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Anit [1.1K]
2 years ago
10

Let A,B, and C be square invertible matrices of the same size. If C has no eigenvalue equal to -1, then (AB + ACB)^-1 is equal t

o:
a. B^-1 (I + C)^-1*A^-1
b. (I + C)^-1*B^-1*A^-1
c. A^-1B*(I + C)^-1
d. (I + C)^-1*BA^-1
Mathematics
1 answer:
Kisachek [45]2 years ago
6 0

<em>AB</em> + <em>ACB</em> = <em>A</em> (<em>B</em> + <em>CB</em>) = <em>A</em> (<em>I</em> + <em>C </em>) <em>B</em>

Taking the inverse gives

(<em>A</em> (<em>I</em> + <em>C </em>) <em>B</em>)⁻¹ = <em>B </em>⁻¹ (<em>I</em> + <em>C</em> )⁻¹ <em>A</em> ⁻¹

so the answer is (A)

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Answer:

H0 is accepted

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Step-by-step explanation:

Given  a sample survey compared 634 randomly chosen Indians aged 15 to 25 with 567 randomly selected Asians in the same age group.

p_{1} = \frac{368}{634} =0.580

It found that 368 of the Indians and 130 Asians listened to rap music every day.

p_{2} = \frac{130}{567} =0.229

Null hypothesis H0: there is no difference between the proportions of Indian and Asian young people who listen to rap every day.

p_{1} = p_{2}

Alternative hypothesis:- p_{1} \neq  p_{2}

Level of significance α = 0.05

The test of statistic

z = \frac{p_{1} -  p_{2}}{\sqrt{pq(\frac{1}{n_{1} }+\frac{1}{n_{2} })  } } }

where p = \frac{n_{1}p_{1} + n_{1}p_{2}}{n_{1}+n_{2}} = \frac{634(0.580)+567(0.229)}{634+567}

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Z = \frac{0.580-0.229}{\sqrt{0.414(0.586)(\frac{1}{634} +\frac{1}{567} } } }

on calculation , we get

z = 0.300 ><1.96 at 95 % level of significance

H0 is accepted

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