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Leviafan [203]
3 years ago
11

Recall that if we have m vectors u1, U2, ..., um in R", then we can form the matrix A whose columns are u1, ... , Um. Let B be t

he echelon form of A. Most questions have a yes no answer, but I am mostly interested in your reasons for the answer. Give full reasons for all answers. Suppose we are given the following matrix B: (3 0 -1 5 0 0 2 -1 10000)
(a) What is n?

(b) What is m?

(c) Are ui, ..., Um linearly independent?

(d) Does {U1, ... , um} span R ?

(e) Looking at B can you write down a subset of the original set {U1, ..., Um} that would be guaranteed to be linearly independent?

(f) Is there a subset of the original set {u1, ... , Um} that would be guaranteed to span R"?

(g) Write down a b e Rn for which Bx = b does not have a solution.

(h) Write down a b e R" for which Bx = b has a unique solution.

(i) Is there a new vector w E Rthat you could add to the set {u1, ..., Um} to guarantee that {U1, ... , um, w} will span R ?

(i) Is there a column of B that is in the span of the rest? If so, find it (k) Put B into reduced echelon form.

Write down a non-zero solution of Ax = 0 if you can. (m) How many free variables are there in the set of solutions to Ax = b when there is a solution? (n) If you erased the last row of zeros in B then would the columns of the resulting matrix be linearly independent?

Can you add rows to B to make the columns of the new matrix linearly independent? If yes, give an example of the new matrix you would construct.
Mathematics
1 answer:
Fed [463]3 years ago
5 0

<em><u>Recall that if we have m vectors u1, U2, ..., um in R", then we can form the matrix A whose columns are u1, ... , Um. Let B be the echelon form of A. Most questions have a yes no answer, but I am mostly interested in your reasons for the answer. Give full reasons for all answers. Suppose we are given the following matrix B: </u></em>

[I'm gonna guess that B really looks like

<em>(3 0 -1 5 </em>

<em> 0 0 2 -1 </em>

0 0 0 0)

and that 1 in 10000 is a typo (as is most of the rest, really).

<u>(a) What is n?</u>

There's actually no n mentioned in the question.  I'll guess the vector space is supposed to be Rⁿ which means A and B are n×m matrices.  So n is the length of each vector, the number of rows in A and B, which I guessed (because the matrix wasn't particularly formatted very well) was three.

Answer: 3

<u>(b) What is m?</u>

m vectors so m columns,

Answer: 4

<u>(c) Are ui, ..., Um linearly independent?</u>

We have a column of all zeros, second one. They can't be linearly independent with a column of zeros because there's always a non-trivial linear combination of the vectors that gives zero.

Answer: NO

<u>(d) Does {U1, ... , um} span R ?</u>

It obviously doesn't span R.  The question probably should be does it span Rⁿ?  All the vectors have a zero as their third element, so so will any linear combination of them.  If we can only get 0 for the last element we can't be spanning the entire 3D space.

In echelon form to span the space we need a pivot in every row, meaning a leading non-zero term.  There's none in the last row.

Answer: NO

<u>(e) Looking at B can you write down a subset of the original set {U1, ..., Um} that would be guaranteed to be linearly independent?</u>

We only have two non-zero rows so our subset has at most two vectors.  We can choose any two of the three non-zero ones, how about

{ (-1,2,0)^T, (5,-1,0)^T }

These are vectors from B. The question is asking for vectors from A, which are what the u's are.   We can't really work backwards to find the vectors from A but we know of the columns are independent from the echelon form they'll be independent in the original A as well.

Answer: { u₃, u₄ }

<u>(f) Is there a subset of the original set {u1, ... , Um} that would be guaranteed to span R"?</u>

Answer: NO

If the full set of vectors doesn't span Rⁿ there's no subset that will either.

<u>(g) Write down a b e Rn for which Bx = b does not have a solution.</u>

Bx is always going to have zero for that last coordinate, no matter what x is.  So

Answer: b=(0,0,2)^T

is impossible.  I wrote ^T because we're after a column vector, the transpose of the vector I typed.

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1+9x14+12=? please help fast
Klio2033 [76]

Answer:

139

Step-by-step explanation:

So we start with PEMDAS. (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction.)

Since we don't have parenthesis or exponents, we can go straight to multiplication. 9 multiplied by 14 is 126.

Now we go to addition. 126+1=127

127+12=139

Hope I helped :)

~Sunflower UwU

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In a valid probability distribution, each probability must be between 0 and 1, inclusive, and the probabilities must add up to 1
lora16 [44]

The value of x is 1/5

Step-by-step explanation:

In a probability distribution, probability should be between 0  and 1. The probabilities add up to one.

Given probabilities are:

\frac{1}{10}, \frac{1}{5}, \frac{1}{2}, x

The sum of these probabilities will be equal to 1.

So,

To find x,

\frac{1}{10}+\frac{1}{5}+\frac{1}{2}+x = 1\\\frac{1+2+5}{10}+x=1\\\frac{8}{10}+x=1\\x= 1- \frac{8}{10}\\x = 1- \frac{4}{5}\\x= \frac{5-4}{5}\\x=\frac{1}{5}

The value of x is 1/5

Keywords: Probability, Distribution

Learn more about probability at:

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Is 19 equal to 4.358
AURORKA [14]

Answer:

No

Step-by-step explanation:

19=19 and 4.358=4.358

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2 years ago
I NEEP HELP ASAP PLEASE AND THANK YOU
sertanlavr [38]

Answer:

The first answer, 3 times the square root of x squared is correct.

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If a seed is planted, it has a 80% chance of growing into a healthy plant.
AlladinOne [14]

Answer:

0.336

Step-by-step explanation:

Use binomial probability:

P = nCr p^r q^(n-r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1-p).

Here, n = 8, r = 7, p = 0.8, and q = 0.2.

P = ₈C₇ (0.8)⁷ (0.2)⁸⁻⁷

P = 0.336

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3 years ago
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