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Find all matrices that commute with the given matrix \(A\). $$A=\left[\begin{array}{rr} 0 & -2 \\ 2 & 0 \end{array}\right]$$

Short Answer

Expert verified
The matrices that commute with A are of the form \(\left[\begin{array}{cc} -z & z \ z & -z \end{array}\right]\), where z is any real number.

Step by step solution

01

Define Commutativity for Matrices

Two matrices, say A and B, are said to commute with each other if AB = BA, where AB represents the matrix product of A and B.
02

Propose a General Form of Matrix B

Let B be any 2x2 matrix that can be written in the form \(B = \left[\begin{array}{cc} x & y \ z & w \end{array}\right]\), where x, y, z, and w are real numbers that we need to find.
03

Calculate the Product AB

Multiply the given matrix A by the general matrix B to get the product AB: \(AB = A \cdot B = \left[\begin{array}{rr} 0 & -2 \ 2 & 0 \end{array}\right] \cdot \left[\begin{array}{cc} x & y \ z & w \end{array}\right] = \left[\begin{array}{cc} -2z & -2w \ 2x & 2y \end{array}\right]\).
04

Calculate the Product BA

Now multiply matrix B by the given matrix A to find the product BA: \(BA = B \cdot A = \left[\begin{array}{cc} x & y \ z & w \end{array}\right] \cdot \left[\begin{array}{rr} 0 & -2 \ 2 & 0 \end{array}\right] = \left[\begin{array}{cc} 2y & -2x \ 2w & -2z \end{array}\right]\).
05

Set up the Equations for Commutativity

For AB to equal BA, the corresponding entries of the matrices from Steps 3 and 4 must be equal. This gives us the equations: \(-2z = 2y, -2w = -2x, 2x = 2y, 2y = -2z\).
06

Solve the Equations

Solving the equations from Step 5, we find that they simplify to: \(z = -y, w = x, x = y, y = -z\). Combining these, we can deduce that \(w = x = y = -z\).
07

General Form of Commuting Matrices

Since all the variables are related and can be expressed through one variable, let's choose z. Hence, the matrices B that commute with A are of the form \(B = \left[\begin{array}{cc} -z & z \ z & -z \end{array}\right]\), where z is any real number.

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Key Concepts

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

Matrix Commutativity
When exploring the notion of matrix commutativity, we delve into the algebraic property detailing how two matrices interact when multiplied. To put it simply, if we have two matrices, let's call them Matrix A and Matrix B, they commute if the result of multiplying A by B is the same as multiplying B by A. This is symbolically represented as AB = BA, which is not always guaranteed for matrices. Unlike numbers, matrices have strict rules governing their multiplication, which means commutativity is a rare and special property in the world of matrices.

For instance, given a specific Matrix A in our exercise, the challenge lies in identifying all possible matrices (Matrix B) that, when multiplied in either order with A, yield the same result. To comprehend this concept, one must understand the arithmetic of matrix multiplication, which will be expanded upon in the next section.
Matrix Multiplication
Diving deeper into the process of matrix multiplication, we can see that it's not as straightforward as multiplying numbers. When we take matrix A and multiply it by matrix B, what we're actually doing is taking the rows of A and computing the dot product with the columns of B. To execute this operation, the number of columns in the first matrix must match the number of rows in the second. The resulting matrix has dimensions determined by the outer dimensions of the matrices we're multiplying.

For example, when we multiplied Matrix A by a general 2x2 Matrix B, we calculated each entry of the product by combining A's rows with B's columns. This means multiplying and summing specific elements according to matrix multiplication rules. If this procedure seems intricate, that's because it is, especially compared to single number multiplication. Despite the complexity, grasping this concept is vital for advancing in linear algebra and understanding the phenomena of matrix commutativity.
Solving Linear Equations
The journey towards solving linear equations in the context of matrix commutativity often involves setting up and solving a system of equations. Essentially, we're using algebraic methods to find the unknowns in the matrices that commute. In our exercise, after computing the products AB and BA, we were faced with finding the values for the unknowns in Matrix B that will satisfy the condition AB = BA.

We set up a system of equations based on the entries of the matrices resulting from AB and BA. In this case, we were looking for the unknowns x, y, z, and w that make the two matrix products equal. After equating the corresponding entries and solving, we deduced relationships among these variables. This method is ubiquitous in algebra and finds great utility in various applications, ranging from basic math problems to complex engineering computations. Understanding how to manipulate equations and apply logical reasoning to solve for unknowns is an indispensable skill when analyzing matrix behavior and interaction.

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Most popular questions from this chapter

For two invertible \(n \times n\) matrices \(A\) and \(B,\) determine which of the formulas stated are necessarily true. \(A^{-1} B\) is invertible, and \(\left(A^{-1} B\right)^{-1}=B^{-1} A\)

Find the matrices of the linear transformations from \(\mathbb{R}^{3}\) to \(\mathbb{R}^{3}\) given in Exercises 19 through \(23 .\) Some of these transformations have not been formally defined in the text. Use common sense. You may assume that all these transformations are linear. The orthogonal projection onto the \(x-y\) -plane.

For the matrices \(A\) compute \(A^{2}=A A, A^{3}=A A A,\) and \(A^{4} .\) Describe the pattern that emerges, and use this pattern to find \(A^{1001}\). Interpret your answers geometrically, in terms of rotations, reflections, shears, and orthogonal projections. $$\frac{1}{\sqrt{2}}\left[\begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array}\right]$$

In the financial pages of a newspaper, one can sometimes find a table (or matrix) listing the exchange rates between currencies. In this exercise we will consider a miniature version of such a table, involving only the Canadian dollar (C\$) and the South African Rand (ZAR). Consider the matrix \(\mathrm{CS} \quad \mathrm{ZAR}\) \\[ A=\left[\begin{array}{cc} 1 & 1 / 8 \\ 8 & 1 \end{array}\right] \begin{array}{c} \mathrm{CS} \\ \mathrm{ZAR} \end{array} \\] representing the fact that \(\mathrm{C} \$ 1\) is worth \(\mathrm{ZAR} 8\) (as of September 2012 ). a. After a trip you have \(\mathrm{C} \$ 100\) and \(\mathrm{ZAR} 1,600\) in your pocket. We represent these two values in the vector \(\vec{x}=\left[\begin{array}{c}100 \\ 1,600\end{array}\right] .\) Compute \(A \vec{x} .\) What is the practical significance of the two components of the vector \(A \vec{x} ?\) b. Verify that matrix \(A\) fails to be invertible. For which vectors \(\vec{b}\) is the system \(A \vec{x}=\vec{b}\) consistent? What is the practical significance of your answer? If the system \(A \vec{x}=\vec{b}\) is consistent, how many solutions \(\vec{x}\) are there? Again, what is the practical significance of the answer?

To determine whether a square matrix \(A\) is invertible, it is not always necessary to bring it into reduced rowechelon form. Instead, reduce \(A\) to (upper or lower) triangular form, using elementary row operations. Show that \(A\) is invertible if (and only if) all entries on the diagonal of this triangular form are nonzero.

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