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3 years ago

If A,B are n×n matrices, then (A + B)^2 = A^2 + 2AB + B^2. Either prove that this statement is true, or explain why it is false.

Mathematics

1 answer:

bonufazy [111]3 years ago

6 0

<h2>Answer:</h2>

The given statement is false.

The statement is necessarily true when two matrix A and B commute.

<h2>Step-by-step explanation:</h2>

This statement is false.

Since, for two matrices A and B , the expansion of the term:

$(A+B)^2$ is given by:

$(A+B)^2=(A+B)(A+B)\\\\i.e.\\\\(A+B)^2=A^2+B\cdot A+A\cdot B+B^2$

Also, we know that when two matrix A and B commute then we have:

$A\cdot B=B\cdot A$

Hence, we get the expression as:

$(A+B)^2=A^2+2A\cdot B+B^2$

But when two matrix A and B do not commute then we need not have:

$(A+B)^2=A^2+2A\cdot B+B^2$

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74th percentile

value of X when Z has a pvalue of 0.74. So X when Z = 0.643.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$0.643 = \frac{X - 220}{2.1974}$

$X - 220 = 0.643*2.1974$

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Value of X when Z has a pvalue of 0.26. So X when Z = -0.643

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$-0.643 = \frac{X - 220}{2.1974}$

$X - 220 = -0.643*2.1974$

$X = 218.59$

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