Problem 56 - Precalculus: Mathematical for Calculus

Question

Let \(A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] .\) Calculate \(A^{2}, A^{3}\) \(A^{4}, \ldots\) until you detect a pattern. Write a general formula for \(A^{n}\)

Step-by-Step Solution

Answer

The general formula for \(A^n\) is \(A^n = 2^{n-1} \begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}.\)

Step 1: Calculate \(A^2\)

To calculate \(A^2\), we multiply matrix \(A\) by itself: \[A \cdot A = \begin{bmatrix}1 & 1 \ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 1 \ 1 & 1\end{bmatrix}.\] Performing the matrix multiplication, we get: \[A^2 = \begin{bmatrix}(1\cdot1 + 1\cdot1) & (1\cdot1 + 1\cdot1) \ (1\cdot1 + 1\cdot1) & (1\cdot1 + 1\cdot1)\end{bmatrix} = \begin{bmatrix}2 & 2 \ 2 & 2\end{bmatrix}.\]

Step 2: Calculate \(A^3\)

Now, let's find \(A^3\) by multiplying \(A^2\) by \(A\): \[A^2 \cdot A = \begin{bmatrix}2 & 2 \ 2 & 2\end{bmatrix} \cdot \begin{bmatrix}1 & 1 \ 1 & 1\end{bmatrix}.\] This gives: \[A^3 = \begin{bmatrix}(2\cdot1 + 2\cdot1) & (2\cdot1 + 2\cdot1) \ (2\cdot1 + 2\cdot1) & (2\cdot1 + 2\cdot1)\end{bmatrix} = \begin{bmatrix}4 & 4 \ 4 & 4\end{bmatrix}.\]

Step 3: Calculate \(A^4\)

Let's find \(A^4\) by multiplying \(A^3\) by \(A\): \[A^3 \cdot A = \begin{bmatrix}4 & 4 \ 4 & 4\end{bmatrix} \cdot \begin{bmatrix}1 & 1 \ 1 & 1\end{bmatrix}.\] Performing the multiplication, we get: \[A^4 = \begin{bmatrix}(4\cdot1 + 4\cdot1) & (4\cdot1 + 4\cdot1) \ (4\cdot1 + 4\cdot1) & (4\cdot1 + 4\cdot1)\end{bmatrix} = \begin{bmatrix}8 & 8 \ 8 & 8\end{bmatrix}.\]

Step 4: Identify the Pattern

Notice the pattern: Each power of matrix \(A\) results in a matrix where all elements are the same power of 2. From this, the elements seem to be doubling with each power: \(A^2 = \begin{bmatrix}2 & 2 \ 2 & 2\end{bmatrix}, A^3 = \begin{bmatrix}4 & 4 \ 4 & 4\end{bmatrix}, A^4 = \begin{bmatrix}8 & 8 \ 8 & 8\end{bmatrix}.\)

Step 5: Write the General Formula

Based on the pattern observed, each element of the matrix \(A^n\) is equal to \(2^{n-1}\). Therefore, the general formula for \(A^n\) is: \[A^n = \begin{bmatrix}2^{n-1} & 2^{n-1} \ 2^{n-1} & 2^{n-1}\end{bmatrix}.\] This pattern holds for all \(n \geq 1\).

Key Concepts

matrix multiplication

Matrix multiplication is a way of combining two matrices to create a new matrix. It's much like multiplying numbers but with a set of rules that respect the arrangement of numbers in rows and columns. In the exercise, we first calculated \(A^2\) by multiplying matrix \(A\) by itself. The formula for matrix multiplication involves taking the sum of the products of the rows of the first matrix and the columns of the second matrix.
For example, for two matrices \(A\) and \(B\), you calculate each element of the result matrix by

• Multiplying elements across the rows of the first matrix by the respective elements down the columns of the second matrix.
• Summing these products together.

In our case, \(A\) is \([1, 1; 1, 1]\). When multiplying \(A\) by itself, each element becomes two because each element follows:

• \(1\times1 + 1\times1 = 2\).

This rule holds and is repeated to calculate higher powers, like \(A^3\) and \(A^4\), by extending the same multiplication process.

pattern recognition in mathematics

Pattern recognition is a fundamental concept in mathematics where we observe regularities or repeating structures. In our exercise, recognizing the pattern in the powers of the matrix \(A\) helps us simplify the solution. With each matrix multiplication leading to elements of 2, 4, 8, and so forth (i.e., powers of two), pattern recognition allows us to predict future results without direct computation.
Patterns simplify solutions and reveal the underlying structure, saving time and effort. For example:

• The first step revealed that every element in \(A\) grows exponentially, each element doubling as we ascend in power (like \(2, 4, 8...\)).
• This indicated a simple rule: each element can be represented by \(2^{n-1}\).

By identifying these patterns, we can understand the behavior of matrix exponentiation without needing to perform infinite multiplications.

general formula for matrices

Creating a general formula means finding a rule that always holds true for any given conditions. In our matrix, the general formula \(A^n = \begin{bmatrix}2^{n-1} & 2^{n-1} \ 2^{n-1} & 2^{n-1}\end{bmatrix}\) is derived from observing the pattern in the powers of the matrix \(A\).
This formula tells us precisely what every power of the matrix looks like, where

• each matrix entry is always \(2^{n-1}\),
• indicating the elements are equal and uniformly distributed as powers of two.

This rule is not just an outcome of the repetitive calculations but an understanding of how the structure of the matrix inherently changes as it is exponentiated.
By utilizing a general formula, we can skip the tedious calculations and directly understand and calculate high powers, like \(A^{100}\) or higher, instantaneously.