbf{1}}=\left(\begin{array}{cc}\mathbf{1} & \boldsymbol{k}-\mathbf{1} \\ 0 & \mathbf{1}\end{array}\right).

(iii)Let’s show that result is true for \boldsymbol{n}=\boldsymbol{k} also.
A^{k}=A^{k-1} A=\left(\begin{array}{cc} 1 & k-1 \\ 0 & 1 \end{array}\right)\left(\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right)=\left(\begin{array}{ll} 1 & k \\ 0 & 1 \end{array}\right)
So from the principle of mathematical induction we can state that the above result is true for all \boldsymbol{k} \in \boldsymbol{N}.
Thus we have A^{n}=\left(\begin{array}{ll}1 & n \\ 0 & 1\end{array}\right) ; \forall n \in N

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