Problem 52
Question
Powers of a Matrix \(\quad\) 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
Verified Answer
The general formula for \(A^n\) is \(\begin{bmatrix} 2^{n-1} & 2^{n-1} \\ 2^{n-1} & 2^{n-1} \end{bmatrix}\).
1Step 1: Calculate \(A^2\)
To find \(A^2\), we multiply matrix \(A\) by itself. So, we have:\[A^2 = A \cdot A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\]Let's compute each element of the resulting matrix:- Top left: \((1\cdot1 + 1\cdot1) = 2\)- Top right: \((1\cdot1 + 1\cdot1) = 2\)- Bottom left: \((1\cdot1 + 1\cdot1) = 2\)- Bottom right: \((1\cdot1 + 1\cdot1) = 2\)Thus, \(A^2 = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}\).
2Step 2: Calculate \(A^3\)
Next, compute \(A^3 = A^2 \cdot A\):\[A^3 = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\]Calculate each element in the new matrix:- Top left: \((2\cdot1 + 2\cdot1) = 4\)- Top right: \((2\cdot1 + 2\cdot1) = 4\)- Bottom left: \((2\cdot1 + 2\cdot1) = 4\)- Bottom right: \((2\cdot1 + 2\cdot1) = 4\)Thus, \(A^3 = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix}\).
3Step 3: Calculate \(A^4\)
Now compute \(A^4 = A^3 \cdot A\):\[A^4 = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}\]Perform the matrix multiplication:- Top left: \((4\cdot1 + 4\cdot1) = 8\)- Top right: \((4\cdot1 + 4\cdot1) = 8\)- Bottom left: \((4\cdot1 + 4\cdot1) = 8\)- Bottom right: \((4\cdot1 + 4\cdot1) = 8\)Thus, \(A^4 = \begin{bmatrix} 8 & 8 \ 8 & 8 \end{bmatrix}\).
4Step 4: Identify the Pattern
Observe the pattern in matrices from \(A^2\) to \(A^4\):- \(A^2 = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix}\)- \(A^3 = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix}\)- \(A^4 = \begin{bmatrix} 8 & 8 \ 8 & 8 \end{bmatrix}\)Each element in the \(n\)th power matrix \(A^n\) is \(2^{n-1}\).
5Step 5: Write the General Formula
From the pattern identified, 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 formula holds for any positive integer \(n\).
Key Concepts
Matrix MultiplicationMatricesPatterns
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra that involves combining two matrices to produce a third one. To multiply matrices, alignment is key. Consider two matrices, A and B, where matrix A has dimensions \( m \times n \) and matrix B has dimensions \( n \times p \). The resulting matrix will have dimensions \( m \times p \). Each element in the resulting matrix is computed by taking the dot product of the corresponding row from the first matrix and the column from the second matrix.
This involves multiplying pairs of elements from the row and column and summing these products.
For example, if we have:
- Top left entry: \( (1 \cdot 5) + (2 \cdot 7) = 19 \)
- Top right entry: \( (1 \cdot 6) + (2 \cdot 8) = 22 \)
- Bottom left entry: \( (3 \cdot 5) + (4 \cdot 7) = 43 \)
- Bottom right entry: \( (3 \cdot 6) + (4 \cdot 8) = 50 \)
So, resulting matrix C = \( \begin{bmatrix} 19 & 22 \ 43 & 50 \end{bmatrix} \)
This principle is similarly applied to matrix powers, where a matrix is multiplied by itself repeatedly.
This involves multiplying pairs of elements from the row and column and summing these products.
For example, if we have:
- Matrix A = \( \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \)
- Matrix B = \( \begin{bmatrix} 5 & 6 \ 7 & 8 \end{bmatrix} \)
- Top left entry: \( (1 \cdot 5) + (2 \cdot 7) = 19 \)
- Top right entry: \( (1 \cdot 6) + (2 \cdot 8) = 22 \)
- Bottom left entry: \( (3 \cdot 5) + (4 \cdot 7) = 43 \)
- Bottom right entry: \( (3 \cdot 6) + (4 \cdot 8) = 50 \)
So, resulting matrix C = \( \begin{bmatrix} 19 & 22 \ 43 & 50 \end{bmatrix} \)
This principle is similarly applied to matrix powers, where a matrix is multiplied by itself repeatedly.
Matrices
Matrices are rectangular arrays of numbers arranged in rows and columns. They are central in various mathematical applications, including systems of linear equations, transformations, and other mathematical operations. The size or dimension of a matrix is defined by the number of its rows and columns. A matrix with \( m \) rows and \( n \) columns is called an \( m \times n \) matrix.
Key concepts related to matrices include:
Key concepts related to matrices include:
- Elements: The individual numerals within a matrix, usually denoted by subscripts indicating their position (e.g., \( a_{ij} \) where \( i \) is the row position and \( j \) is the column position).
- Square Matrix: A matrix with an equal number of rows and columns (e.g., a 2x2 matrix).
- Identity Matrix: A special kind of square matrix where all elements are zero except for the principal diagonal, which is composed of ones.
- Transpose: Obtained by flipping a matrix over its diagonal, switching the row and column indices.
- Determinant: A unique number associated with a square matrix that has important properties in linear transformations.
Patterns
Patterns in matrices, especially when dealing with matrix powers, reveal interesting mathematical properties. Detecting and understanding these patterns can simplify complex matrix operations. When a matrix is repeatedly multiplied by itself, certain forms and sequences emerge.
To see a pattern, let's revisit the example matrix:
- \( A^2 = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \)
- \( A^3 = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix} \)
- \( A^4 = \begin{bmatrix} 8 & 8 \ 8 & 8 \end{bmatrix} \)
The pattern reveals that each element of the matrix multiplies by 2 with each power. Thus, the nth power of the matrix has all of its elements as \( 2^{n-1} \). Understanding such patterns allows for the derivation of general formulas like \( A^n = \begin{bmatrix} 2^{n-1} & 2^{n-1} \ 2^{n-1} & 2^{n-1} \end{bmatrix} \), which provide a quick way to calculate large powers of the matrix.
To see a pattern, let's revisit the example matrix:
- Suppose \( A = \begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix} \)
- \( A^2 = \begin{bmatrix} 2 & 2 \ 2 & 2 \end{bmatrix} \)
- \( A^3 = \begin{bmatrix} 4 & 4 \ 4 & 4 \end{bmatrix} \)
- \( A^4 = \begin{bmatrix} 8 & 8 \ 8 & 8 \end{bmatrix} \)
The pattern reveals that each element of the matrix multiplies by 2 with each power. Thus, the nth power of the matrix has all of its elements as \( 2^{n-1} \). Understanding such patterns allows for the derivation of general formulas like \( A^n = \begin{bmatrix} 2^{n-1} & 2^{n-1} \ 2^{n-1} & 2^{n-1} \end{bmatrix} \), which provide a quick way to calculate large powers of the matrix.
Other exercises in this chapter
Problem 51
Use a graphing device to graph both lines in the same viewing rectangle. (Note that you must solve for \(y\) in terms of \(x\) before graphing if you are using
View solution Problem 52
A man and his daughter manufacture unfinished tables and chairs. Each table requires 3 hours of sawing and 1 hour of assembly. Each chair requires 2 hours of sa
View solution Problem 52
Solve for \(x\) $$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & 1 & x \\ x & 1 & x \end{array}\right|=0$$
View solution Problem 52
Use a graphing device to graph both lines in the same viewing rectangle. (Note that you must solve for \(y\) in terms of \(x\) before graphing if you are using
View solution