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Find \(^{7} A^{2}=A \cdot A, A^{3}=A \cdot A \cdot A, A^{4}\), and \(A^{100}\), given that $$ A=\left[\begin{array}{llll} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$

Short Answer

Expert verified
The results for the given matrix powers are: \(A^2 = \left[\begin{array}{llll} 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\), \(A^3 = \left[\begin{array}{llll} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\), \(A^4 = \left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\), and for any power of A with exponent greater than or equal 4, such as \(A^{100}\), the result is a null matrix: \(A^{100} = \left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\).

Step by step solution

01

Find \(A^2\) by multiplying A by itself

We are given the matrix A as follows: $$ A=\left[\begin{array}{llll} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ To find square of matrix A i.e. \(A^2\), multiply the matrix A by itself: $$ A^2 =A \cdot A = \left[\begin{array}{llll} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \cdot \left[\begin{array}{llll} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ Now, perform matrix multiplication: $$ A^2 = \left[\begin{array}{llll} 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$
02

Find \(A^3\) by multiplying \(A^2\) by A

Now, we need to find the cube of matrix A i.e. \(A^3\). We already know \(A^{2}\) from step 1. Multiply \(A^{2}\) with A to find \(A^3\): $$ A^3 = A^2 \cdot A = \left[\begin{array}{llll} 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \cdot \left[\begin{array}{llll} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ Perform matrix multiplication: $$ A^3 = \left[\begin{array}{llll} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$
03

Find \(A^4\) by multiplying \(A^3\) by A

Now find the fourth power of matrix A i.e. \(A^4\). We already know \(A^3\) from step 2. Multiply \(A^3\) with A to find \(A^4\): $$ A^4 = A^3 \cdot A = \left[\begin{array}{llll} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \cdot \left[\begin{array}{llll} 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ Perform matrix multiplication: $$ A^4 = \left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$
04

Find \(A^{100}\) using \(A^4\)

Now, here's an observation: since \(A^4\) is a matrix with all elements equal to 0 (a null matrix), then \(A^4 \cdot A = A^5\) will also be a null matrix, and so will any product of A with powers greater than or equal to 4 (e.g., \(A^6, A^7, \dots, A^{100}\)). Thus, any power of A with exponent 4 or higher will result in a null matrix. So, we can directly write: $$ A^{100} = \left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$

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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 fundamental operation where two matrices, say matrix "A" and matrix "B", are multiplied to yield a new matrix, typically called matrix "C". The resulting matrix is formed using a specified method:

* The number of columns in the first matrix (A) must match the number of rows in the second matrix (B) for multiplication to be possible.
* Each cell in the resulting matrix is computed as the sum of products. This involves taking the respective row from the first matrix and the column from the second matrix, multiplying these components and then summing them up.

For example, if we have matrices: \( A = [a_{ij}] \, \text{and} \, B = [b_{jk}] \) the multiplied matrix \( C \) has elements \( c_{ik} \). Each is calculated as \( c_{ik} = a_{i1}b_{1k} + a_{i2}b_{2k} + \.\.\. + a_{im}b_{mk} \).

This method, although appearing simple, is computationally intensive, especially with increasing matrix size, but is crucial for complex operations in linear algebra including matrix exponentiation, where we repeatedly multiply a matrix by itself.
Zero Matrix
A zero matrix, also referred to as a null matrix, is a special type of matrix where all of its entries are zero. This matrix serves various roles in matrix algebra:

* It acts as the additive identity. When any matrix is added to a zero matrix of compatible dimensions, the original matrix remains unchanged.
* It is critical in processes like matrix exponentiation. As observed in your exercise, when a matrix raised to a particular power yields a zero matrix, subsequently raising it to any higher power will still result in a zero matrix.

For example, the zero matrix denoted as \( O_{m\times n} \) for an \( m \times n \) matrix, is represented as \[ \begin{bmatrix} 0 & 0 & ... & 0 \ 0 & 0 & ... & 0 \ ... & ... & ... & ... \ 0 & 0 & ... & 0 \end{bmatrix} \].

In the context of the exercise, knowing that \( A^4 \) results in a zero matrix implies \( A^{100} \) will also be a zero matrix, signaling an extinction of any non-zero values through repetitive multiplication.
Upper Triangular Matrix
An upper triangular matrix is a type of square matrix where all elements below the main diagonal are zero. In mathematical terms, matrix \( A = [a_{ij}] \) is upper triangular if \( a_{ij} = 0 \) for any row index \( i > j \). They have particular significance due to their structured form:

* Computational Efficacy: In calculations involving determinants and inverses, upper triangular matrices make operations straightforwardly and computationally efficient.
* Simplifying Solutions: They simplify solving systems of linear equations through back substitution.

An example of a basic upper triangular matrix is:
\[\begin{bmatrix} a & b & c \ 0 & e & f \ 0 & 0 & i \end{bmatrix}\]

In the exercise, matrix \( A \) shares properties of an upper triangular matrix, with its primary non-zero values situated above the main diagonal. This structure is pertinent in the matrix exponentiation solution, influencing multiplication outcomes uniquely due to the sequential zeroing process.

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Most popular questions from this chapter

C o s t s ~ M i c r o b u c k s ~ C o m p u t e r ~ C o . ~ m a k e s ~ t w o ~ c o m p u t e r s , ~ t h e ~ Pomegranate II and the Pomegranate Classic. The Pom II requires 2 processor chips, 16 memory chips, and 20 vacuum tubes, while the Pom Classic requires 1 processor chip, 4 memory chips, and 40 vacuum tubes. There are two companies that can supply these parts: Motorel can supply them at \(\$ 100\) per processor chip, \(\$ 50\) per memory chip, and \(\$ 10\) per vacuum tube, while Intola can supply them at \(\$ 150\) per processor chip, \(\$ 40\) per memory chip, and \(\$ 15\) per vacuum tube. Write down all of this data in two matrices, one showing the parts required for each model computer, and the other showing the prices for each part from each supplier. Then show how matrix multiplication allows you to compute the total cost for parts for each model when parts are bought from either supplier.

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Is it possible for \(a 2 \times 3\) matrix to equal a \(3 \times 2\) matrix? Explain.

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