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Let `A= [[1,1,1],[0,1,1],[0,0,1]]` Use the Principle of Mathematical Introduction to Show that `A^N [[1,N,N(N+1)//2],[0,1,1],[0,0,1]]` for Every Position Integer N. - Mathematics

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Question

Let `A= [[1,1,1],[0,1,1],[0,0,1]]` Use the principle of mathematical introduction to show  that `A^n [[1,n,n(n+1)//2],[0,1,1],[0,0,1]]` for every position integer n.

Sum

Solution

We shall prove the result by the principle of

Step 1: If = 1, by definition of integral power of a matrix, we have
mathematical induction on n.\[A^1 = \begin{bmatrix}1 & 1 & 1\left( 1 + 1 \right)/2 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} = A\]

Thus, the result is true for n = 1.

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

\[A^m = \begin{bmatrix}1 & m & m\left( m + 1 \right)/2 \\ 0 & 1 & m \\ 0 & 0 & 1\end{bmatrix}\]

Now, we shall show that the result is true for

\[n = m + 1\]

 Here,

\[A^{m + 1} = \begin{bmatrix}1 & m + 1 & m + 1\left( m + 1 + 1 \right)/2 \\ 0 & 1 & m + 1 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}1 & m + 1 & \left( m + 1 \right)\left( m + 2 \right)/2 \\ 0 & 1 & m + 1 \\ 0 & 0 & 1\end{bmatrix}\]

By definition of integral power of matrix, we have

\[A^{m + 1} = A^m A\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}1 & m & m\left( m + 1 \right)/2 \\ 0 & 1 & m \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix} \left[ \text{From eq .} \left( 1 \right) \right]\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}1 + 0 + 0 & 1 + m + 0 & 1 + m + m\left( m + 1 \right)/2 \\ 0 + 0 + 0 & 0 + 1 + 0 & 0 + 1 + m \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 1\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}1 & 1 + m & \left( 2 + 2m + m^2 + m \right)/2 \\ 0 & 1 & 1 + m \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}1 & 1 + m & \left( m^2 + 3m + 2 \right)/2 \\ 0 & 1 & 1 + m \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^{m + 1} = \begin{bmatrix}1 & 1 + m & \left( m + 1 \right)\left( m + 2 \right)/2 \\ 0 & 1 & 1 + m \\ 0 & 0 & 1\end{bmatrix}\]
\[\]
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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Chapter 5: Algebra of Matrices - Exercise 5.3 [Page 46]

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RD Sharma Mathematics [English] Class 12
Chapter 5 Algebra of Matrices
Exercise 5.3 | Q 60 | Page 46

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