If A= a&b 0&1 , prove that A^n= a^n&b ( a^n-1 a-1 ) 0&1 for every positive integer n.

We shall prove the result by the principle of mathematical induction on n. Step 1: If n = 1, by definition of integral power of a matrix, we have A^1= a^1&b (a^1-1) a-1 0&1 = a&b 0&1 =A So, the result is true for n = 1. Step 2: Let the result be true for n = m. Then, A^ m = a^ m &b (a^ m -1) a-1 0&1 (1) Now, we shall show that the result is true for n = m + 1. Here, A^ m+1 = a^ m+1 &b (a^ m+1 -1) a-1 0&1 By definition of integral power of a matrix, we have A^ m+1 =A^mA ^ m+1 = a^m&b (a^m-1) a-1 0&1& a&b 0&1 [From eq. (1)] ^ m+1 = a^ma+0& a^mb+b(a^m-1) a-1 0+0&0+1 ^ m+1 = a^ m+1 & (a^ m+1 b-a^mb+a^mb-b) a-1 0&1 ^ m+1 = a^ m+1 &b (a^ m+1 -1) a-1 0&1 This shows that when the result is true for n = m, it is also true for n = m + 1. Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

If A= a&b 0&1 , prove that A^n= a^n&b ( a^n-1 a-1 ) 0&1 for every…

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Question

If $\text{A}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix},$ prove that $\text{A}^\text{n}=\begin{bmatrix}\text{a}^\text{n}&\text{b}\Big(\frac{\text{a}^\text{n}-1}{\text{a}-1}\Big)\\0&1\end{bmatrix}$ for every positive integer n.

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Answer

We shall prove the result by the principle of mathematical induction on n.
Step 1: If n = 1, by definition of integral power of a matrix, we have
$\text{A}^1=\begin{bmatrix}\text{a}^1&\text{b}\frac{(\text{a}^1-1)}{\text{a}-1}\\0&1\end{bmatrix}=\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix}=\text{A}$
So, the result is true for n = 1.

Step 2: Let the result be true for n = m. Then,

$\text{A}^{\text{m}}=\begin{bmatrix}\text{a}^{\text{m}}&\text{b}\frac{(\text{a}^{\text{m}}-1)}{\text{a}-1}\\0&1\end{bmatrix}\ \dots(1)$

Now, we shall show that the result is true for n = m + 1.

Here,

$\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\text{b}\frac{(\text{a}^{\text{m}+1}-1)}{\text{a}-1}\\0&1\end{bmatrix}$

By definition of integral power of a matrix, we have

$\text{A}^{\text{m}+1}=\text{A}^\text{m}\text{A}$

$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^\text{m}&\text{b}\frac{(\text{a}^\text{m}-1)}{\text{a}-1}\\0&1&\end{bmatrix}\begin{bmatrix}\text{a}&\text{b}\\0&1\end{bmatrix}$ [From eq. (1)]

$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^\text{m}\text{a}+0&\frac{\big\{\text{a}^\text{m}\text{b}+\text{b}(\text{a}^\text{m}-1)\big\}}{\text{a}-1}\\0+0&0+1\end{bmatrix}$

$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\frac{(\text{a}^{\text{m}+1}\text{b}-\text{a}^\text{m}\text{b}+\text{a}^\text{m}\text{b}-\text{b})}{\text{a}-1}\\0&1\end{bmatrix}$

$ \Rightarrow\text{A}^{\text{m}+1}=\begin{bmatrix}\text{a}^{\text{m}+1}&\text{b}\frac{(\text{a}^{\text{m}+1}-1)}{\text{a}-1}\\0&1\end{bmatrix}$

This shows that when the result is true for n = m, it is also true for n = m + 1.

Hence, by the principle of mathematical induction, the result is valid for any positive integer n.

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