Question

If $I$ is the identity matrix of order $2$ and $A=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$ , then for $n\ge 1$, mathematical induction gives
(a) ${{A}^{n}}=nA-(n-1)I$
(b) ${{A}^{n}}=nA+(n-1)I$
(c) ${{A}^{n}}={{2}^{n}}A-(n+1)I$
(d) ${{A}^{n}}={{2}^{n-1}}A-(n-1)I$

Answer 1

Hint: The principle of mathematical induction states that a statement $P(n)$ is true for $n\in \mathbb{N}$ if $P(1)$ is true and assuming $P(k)$ is true, $P(k+1)$ can be proven to be true. We will first formulate a statement $P(n)$. Then we will prove that $P(n)$ is true for $n\in \mathbb{N}$ using the principle of mathematical induction.

Complete step-by-step answer:
The statement of the principle of mathematical induction is as follows,
Suppose there is a given statement $P(n)$ involving $n\in \mathbb{N}$ such that
(i) The statement is true for $n=1$, that is, $P(1)$ is true, and
(ii) If the statement is true for $n=k$ where $k$ is some positive integer, then the statement is also true for $n=k+1$. That is, the truth of $P(k)$ implies the truth of $P(k+1)$.
Then $P(n)$ is true for all $n\in \mathbb{N}$.
We have $A=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$ . Let us look at the square of this matrix,
${{A}^{2}}=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]$ .
Now, we consider the next power of $A$,
${{A}^{3}}=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]=3~\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]-(3-1)\left[ \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right]$ .
Looking at the pattern, we can define a statement $P(n)$ in the following manner,
$P(n):={{A}^{n}}=nA-(n-1)I$
For the base case, $n=1$ , we have
$A=\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]=1\cdot A-0\cdot I$ .
Therefore, $P(1)$ is true. Now, let us assume that $P(k)$ is true. Therefore we have, ${{A}^{k}}=kA-(k-1)I$.
Now, consider $P(k+1)$. We will first calculate the LHS,
$\begin{align}
  & LHS={{A}^{k+1}} \\
 & ={{A}^{k}}\cdot A \\
 & =\left[ \begin{matrix}
   1 & k \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   1 & k+1 \\
   0 & 1 \\
\end{matrix} \right]
\end{align}$
 Calculating the RHS, we get
$\begin{align}
  & RHS=(k+1)A-(k+1-1)I \\
 & =\left[ \begin{matrix}
   k+1 & k+1 \\
   0 & k+1 \\
\end{matrix} \right]-\left[ \begin{matrix}
   k & 0 \\
   0 & k \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   1 & k+1 \\
   0 & 1 \\
\end{matrix} \right]
\end{align}$
Hence, we have $LHS=RHS$ and $P(k+1)$ is true for any $k$. Hence, $P(n)$ is true for $n\in \mathbb{N}$.

So, the correct answer is “Option A”.

Note: We have to be careful while formulating the statement for the principle of mathematical induction. The matrix multiplication needs to be done properly, as there is a possibility to make minor mistakes which can lead to incorrect answers. Proving $P(k+1)$ is true is the tricky part in the mathematical induction.