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In order that $$ A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] $$ will commute with $$ B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 1 \end{array}\right] $$ what conditions must be satisfied by \(a, b, c\), and \(d ?\)

Short Answer

Expert verified
The conditions are: \(a = d\) and \(b = c\).

Step by step solution

01

- Understand matrix commutation

Two matrices A and B commute if and only if their product is the same in both orders, i.e., AB = BA.
02

- Compute AB

First, compute the product AB:\[A = \left[\begin{array}{ll}a & b \c & d\end{array}\right], B = \left[\begin{array}{ll}1 & 2 \2 & 1\end{array}\right]\]Then, calculate AB:\[AB = \left[\begin{array}{ll}a & b \c & d\end{array}\right] \left[\begin{array}{ll}1 & 2 \2 & 1\end{array}\right] = \left[\begin{array}{ll}a(1)+b(2) & a(2)+b(1) \c(1)+d(2) & c(2)+d(1)\end{array}\right] = \left[\begin{array}{ll}a+2b & 2a+b \c+2d & 2c+d\end{array}\right]\]
03

- Compute BA

Now, compute the product BA:\[BA = \left[\begin{array}{ll}1 & 2 \2 & 1\end{array}\right] \left[\begin{array}{ll}a & b \c & d\end{array}\right] = \left[\begin{array}{ll}1(a)+2(c) & 1(b)+2(d) \2(a)+1(c) & 2(b)+1(d)\end{array}\right] = \left[\begin{array}{ll}a+2c & b+2d \2a+c & 2b+d\end{array}\right]\]
04

- Set AB equal to BA

Equate the corresponding elements of matrices AB and BA:\[\left[\begin{array}{ll}a+2b & 2a+b \c+2d & 2c+d\end{array}\right] = \left[\begin{array}{ll}a+2c & b+2d \2a+c & 2b+d\end{array}\right]\]
05

- Solve the system of equations

From step 4, derive the following system of equations:1. \(a + 2b = a + 2c\)2. \(2a + b = b + 2d\)3. \(c + 2d = 2a + c\)4. \(2c + d = 2b + d\)Now, simplify these: - From equation 1: \(2b = 2c \Rightarrow b = c\) - From equation 2: \(2a = 2d \Rightarrow a = d\)

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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 in linear algebra that involves combining two matrices to produce a third. Unlike regular multiplication, it is not commutative, meaning that the order in which you multiply matters.
In mathematical terms, for two matrices A and B to be multiplied, the number of columns in A must be equal to the number of rows in B. The element in the resulting matrix at position (i, j) is calculated as the sum of the products of the corresponding elements of the i-th row of matrix A and the j-th column of matrix B.
Here is the general formula for the multiplication of two matrices:
If A is an m x n matrix and B is an n x p matrix, their product AB is an m x p matrix where each element is given by:
$$ (AB)_{ij} = \sum_{k=1}^{n} A_{ik}B_{kj} $$
Using this, let's multiply matrices A and B from the original problem: A = \[ \begin{array}{ll} a & b \ c & d \end{array} \], B = \[ \begin{array}{ll} 1 & 2 \ 2 & 1 \end{array} \].
Calculating AB:
$$ AB = \[ \begin{array}{ll} a & b \ c & d \end{array} \] \[ \begin{array}{ll} 1 & 2 \ 2 & 1 \end{array} \] = \[ \begin{array}{ll} a+2b & 2a+b \ c+2d & 2c+d \end{array} \] $$
Linear Algebra
Linear Algebra is the branch of mathematics concerning vector spaces and linear mappings between such spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.
One of the essential tools in linear algebra is the matrix, especially when it comes to solving systems of linear equations, performing transformations, and optimizing operations. Mathematically, linear algebra deals with structures such as:
  • Vectors and vector spaces
  • Linear transformations
  • Matrices and determinants
  • Eigenvalues and eigenvectors
Considering our matrices A and B from the problem, we want to find conditions where the matrices commute, meaning that: $$ AB = BA $$
This is important in many applications, including solving systems of linear equations, computer graphics, and more. In this case, proving that two matrices commute involves performing matrix multiplication and equating the resulting matrices.
Solving these systematically leads us to a deeper understanding of linear independence, eigenvectors, and other critical concepts in linear algebra.
System of Equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. Such systems can be solved using various methods such as substitution, elimination, and matrix operations like Gaussian elimination.
In our problem, to determine the values of a, b, c, and d for which the matrices A and B commute, we created a system of equations by equating AB to BA and then comparing corresponding elements:
\[ \begin{array}{ll} a+2b = a+2c \ 2a+b = b+2d \ c+2d = 2a+c \ 2c+d = 2b+d \end{array} \]
Simplifying, we get:
  • From equation 1: \(2b = 2c \Rightarrow b = c \)
  • From equation 2: \(2a = 2d \Rightarrow a = d \)

