True or False: For n xx n matrices A and B, define A ox B = AB − BA. The operator ox is not associative or commutative.

Question
Commutative Algebra
True or False:
For $$\displaystyle{n}\times{n}$$ matrices A and B, define $$\displaystyle{A}\otimes{B}={A}{B}−{B}{A}$$. The operator ox is not associative or commutative.

2020-10-28
Let A, B, C are all in matrices.
Then,
$$\displaystyle{A}\otimes{\left({B}\otimes{C}\right)}={A}\otimes{\left({B}{C}-{C}{B}\right)}$$
$$\displaystyle={A}{\left({B}{C}-{C}{B}\right)}-{\left({B}{C}-{C}{B}\right)}$$
$$\displaystyle={A}{B}{C}-{A}{C}{B}-{B}{C}{A}+{C}{B}{A}\ldots{\left({1}\right)}$$
$$\displaystyle{\left({A}\otimes{B}\right)}\otimes{C}={\left({A}{B}-{B}{A}\right)}\otimes{C}$$
$$\displaystyle={\left({A}{B}-{B}{A}\right)}{C}-{C}{\left({A}{B}-{B}{A}\right)}$$
$$\displaystyle={A}{B}{C}-{B}{A}{C}-{C}{A}{B}+{C}{B}{A}\ldots{\left({2}\right)}$$
Since (1) and (2) are not equal therefore
$$\displaystyle{A}\otimes{\left({B}\otimes{C}\right)}\ne{\left({A}\otimes{B}\right)}\otimes{C}$$
Hence ox id not associative
Now, $$\displaystyle{A}\otimes{B}={A}{B}-{B}{A}$$
$$\displaystyle{B}\otimes{A}={B}{A}-{A}{b}$$
$$\displaystyle=-{\left({A}{B}-{B}{A}\right)}$$
Therefore $$\displaystyle{A}\otimes{B}\ne{B}\otimes{A}$$
Hence $$\displaystyle\otimes$$ is not commutative.

Relevant Questions

Let A be nonepty set and P(A) be the power set of A. Recall the definition of power set:
$$\displaystyle{P}{\left({A}\right)}={\left\lbrace{x}{\mid}{x}\subseteq{A}\right\rbrace}$$
Show that symmetric deference operation on P(A) define by the formula
$$\displaystyle{x}\oplus{y}={\left({x}\cap{y}^{{c}}\right)}\cup{\left({y}\cap{x}^{{c}}\right)},{x}\in{P}{\left({A}\right)},{y}\in{p}{\left({A}\right)}$$
(where $$\displaystyle{y}^{{c}}$$ is the complement of y) the following statement istrue:
The algebraic operation o+ is commutative and associative on P(A).

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Suppose that a and b belong to a commutative ring and ab is a zero-divisor. Show that either a or b is a zero-divisor.

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
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At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
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$$\mu_1 - \mu_2$$.
lower limit
upper limit
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Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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