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mrs_skeptik [129]
2 years ago
8

If the null space of a 4 x 9 matrix is 5-dimensional, what is the dimension of the column space of the matrix? Explain your answ

er b) (3 points) If A is a 7 x 3 matrix, what is the smallest possible dimension of Nul A? Explain your answer
Mathematics
1 answer:
Naily [24]2 years ago
7 0

Answer : Dimension of column A is also be 4 whereas the two vector basis lie in R⁴.

The smallest possible dimension of Nul A would be zero.

Step-by-step explanation:

Since we have given that

A is matrix of 4 x 9 .

so, Number of rows = 4

Number of columns = 9

Nul A = 5

It means that Rank of A would be 9 - 5 =4

So, rank A = 4

Thus, dimension of column A is also be 4 whereas the four vector basis lie in R⁴.

So, dim Col A = 4

If A is 7 x  3 matrix.

So, we know that

rank A + dim (null A) = 3

so, it is possible to have rank A = 3 so the dim col A should be 3

Then the smallest possible dimension of Nul A would be zero.

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How would you choose to reduce the system shown to a 2 × 2? Explain why you would choose this approach. –3x + y – 2z = 10 (1) 5x
gladu [14]
-3x + y - 2z = 10      |* -1
3x  -  y  +2z = -10
5x  -2y -2z =  12 
---------------------------      I add these equations   term by term 
8x  - 3y  = 2 

-3x  + y - 2z =10                       ⇒  -3x  + y   - 2z =10
x     -y    +z = 23         | *2              2x   - 2y + 2z = 46
                                                   -----------------------------  I add these eq.
                                                       -x  -y  = 56

8x  - 3y  = 2 
-x   -y    = 56

this is the system after i reduce it ( it has only two variables x and y)




4 0
2 years ago
Read 2 more answers
LESSON 1 SESSION 1
denpristay [2]

Answer:

  • <em>Between which two tens does it fall?</em><em> </em><u>Between 25 and 26 tens</u>

<em><u /></em>

  • <em>Between which two hundreds does it fall?</em> <u>Between 2 and 3 hundreds</u>

Explanation:

The place-value chart is:

Hundreds         Tens      Ones

       2                   5             3

<em><u /></em>

<em><u>a)  Between which two tens does it fall? </u></em>

Using the place values you can write 253 = 25 × 10 + 3, i.e. 25 tens and 3 ones.

From that you can write:

  • 250 < 253 < 260
  • 250 = 25 × 10 = 25 tens
  • 260 = 26 × 10 = 26 tens

Then, you conclude that 253 is between 25 and 26 tens.

<u><em>b) Between which two hundreds does it fall?</em></u>

Using the same reasoning:

  • 253 = 2 × 100 + 5 × 10 + 3 = 253

  • 200 < 253 < 300
  • 200 = 2 hundreds
  • 300 = 3 hundreds

Conclusion: 253 is between 2 hundreds and 3 hundreds.

3 0
2 years ago
Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
Harlamova29_29 [7]

Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

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Answer:

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