Determinants and Matrices — Maths STD 11 Science — Question

Let $P(n) \equiv A^n=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$, for all $n \in N$.

Step 1: Put n $=1$

$\begin{aligned} \text { R.H.S. } & =\left[\begin{array}{cc}1+2(1) & -4(1) \\ 1 & 1-2(1)\end{array}\right] \\ & =\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right] \\ & =\mathrm{A}=\text { L.H.S. }\end{aligned}$

$\therefore \quad \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$.

Step 2: Let us consider that $P(n)$ is true for $n=k$.

$\therefore \quad A^k=\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]$

...(i)

Step 3: We have to prove that $P(n)$ is true for

$\mathrm{n}=\mathrm{k}+1$

i.e., to prove that

$\begin{aligned} \mathrm{A}^{\mathrm{k}+1} & =\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\end{array}\right] \\ \text { R.H.S. } & =\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\end{array}\right] \\ & =\left[\begin{array}{cc}1+2 k+2 & -4 k-4 \\ k+1 & 1-2 k-2\end{array}\right] \\ & =\left[\begin{array}{cc}3+2 k & -4 k-4 \\ k+1 & -2 k-1\end{array}\right]\end{aligned}$

$\begin{aligned} \text { L.H.S. } & =A^{k+1} \\ & =A^k \cdot A\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right] \ldots[\text { From (i)] } \\ & =\left[\begin{array}{cc}3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k\end{array}\right] \\ & =\left[\begin{array}{cc}3+2 k & -4 k-4 \\ k+1 & -2 k-1\end{array}\right]=\text { R.H.S. }\end{aligned}$

$\therefore \quad P(n)$ is true for $n=k+1$.

Step 4: From all steps above, by the principle of

Mathematical induction $P(n)$ is true for all

$\mathbf{n} \in \mathbf{N}$.

$\therefore \quad A^n=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$ for all $n \in N$.