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GRE Subject Test: Math : Linear Algebra

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Example Questions

Example Question #61 : Vectors

What is the vector form of -6j ?

Possible Answers:

$\left \langle 0,-6,0 \right \rangle$

$\left \langle 0,0,-6 \right \rangle$

$\left \langle 6, 0,0 \right \rangle$

$\left \langle 0,6,0 \right \rangle$

$\left \langle -6, 0,0 \right \rangle$

Correct answer:

$\left \langle 0,-6,0 \right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for -6j , we can derive the vector form $\left \langle 0,-6,0 \right \rangle$ .

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Example Question #82 : Linear Algebra

Given points (0,1,3) and (5,7,4) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 5,-6,1\right \rangle$

$\left \langle 5,6,1\right \rangle$

$\left \langle -5,-6,-1\right \rangle$

$\left \langle -5,6,1\right \rangle$

$\left \langle -5,-6,1\right \rangle$

Correct answer:

$\left \langle 5,6,1\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point

$a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ ,

the distance is the vector

$v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (0,1,3) and (5,7,4) , we get:

$v=\left \langle 5-0,7-1,4-3\right \rangle$

$v=\left \langle 5,6,1\right \rangle$

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Example Question #63 : Vectors & Spaces

Given points (2,4,5) and (9,1,0) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 7,3,5\right \rangle$

$\left \langle -7,-3,-5\right \rangle$

$\left \langle 7,-3,-5\right \rangle$

$\left \langle -7,3,-5\right \rangle$

$\left \langle 7,3,-5\right \rangle$

Correct answer:

$\left \langle 7,-3,-5\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point

$a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ ,

the distance is the vector

$v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (2,4,5) and (9,1,0) , we get:

$v=\left \langle 9-2,1-4,0-5\right \rangle$

$v=\left \langle 7,-3,-5\right \rangle$

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Example Question #61 : Vector Form

Given points (0,0,-4) and (4,9,-1) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle -4,9,-3\right \rangle$

$\left \langle 4,9,-3\right \rangle$

$\left \langle 4,9,3\right \rangle$

$\left \langle 4,-9,3\right \rangle$

$\left \langle -4,9,3\right \rangle$

Correct answer:

$\left \langle 4,9,3\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points. That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (0,0,-4) and (4,9,-1) , we get:

$v=\left \langle 4-0,9-0,-1+4\right \rangle$

$v=\left \langle 4,9,3\right \rangle$

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Example Question #61 : Vector Form

What is the derivative of the vector function

$r(t)=(\cos t, \sin t)$ ?

Possible Answers:

$r'(t) = (\sin t,\cos t)$

$r'(t) = (\sin t,-\cos t)$

$r'(t) = (-\sin t,\cos t)$

$r'(t) = (-\cos t,\sin t)$

Correct answer:

$r'(t) = (-\sin t,\cos t)$

Explanation:

The derivative of a vector function has its components consisting of the derivatives of its components:

$r'(t) = (-\sin t,\cos t)$

since the derivatives of $\cos t, \sin t$ are $-\sin t, \cos t$ , respectively.

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Example Question #442 : Gre Subject Test: Math

Given the vector function

r(t)=(t,t^2,t^3)

what is the derivative of the vector function?

Possible Answers:

r'(t)=(1,2t,3t^2)

r'(t)=(t^2/2,t^3/3,t^4/4)

r'(t)=(3t^2,2t,1)

r'(t)=(0,2,6t)

Correct answer:

r'(t)=(1,2t,3t^2)

Explanation:

The derivative of a vector function is just the derivative of components:

r'(t)=(f_1'(t),f_2'(t),f_3'(t))=(1,2t,3t^2)

The derivative of each component was found using the power rule which states,

$x^n dx=nx^{n-1}$ .

Thus,

$t\ dt=1\cdot t^{1-1}=1\cdot t^0=1$ ,

$t^2dt=2t^{2-1}=2t^1=2t$ ,

$t^3dt=3t^{3-1}=3t^2$ .

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Example Question #61 : Vectors & Spaces

What is not a way of describing the circle x^2+y^2=1 in parametric form? Use the time interval $t\in[0,2\pi]$ .

Possible Answers:

$r(t)=(-\cos t,-\sin -t)$

$r(t)=(-\cos t,\sin -t)$

All of the other answers describe the circle mentioned in the question.

$r(t) = (\sin t,\cos t)$

Correct answer:

All of the other answers describe the circle mentioned in the question.

Explanation:

All of the vector forms above describe the circle x^2+y^2=1 since if we square their components and add them, we get (using trig. identities):

$(x,y)\rightarrow (cos(x), sin(x))$

Therefore we get,

$x^2+y^2=\sin^2+\cos^2=1$ .

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Example Question #62 : Vector

What is the vector form of -2j+k ?

Possible Answers:

$\left \langle -2,0,1\right \rangle$

$\left \langle 0,1,-2\right \rangle$

$\left \langle 1,0,-2\right \rangle$

$\left \langle 0,-2,1\right \rangle$

$\left \langle 1,-2,0\right \rangle$

Correct answer:

$\left \langle 0,-2,1\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for -2j+k , we can derive the vector form $\left \langle 0,-2,1\right \rangle$ .

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Example Question #261 : Algebra

Given points (3,0,4) and (4,6,-2) , what is the vector form of the distance between the points?

Possible Answers:

$\left \langle 1,-6,6\right \rangle$

$\left \langle -1,-6,-6\right \rangle$

$\left \langle -1,6,6\right \rangle$

$\left \langle 1,-6,-6\right \rangle$

$\left \langle 1,6,-6\right \rangle$

Correct answer:

$\left \langle 1,6,-6\right \rangle$

Explanation:

In order to derive the vector form of the distance between two points, we must find the difference between the , , and elements of the points.

That is, for any point $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , the distance is the vector $v=\left \langle b_1-a_1,b_2-a_2,b_3-a_3\right \rangle$ .

Subbing in our original points (3,0,4) and (4,6,-2) , we get:

$v=\left \langle 4-3,6-0,-2-4\right \rangle$

$v=\left \langle 1,6,-6\right \rangle$

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Example Question #69 : Vectors

What is the vector form of -5i ?

Possible Answers:

$\left \langle 0,0,-5\right \rangle$

$\left \langle 5,0,0\right \rangle$

$\left \langle -5,0,0\right \rangle$

$\left \langle 0,-5,0\right \rangle$

$\left \langle 0,0,5\right \rangle$

Correct answer:

$\left \langle -5,0,0\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for -5i , we can derive the vector form $\left \langle -5,0,0\right \rangle$ .

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GRE Subject Test: Math : Linear Algebra

Study concepts, example questions & explanations for GRE Subject Test: Math

All GRE Subject Test: Math Resources

Example Questions

Example Question #61 : Vectors

Example Question #82 : Linear Algebra

Example Question #63 : Vectors & Spaces

Example Question #61 : Vector Form

Example Question #61 : Vector Form

Example Question #442 : Gre Subject Test: Math

Example Question #61 : Vectors & Spaces

Example Question #62 : Vector

Example Question #261 : Algebra

Example Question #69 : Vectors

All GRE Subject Test: Math Resources

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