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2 years ago

WILL MARK BRAINIEST

Mathematics

1 answer:

antiseptic1488 [7]2 years ago

3 0

A) (0,1), (1,-2), (-1,4), (2,-5)
b) (0,4), (-1,2), (-2,0), (1,6)
c) (-2,-7), (-1, -7), (0,-7), (1, -7)

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Write in slope-intercept form the equation of the line passing through the given point and perpendicular to the given line.

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2 years ago

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3 years ago

Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.

Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]$

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]$

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

$\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]$

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

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2 years ago