Question

If matrix \(\mathrm{A}=\left|\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right|\) and \(\mathrm{I}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|\) then which one of the following holds for all \(n \in N\). (use principle of mathematical Induction) (a) \(\mathrm{A}^{\mathrm{n}}=\mathrm{n} \cdot \mathrm{A}-(\mathrm{n}-1) \mathrm{I}\) (b) \(\mathrm{A}^{\mathrm{n}}=2^{\mathrm{n}-1} \cdot \mathrm{A}+(\mathrm{n}-1) \mathrm{I}\) (a) \(\mathrm{A}^{\mathrm{n}}=\mathrm{n} \cdot \mathrm{A}+(\mathrm{n}-1) \mathrm{I}\) (b) \(A^{n}=2^{n-1} \cdot A-(n-1) I\)

Short answer

The correct statement is: \[\boxed{A^{n} = n \cdot A - (n-1) I}\]

Step-by-step solution

Base case: n = 1

For n=1, we should test the given statements to see if any of them holds true. Let's check them one by one. (a) For n = 1, \[A^1 = 1 \cdot A - (1-1)I\Leftrightarrow A = A - 0I\Leftrightarrow A = A\] This expression holds true. (b) For n = 1, \[A^1 = 2^{1-1} \cdot A + (1-1)I \Leftrightarrow A = A + 0I\Leftrightarrow A = A\] This expression also holds true. Consequently, at this point, we cannot eliminate any option. Let's proceed with the induction step. Step 2: Induction hypothesis and step

Induction hypothesis

Assume that one of the given statements is true for some arbitrary n ∈ N. Let's denote this arbitrary expression as \(\text{EXPR}(n)\), where EXPR can be any of the given statements.

Induction step

We need to show that EXPR(n+1) is true if our inductive hypothesis holds. Using the given matrix A and I, we will check which of the statements holds for n+1 when their n-th version holds true. (a) For n+1: \[A^{n+1} = A^n \cdot A\] Assuming the induction hypothesis for this option, \[A^n = nA - (n-1)I\] Now, let's find \(A^{n+1}\): \[A^{n+1} = (nA - (n-1)I) \cdot A = nA^2 - (n-1)AI\] However, we need to find A^2 and AI: \[A^2 = A\cdot A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\] \[AI = A\cdot I = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = A\] Therefore, \[A^{n+1} = nA^2 - (n-1)AI = n\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} - (n-1)\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = (n+1)A - nI\] which is the same as option (a) for n+1. (b) For n+1 \[A^{n+1} = A^n \cdot A\] Assuming the induction hypothesis for this option, \[A^n = 2^{n-1} A + (n-1)I\] Now, let's find \(A^{n+1}\): \[A^{n+1} = (2^{n-1} A + (n-1)I) \cdot A = 2^{n-1} A^2 + (n-1)AI\] We already found A^2 and AI in the previous step, so we can substitute these values: \[A^{n+1} = 2^{n-1}\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} + (n-1)\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2^n & 0 \\ 2^n & 2^{n-1} \end{bmatrix} + \begin{bmatrix} n-1 & 0 \\ n-1 & n-1 \end{bmatrix} = 2^nA + nI\] This expression is not the same as any of the given options for n+1. Since only statement (a) holds true for all n ∈ N, we have found the answer. #Answer# The correct statement is: \[\boxed{A^{n} = n \cdot A - (n-1) I}\]

Key concepts

Mathematical Induction

Understanding mathematical induction is much like learning the domino effect. Think of it as a way to prove that something is true for all natural numbers by creating a chain reaction.

Initially, you need to prove your case for the first 'domino', or the base case (usually when n=1), and ensure it tumbles as expected. That's our first step, verifying that the theorem holds for the starting number, laying down the groundwork for the induction.

• The base case for our problem was showing that both expressions (a) and (b) hold true with n=1.

Next, you assume that if one 'domino' falls (the induction hypothesis), it will knock over the next one (the induction step). This is where the true power of induction lies, as it allows us to extend the truth from one instance to all natural numbers.

• We assumed the validity for an arbitrary natural number n, then pushed the dominos to show that it must also be true for n+1.

Matrix Algebra

Matrix algebra is quite the handy tool when it comes to organizing and manipulating data. It’s like a mathematical grid where each slot has a value, and whole grids can interact in predictable ways.

With matrices, you can solve systems of equations, transform geometric figures, and handle vast amounts of data with ease.

• In our exercise, we worked with matrix A to understand how multiplying it by itself or by the identity matrix I affects the result. Matrix multiplication is not commutative, meaning \( AB \) may not equal \( BA \), which is crucial to remember during calculations.

Furthermore, the identity matrix, denoted as I, acts as the number 1 in matrix algebra; any matrix multiplied by I will result in itself, just as when any number is multiplied by 1.

Matrix Powers

If you've dealt with squaring numbers, then matrix powers are in the same ballpark. Here, instead of multiplying a number by itself, we multiply a matrix by itself a certain number of times, indicated by the exponent.

For instance, \( A^2 \) means \( A \) times \( A \) and \( A^3 \) means \( A \) times \( A \) times \( A \), and so on. But remember, since we're dealing with matrices, the order is crucial, and \( A^n \) involves n-1 multiplications of \( A \) by itself.

• During the solution, calculating \( A^2 \) and \( AI \) was essential for advancing the problem in the induction step—the power of the matrix underpinned the rest of our calculations.

Ultimately, grasping matrix powers simplifies operations such as finding the result of \( A^n \) for any natural number n, which is particularly useful in fields like computer graphics, quantum mechanics, and computer science algorithms.