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

Short Answer

Expert verified
The pattern is that \( A^n = 2^{n-1} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \).

Step by step solution

01

Compute A虏

First, we need to multiply matrix A by itself to find A虏. Given matrix A\[A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\]the multiplication A * A is\[A^{2} = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} (1 \times 1 + 1 \times 1) & (1 \times 1 + 1 \times 1) \ (1 \times 1 + 1 \times 1) & (1 \times 1 + 1 \times 1) \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}\]
02

Compute A鲁

Multiply A虏 with matrix A to get A鲁.\[A^{3} = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} (2 \times 1 + 2 \times 1) & (2 \times 1 + 2 \times 1) \ (2 \times 1 + 2 \times 1) & (2 \times 1 + 2 \times 1) \end{bmatrix} = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix}\]
03

Compute A鈦

Multiply A鲁 with matrix A to get A鈦.\[A^{4} = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix} \times \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} (4 \times 1 + 4 \times 1) & (4 \times 1 + 4 \times 1) \ (4 \times 1 + 4 \times 1) & (4 \times 1 + 4 \times 1) \end{bmatrix} = \begin{bmatrix} 8 & 8 \ 8 & 8 \end{bmatrix}\]
04

Observe the Pattern

From steps 1 to 3, we notice that A becomes a matrix where every element is a power of 2. Specifically, \( A^{n} \) is\[A^{n} = 2^{n-1} \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\]where each entry of \( A^{n} \) is \( 2^{n-1} \).
05

General Formula for A鈦

Based on the observed pattern, the general formula for any positive integer \( n \) is\[A^{n} = 2^{n-1} \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\]This represents that all elements of \( A^{n} \) are simply \( 2^{n-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Multiplication
Matrix multiplication is a process that involves multiplying two matrices to produce a third matrix. This operation is fundamental in linear algebra and is used in many applications across different fields, such as physics, computer graphics, and systems engineering. To multiply two matrices, say matrix A and matrix B, each element in the resulting matrix is obtained by taking the dot product of the rows from matrix A with the columns of matrix B.

For example, when multiplying matrix A by itself to find its square, \( A^2 \), consider matrix A given by:
  • A = \( \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \)
The element at the first row and first column of \( A^2 \) is found by multiplying the elements of the first row of A with the first column of A, resulting in \( 1 \times 1 + 1 \times 1 = 2 \). This process is repeated for each element to fill the resulting matrix. Since each operation involves addition and multiplication, it's essential to follow the order of operations closely to achieve accurate results.

Matrix multiplication is not commutative, meaning that \( A \times B eq B \times A \) in general. However, it is associative and distributive鈥攊mportant properties when working with multiple matrices.
Patterns in Matrices
Finding patterns in matrices can reveal insights about their properties, simplifying computation. In this exercise, repeatedly multiplying the matrix A reveals a pattern. Each time the computation for the next power of A is performed, a consistent matrix emerges.

Starting with the matrix A:
  • A = \( \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \)
Computing higher powers, \( A^2 = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \), \( A^3 = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix} \), and \( A^4 = \begin{bmatrix} 8 & 8 \ 8 & 8 \end{bmatrix} \), indicates that each resulting matrix maintains a structure where all entries are identical. More specifically, all entries are powers of 2.

Observing these patterns can simplify calculations and help in predicting future results without performing exhaustive matrix multiplication again. Understanding such a pattern allows for a more straightforward approach to determining any power of the given matrix.
General Formula for Powers of a Matrix
When a pattern in the matrix powers is established, a general formula can often be developed to represent these findings for any positive integer power. This requires understanding the consistency in the transformations of the matrix across computations.

For the matrix A, the observed pattern leads to a general formula:
  • \( A^n = 2^{n-1} \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \)
This formula shows that for any power \( n \), every element in the matrix is simply \( 2^{n-1} \). Knowing this formula allows one to compute \( A^n \) easily without multiplying matrices repeatedly, as each element of the result is derived directly from the power of 2.

Creating such a formula requires careful observation of the formed matrix powers and understanding the arithmetic progression of results as matrices are modified over multiplication. By using this formula, computational efficiency is significantly enhanced while addressing the original matrix and its repeated applications.

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Most popular questions from this chapter

Shading Unwanted Regions To graph the solution of a system of inequalities, we have shaded the solution of each inequality in a different color; the solution of the system is the region where all the shaded parts overlap. Here is a different method: For each inequality, shade the region that does not satisfy the inequality. Explain why the part of the plane that is left unshaded is the solution of the system. Solve the following system by both methods. Which do you prefer? Why? $$ \left\\{\begin{aligned} x+2 y &>4 \\\\-x+y &<1 \\ x+3 y &<9 \\ x &<3 \end{aligned}\right. $$

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