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Let \(A=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right] .\) Find \(A^{2}\) and \(A^{3}\).

Short Answer

Expert verified
\(A^2\) is the identity matrix; \(A^3 = A\).

Step by step solution

01

Understanding the Matrix

Given the matrix \[ A=\left[\begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 &0\end{array}\right]. \] The task requires us to find \(A^2\) and \(A^3\). This involves matrix multiplication.
02

Finding \(A^2\)

To find \(A^2\), we calculate \(A \times A\). This involves multiplying each row of the first matrix by each column of the second matrix:\[A^2 = \left[\begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0\end{array}\right] \times \left[\begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right].\] This is the identity matrix of size 3x3.
03

Finding \(A^3\)

Next, we need to find \(A^3\) by multiplying \(A^2\) with \(A\):\[A^3 = A^2 \times A = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}\right] \times \left[\begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0\end{array}\right].\]We find the result is the same as the matrix \(A\).
04

Conclusion on Powers of A

The computations show \(A^2\) is the identity matrix, and \(A^3 = A\). This implies \(A^3\) and \(A\) are identical.

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Key Concepts

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

Power of a Matrix
The power of a matrix refers to the operation of multiplying a matrix by itself a certain number of times. More specifically, if we have a matrix \( A \), raising it to the power \( n \) involves multiplying \( A \) by itself \( n-1 \) times, denoted as \( A^n \).
Given the matrix \( A = \left[\begin{array}{ccc} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{array}\right] \), calculating its power involves repeated matrix multiplication. For example, to find \( A^2 \), we simply multiply \( A \) with itself:
  • Take the first row of \( A \) and multiply each element by the corresponding element in each column of the second \( A \). Sum these products to find the resulting element for the first row and column.
  • Repeat this for the other rows and columns to get \( A^2 \).
The process continues similarly for \( A^3 \), using the result from \( A^2 \) multiplied by \( A \) again.
Understanding the power of a matrix helps in fields like computer graphics and systems of linear equations.
Identity Matrix
The identity matrix plays a crucial role in matrix multiplication, acting similarly to the number 1 in regular algebra. The identity matrix, often denoted as \( I \), is a square matrix that has 1's on the diagonal and 0's elsewhere. For a matrix of size 3x3, it looks like this:
\[I = \left[\begin{array}{ccc} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{array}\right]\]
When you multiply any square matrix by its corresponding identity matrix, the result is the original matrix itself. This is because each element of the matrix retains its position and value during the multiplication process.
In the example given, after the matrix \( A \) was raised to the power of 2, the resulting matrix was \( I \), indicating that \( A^2 \) is the identity matrix. This means \( A \) is a permutation matrix, which when squared yields the identity matrix.
Matrix Algebra
Matrix algebra is the branch of mathematics that deals with matrices and operations on them. This involves addition, subtraction, multiplication, and more complex processes like finding determinants and inverses. Matrix multiplication, as used in this exercise, is a fundamental operation with specific rules:
  • The number of columns in the first matrix must equal the number of rows in the second matrix.
  • Each element in the resulting matrix is calculated as the sum of products from the corresponding row and column pair of the original matrices.
One interesting point in matrix algebra is how matrix multiplication is non-commutative. This means \( A \times B eq B \times A \).
In our exercise, multiplying matrix \( A \) by itself to find \( A^2 \) highlights these properties and rules. It's a great way to deepen one's understanding of matrix operations, paving the way for more advanced studies in fields such as quantum mechanics, computer science, and economics.

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Most popular questions from this chapter

A matrix \(A\) and vectors \(\vec{b}, \bar{u}\) and \(\vec{v}\) are given. Verify that \(A \vec{u}=\overrightarrow{0}, A \vec{v}=\vec{b}\) and \(A(\vec{u}+\vec{v})=\vec{b}\).$$ \begin{array}{l} A=\left[\begin{array}{ccc} 2 & -2 & -1 \\ -1 & 1 & -1 \\ -2 & 2 & -1 \end{array}\right] \\ \vec{b}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \vec{u}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \vec{v}=\left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ A=\left[\begin{array}{cc} 1 & 0 \\ 5 & -4 \end{array}\right], \vec{b}=\left[\begin{array}{l} -2 \\ -1 \end{array}\right] $$

A matrix \(A\) and a vector \(\vec{x}\) are given. Find the product \(A \vec{x}\). $$ A=\left[\begin{array}{ccc} -2 & 0 & 3 \\ 1 & 1 & -2 \\ 4 & 2 & -1 \end{array}\right], \quad \vec{x}=\left[\begin{array}{l} 4 \\ 3 \\ 1 \end{array}\right] $$

Vectors \(\vec{x}\) and \(\vec{y}\) are given. Sketch \(\vec{x}, \vec{y}, \vec{x}+\vec{y},\) and \(\vec{x}-\vec{y}\) on the same Cartesian axes. $$ \vec{x}=\left[\begin{array}{c} -1 \\ 1 \end{array}\right], \vec{y}=\left[\begin{array}{c} -2 \\ 2 \end{array}\right] $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ \begin{array}{l} A=\left[\begin{array}{ccccc} 3 & 0 & -2 & -4 & 5 \\ 2 & 3 & 2 & 0 & 2 \\ -5 & 0 & 4 & 0 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} -1 \\ -5 \\ 4 \end{array}\right] \end{array} $$

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