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MAVERICK [17]
3 years ago
8

Help please and ty ! (no its not 388)

Mathematics
2 answers:
jeka943 years ago
5 0
It is 484 I knwonir
Yakvenalex [24]3 years ago
5 0

Answer:

It is 484

Step-by-step explanation:

Happy Holidays❄️☃️

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Can someone help me 19 to 32
andreev551 [17]

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Step-by-step explanation:

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Answer:

17.5  yds =r

Step-by-step explanation:

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* Let S = Span {(2,-1, 1), (3, 1, 1), (1, 2, 0)}. (i) Calculate the dimension of S.
Sholpan [36]

The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.

Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars c_1,c_2,c_3 (not all zero) such that

c_1 (2,-1,1) + c_2 (3,1,1) + c_3 (1,2,0) = (0,0,0)

which leads to the system of linear equations,

\begin{cases} 2c_1 + 3c_2 + c_3 = 0 \\ -c_1 + c_2 + 2c_3 = 0 \\ c_1 + c_2 = 0 \end{cases}

From the third equation, we have c_1=-c_2, and substituting this into the second equation gives

-c_1 + c_2 + 2c_3 = 2c_2 + 2c_3 = 0 \implies c_2 + c_3 = 0 \implies c_2 = -c_3

and in turn, c_1=c_3. Substituting these into the first equation gives

2c_1 + 3c_2 + c_3 = 2c_3 - 3c_3 + c_3 = 0 \implies 0=0

which tells us that any value of c_3 will work. If c_3 = t, then c_1=t and c_2 = -t. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.

Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.

3 0
2 years ago
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