So, the conditions for which the matrices A and B commute are that b must equal c, and a must equal d. These solutions show how interconnected variables must align to meet specific criteria in systems involving matrices, significantly aiding solving linear algebra problems.

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Most popular questions from this chapter

Produce two square matrices \(A\) and \(B\) of order 2 for which \(A B=O\), but \(A\) and \(B\) have all nonzero elements. Can either of the two matrices be nonsingular?

When \(A\) is a symmetric, positive definite matrix, it can be factored in the form $$ A=L L^{T} $$ for some lower triangular matrix \(L\) that has nonzero diagonal elements, usually taken to be positive. This is called the Cholesky factorization. Recall that a symmetric matrix \(A\) is said to be positive definite if for any \(x \neq 0, x^{T} A x>0\) (a) For $$ A=\left[\begin{array}{rr} 1 & -1 \\ -1 & 5 \end{array}\right] $$ let $$ L=\left[\begin{array}{cc} l_{11} & 0 \\ l_{21} & l_{22} \end{array}\right] $$ Find \(L\) such that \(L L^{T}=A\). (b) Repeat this process for $$ A=\left[\begin{array}{crc} 2.25 & -3 & 4.5 \\ -3 & 5 & -10 \\ 4.5 & -10 & 34 \end{array}\right] $$ finding a lower triangular matrix \(L\) for which \(L L^{T}=A\).

Define \(B=w w^{T}\), with \(w\) a column vector of length \(n\). (a) Give an operations count for forming \(B .\) Be as efficient as possible. (b) Let \(A\) be an arbitrary matrix of order \(n \times n\). Give the additional number of operations needed to form the product \(A\) and \(B\), using the matrix \(B\) formed in (a). (c) Give an alternative and less costly way, in operations, to form the product \(A B=A\left(w w^{T}\right)\)

By using Theorem \(6.2 .6\) and assuming \(\operatorname{det}(A)=0\), then there is at least one vector \(x \neq 0\) for which \(A x=0\). For the following singular matrix, find such an \(x\) : $$ \left[\begin{array}{rrrr} 0 & 4 & 1 & 1 \\ 4 & 0 & 1 & 1 \\ 1 & 1 & -1 & 2 \\ 1 & 1 & 2 & -1 \end{array}\right] $$

Consider the linear system $$ \begin{array}{r} x_{1}+4 x_{2}=1 \\ 4 x_{1}+x_{2}=0 \end{array} $$ The true solution is \(x_{1}=-1 / 15, x_{2}=4 / 15\). Apply the Jacobi and Gauss-Seidel methods with \(x^{(0)}=[0,0]^{T}\) to the system and find out which methods diverge more rapidly. Next, interchange the two equations to write the system as $$ \begin{array}{r} 4 x_{1}+x_{2}=0 \\ x_{1}+4 x_{2}=1 \end{array} $$ and apply both methods with \(x^{(0)}=[0,0]^{T}\). Iterate until \(\left\|x-x^{(k)}\right\| \leq 10^{-5}\). Which method converges faster?

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