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USPshnik [31]
3 years ago
9

Suppose v1 , v2 , v3 ,v4 are vectors in R3.

Mathematics
1 answer:
mestny [16]3 years ago
5 0

Answer:

a. These four vectors are dependent because there are columns of 3 by 4 matrix with one free variable.

b. If one is a multiple of other

c. c1v1 + c20 = 0 has nontrivial solution.

Step-by-step explanation:

Any set of 4 or more vectors must be linearly dependent. The non trivial combination of vector may produce zero as the set is linearly dependent. The vector v1 and v2 will be dependent if one is the multiple of the other.

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Step-by-step explanation:

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2 years ago
The sum of three numbers is 8. The third is 9
Ierofanga [76]

Answer:

→<u> </u><u>First</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>1</u>

→<u> </u><u>Second</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>2</u>

→<u> </u><u>Third</u><u> </u><u>value</u><u> </u><u>is</u><u> </u><u>5</u>

Step-by-step explanation:

• let numbers be x, y and z

{ \tt{z = 8y - 9 -  -  - (eqn \: 1)}} \\  \\ { \tt{10x = 8y - 7 -  -  - (eqn \: 2)}} \\  \\ { \tt{x + y + z = 8 -  -  - (eqn \: 3)}}

• from eqn 2, make x the subject:

{ \tt{x =  \frac{8y - 7}{10} }} \\

• substitute all variables in eqn 3:

{ \tt{ \frac{8y - 7}{10}  + y + 8y - 9 = 8}} \\  \\ { \tt{8y - 7 + 10y + 80y - 90 = 80}} \\  \\ { \tt{98y = 177}} \\  \\ { \boxed{ \tt{ \: y = 1.8}}}

• find z

{ \tt{z = 8y - 9}} \\  \\ { \tt{z = 8(1.8) - 9}} \\  \\ { \tt{z = 14.4 - 9}} \\  \\ { \boxed{ \tt{ \: z = 5.4 \: }}}

• find x:

{ \tt{x =  \frac{8(1.8) - 7}{10} }} \\  \\ { \tt{x =  \frac{7.4}{10} }} \\  \\ { \boxed{ \tt{ \: x = 0.74 \: }}}

Rounding to nearest value:

{ \boxed{ \rm{x = 1}}} \\ { \boxed{ \rm{y =2 }}} \\ { \boxed{  \rm{z = 5}}}

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Answer:

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Step-by-step explanation:

26- (-23)

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3 years ago
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