If A= [ ll a & 0 0 & b ], prove that A^n= [ cc a^n & 0 0 & b^n ] for all n N.

Given A= [ ll a & 0 0 & b ] We prove A^n= [ cc a^n & 0 0 & b^n ] for all n N using mathematical induction Let P(n) be = [ cc a^n & 0 0 & b^n ] for n N. To prove that P(n) is true for n=1 P(1) is A^1=A= [ ll a & 0 0 & b ] P(1) is true. Assume that P(K) is true for some K N That is P(K) is A^k= [ cc a^K & 0 0 & b^K ] To prove that P(K) P(K+1) is true consider L.H.S. of P(K+1) That is A^ k+1 & =A^k A & = [ cc a^K & 0 0 & b^K ] [ ll a & 0 0 & b ]= [ cc a^ K+1 +0 & 0+0 0+0 & b^ K+1 +0 ] & = [ cc a^ K+1 & 0 0 & b^ K+1 ]= R.H.S of P(K+1) Hence P(K+1) is true. P(K) P(K+1) for all K N Hence by principle of mathematical induction, the statement P(n) is true for all n N. That is P(n) is true P(2) is true P(3) is true and so on P(n) is true, n N. =A^n [ cc a^n & 0 0 & b^n ] for all n N.

If A= [ ll a & 0 0 & b ], prove that A^n= [ cc a^n & 0 0 & b^n ] …

Download App

Question

If $\mathrm{A}=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]$, prove that $A^n=\left[\begin{array}{cc}a^n & 0 \\ 0 & b^n\end{array}\right]$ for all $n \in N$.

✓

Answer

Given $\mathrm{A}=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right]$
We prove $A^n=\left[\begin{array}{cc}a^n & 0 \\ 0 & b^n\end{array}\right]$ for all $\mathrm{n} \in \mathrm{N}$ using mathematical induction
Let $\mathrm{P}(\mathrm{n})$ be $=\left[\begin{array}{cc}a^n & 0 \\ 0 & b^n\end{array}\right]$ for $\mathrm{n} \in \mathrm{N}$.
To prove that $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$
$\mathrm{P}(1)$ is $A^1=\mathrm{A}=\left[\begin{array}{ll}a & 0 \\ 0 & b\end{array}\right] \quad \therefore \mathrm{P}(1)$ is true.
Assume that $\mathrm{P}(\mathrm{K})$ is true for some $\mathrm{K} \in \mathrm{N}$
That is $\mathrm{P}(\mathrm{K})$ is $A^k=\left[\begin{array}{cc}a^K & 0 \\ 0 & b^K\end{array}\right]$
To prove that $\mathrm{P}(\mathrm{K}) \rightarrow \mathrm{P}(\mathrm{K}+1)$ is true consider L.H.S. of $\mathrm{P}(\mathrm{K}+1)$
That is $A^{k+1}$
$
\begin{aligned}
& =A^k \cdot \mathrm{A} \\
& =\left[\begin{array}{cc}
a^K & 0 \\
0 & b^K
\end{array}\right]\left[\begin{array}{ll}
a & 0 \\
0 & b
\end{array}\right]=\left[\begin{array}{cc}
a^{K+1}+0 & 0+0 \\
0+0 & b^{K+1}+0
\end{array}\right] \\
& =\left[\begin{array}{cc}
a^{K+1} & 0 \\
0 & b^{K+1}
\end{array}\right]=\text { R.H.S of } \mathrm{P}(\mathrm{K}+1)
\end{aligned}
$
Hence $\mathrm{P}(\mathrm{K}+1)$ is true.
$\therefore \mathrm{P}(\mathrm{K}) \Rightarrow \mathrm{P}(\mathrm{K}+1)$ for all $\mathrm{K} \in \mathrm{N}$
Hence by principle of mathematical induction, the statement $P(n)$ is true for all $n \in N$.

That is $\mathrm{P}(\mathrm{n})$ is true $\rightarrow \mathrm{P}(2)$ is true $\rightarrow \mathrm{P}(3)$ is true and so on $\rightarrow P(n)$ is true, $n \in N$.
$\therefore=\mathrm{A}^n\left[\begin{array}{cc}a^n & 0 \\ 0 & b^n\end{array}\right] \quad$ for all $\mathrm{n} \in \mathrm{N}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "Solve the Following Question.(5 Marks)" from Determinants and Matrices→Determinants and Matrices — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Determinants and Matrices — Maths STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions