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Linear Algebra : Linear Algebra

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4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Norms

Find the norm of the vector $\vec{v}= (4, 2, -5 )$ $\displaystyle \vec{v}= (4, 2, -5 )$

Possible Answers:

$5 \sqrt{2}$ $\displaystyle 5 \sqrt{2}$

$3 \sqrt{5}$ $\displaystyle 3 \sqrt{5}$

$\displaystyle 5$

$\displaystyle 9$

$\displaystyle 3$

Correct answer:

$3 \sqrt{5}$ $\displaystyle 3 \sqrt{5}$

Explanation:

$|| \vec{v} || = \sqrt{ 4^2 + 2^2 + (-5)^2 } = \sqrt{16 + 4 + 25 } = \sqrt{45}$ $\displaystyle || \vec{v} || = \sqrt{ 4^2 + 2^2 + (-5)^2 } = \sqrt{16 + 4 + 25 } = \sqrt{45}$

This can be simplified:

$\sqrt{3 \cdot 3 \cdot 5 } = 3 \sqrt {5}$ $\displaystyle \sqrt{3 \cdot 3 \cdot 5 } = 3 \sqrt {5}$

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Example Question #1 : Norms

Find the norm of the vector $\vec{v} = (2, -3, -6 )$ $\displaystyle \vec{v} = (2, -3, -6 )$

Possible Answers:

$\sqrt{23}$ $\displaystyle \sqrt{23}$

$\displaystyle 6$

$\displaystyle 7$

$\sqrt{41 }$ $\displaystyle \sqrt{41 }$

Correct answer:

$\displaystyle 7$

Explanation:

$|| \vec{v} || = \sqrt{ 2^2 + (-3)^2 + (-6)^2 } = \sqrt{4 + 9 + 36 } = \sqrt{49} = 7$ $\displaystyle || \vec{v} || = \sqrt{ 2^2 + (-3)^2 + (-6)^2 } = \sqrt{4 + 9 + 36 } = \sqrt{49} = 7$

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Example Question #221 : Linear Algebra

Find the norm of the vector $\vec{v} = (3, -2, -2)$ $\displaystyle \vec{v} = (3, -2, -2)$

Possible Answers:

$\displaystyle 1$

$\sqrt{17}$ $\displaystyle \sqrt{17}$

$\sqrt{15}$ $\displaystyle \sqrt{15}$

$\sqrt{7}$ $\displaystyle \sqrt{7}$

Correct answer:

$\sqrt{17}$ $\displaystyle \sqrt{17}$

Explanation:

$|| \vec{v} || = \sqrt{3^2 + (-2)^2 + (-2)^2 } = \sqrt{9+4+4} = \sqrt{17}$ $\displaystyle || \vec{v} || = \sqrt{3^2 + (-2)^2 + (-2)^2 } = \sqrt{9+4+4} = \sqrt{17}$

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Example Question #222 : Linear Algebra

Find the norm of the vector $\vec{v} = (4, -1, 3 )$ $\displaystyle \vec{v} = (4, -1, 3 )$

Possible Answers:

$\sqrt{26}$ $\displaystyle \sqrt{26}$

$\sqrt{24}$ $\displaystyle \sqrt{24}$

$\sqrt{6}$ $\displaystyle \sqrt{6}$

$\displaystyle 5$

Correct answer:

$\sqrt{26}$ $\displaystyle \sqrt{26}$

Explanation:

$|| \vec{v} || = \sqrt{ 4^2 + (-1)^2 + 3^2 } = \sqrt{16+1+9 } = \sqrt{26}$ $\displaystyle || \vec{v} || = \sqrt{ 4^2 + (-1)^2 + 3^2 } = \sqrt{16+1+9 } = \sqrt{26}$

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Example Question #223 : Linear Algebra

Find the norm of the vector $\vec{v} = (3, 4, 5 )$ $\displaystyle \vec{v} = (3, 4, 5 )$

Possible Answers:

$5 \sqrt{2}$ $\displaystyle 5 \sqrt{2}$

$3 \sqrt{2}$ $\displaystyle 3 \sqrt{2}$

$2\sqrt{5}$ $\displaystyle 2\sqrt{5}$

$2 \sqrt{3}$ $\displaystyle 2 \sqrt{3}$

Correct answer:

$5 \sqrt{2}$ $\displaystyle 5 \sqrt{2}$

Explanation:

$|| \vec{v} || = \sqrt{ 3^2 + 4^2 + 5^2 } = \sqrt{9+16+25} = \sqrt{50 }$ $\displaystyle || \vec{v} || = \sqrt{ 3^2 + 4^2 + 5^2 } = \sqrt{9+16+25} = \sqrt{50 }$

This can be simplified:

$\sqrt{2 \cdot 5 \cdot 5 } = 5 \sqrt{2 }$ $\displaystyle \sqrt{2 \cdot 5 \cdot 5 } = 5 \sqrt{2 }$

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Example Question #144 : Operations And Properties

Let $\textbf{u} = (2, 3, a, 1 )$ $\displaystyle \textbf{u} = (2, 3, a, 1 )$ for some real number $\displaystyle a$.

Give $\displaystyle a$ such that $\left \| \textbf{u} \right \| = 4$ $\displaystyle \left \| \textbf{u} \right \| = 4$.

Possible Answers:

$\displaystyle -2$

$\pm \sqrt{2}$ $\displaystyle \pm \sqrt{2}$

$\pm \frac{16}{13}$ $\displaystyle \pm \frac{16}{13}$

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\pm \frac{4 \sqrt{13}}{13}$ $\displaystyle \pm \frac{4 \sqrt{13}}{13}$

Correct answer:

$\pm \sqrt{2}$ $\displaystyle \pm \sqrt{2}$

Explanation:

$\left \| \textbf{u} \right \|$ $\displaystyle \left \| \textbf{u} \right \|$, the norm, or length, of vector $\textbf{u}$ $\displaystyle \textbf{u}$, is equal to the square root of the sum of the squares of its elements. Therefore,

$\left \| \textbf{u} \right \| = \sqrt{2^{2}+3^{2}+a^{2}+1^{2}}$ $\displaystyle \left \| \textbf{u} \right \| = \sqrt{2^{2}+3^{2}+a^{2}+1^{2}}$

$= \sqrt{4+9+a^{2}+1}$ $\displaystyle = \sqrt{4+9+a^{2}+1}$

$= \sqrt{ a^{2}+14}$ $\displaystyle = \sqrt{ a^{2}+14}$

Set this equal to 4:

$\sqrt{ a^{2}+14} = 4$ $\displaystyle \sqrt{ a^{2}+14} = 4$

$a^{2}+14 = 16$ $\displaystyle a^{2}+14 = 16$

$a^{2}= 2$ $\displaystyle a^{2}= 2$

$a = \pm \sqrt{2}$ $\displaystyle a = \pm \sqrt{2}$

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Example Question #145 : Operations And Properties

$\textbf{u}= (-3, 2, x, x)$ $\displaystyle \textbf{u}= (-3, 2, x, x)$,

where $\displaystyle x$ is a real number.

In terms of $\displaystyle x$, give $\left \| \textbf{u} \right \|$ $\displaystyle \left \| \textbf{u} \right \|$.

Possible Answers:

$\sqrt{2x-1}$ $\displaystyle \sqrt{2x-1}$

$x \sqrt{6}$ $\displaystyle x \sqrt{6}$

$\sqrt{2 x ^{2} +13}$ $\displaystyle \sqrt{2 x ^{2} +13}$

$\sqrt{2x+5}$ $\displaystyle \sqrt{2x+5}$

$\sqrt{2 x ^{2} -5}$ $\displaystyle \sqrt{2 x ^{2} -5}$

Correct answer:

$\sqrt{2 x ^{2} +13}$ $\displaystyle \sqrt{2 x ^{2} +13}$

Explanation:

$\left \| \textbf{u} \right \|= \sqrt{(-3)^{2}+ 2 ^{2}+ x ^{2} +x ^{2}}$ $\displaystyle \left \| \textbf{u} \right \|= \sqrt{(-3)^{2}+ 2 ^{2}+ x ^{2} +x ^{2}}$

$=\sqrt{9+4+ x ^{2} +x ^{2}}$ $\displaystyle =\sqrt{9+4+ x ^{2} +x ^{2}}$

$=\sqrt{2 x ^{2} +13}$ $\displaystyle =\sqrt{2 x ^{2} +13}$

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Example Question #146 : Operations And Properties

$\bold{v} = \left ( \frac{1}{3}, - \frac{1}{3},\frac{1}{3} \right )$ $\displaystyle \bold{v} = \left ( \frac{1}{3}, - \frac{1}{3},\frac{1}{3} \right )$

True or false: $\textbf{v}$ $\displaystyle \textbf{v}$ is an example of a unit vector.

Possible Answers:

True

False

Correct answer:

False

Explanation:

$\textbf{v}$ $\displaystyle \textbf{v}$ is a unit vector if and only if its norm, or length, $\left \| \textbf{v} \right \|$ $\displaystyle \left \| \textbf{v} \right \|$ - the square root of the sum of the squares of its elements - is equal to 1. Find the length using this definition:

$\left \| \textbf{v} \right \| = \sqrt{\left ( \frac{1}{3} \right )^{2}+ \left ( -\frac{1}{3} \right )^{2}+ \left ( \frac{1}{3} \right )^{2}}$ $\displaystyle \left \| \textbf{v} \right \| = \sqrt{\left ( \frac{1}{3} \right )^{2}+ \left ( -\frac{1}{3} \right )^{2}+ \left ( \frac{1}{3} \right )^{2}}$

$= \sqrt{\frac{1}{9}+ \frac{1}{9} + \frac{1}{9} }$ $\displaystyle = \sqrt{\frac{1}{9}+ \frac{1}{9} + \frac{1}{9} }$

$= \sqrt{\frac{1}{3} }$ $\displaystyle = \sqrt{\frac{1}{3} }$

$\left \| \textbf{v} \right \| \ne 1$ $\displaystyle \left \| \textbf{v} \right \| \ne 1$, so $\textbf{v}$ $\displaystyle \textbf{v}$ is not a unit vector.

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Example Question #14 : Norms

$\bold{v} = \left ( \frac{1}{2},\frac{ \sqrt{2}}{2}, -\frac{1}{2} \right )$ $\displaystyle \bold{v} = \left ( \frac{1}{2},\frac{ \sqrt{2}}{2}, -\frac{1}{2} \right )$

True or false: $\textbf{v}$ $\displaystyle \textbf{v}$ is an example of a unit vector.

Possible Answers:

False

True

Correct answer:

True

Explanation:

$\left \| \textbf{v} \right \| = \sqrt{\left ( \frac{1}{2} \right )^{2}+ \left ( \frac{\sqrt{2}}{2} \right ) ^{2}+ \left ( -\frac{1}{2} \right ) ^{2}}$ $\displaystyle \left \| \textbf{v} \right \| = \sqrt{\left ( \frac{1}{2} \right )^{2}+ \left ( \frac{\sqrt{2}}{2} \right ) ^{2}+ \left ( -\frac{1}{2} \right ) ^{2}}$

$=\sqrt{ \frac{1}{4} + \frac{2}{4} + \frac{1}{4} }$ $\displaystyle =\sqrt{ \frac{1}{4} + \frac{2}{4} + \frac{1}{4} }$

$=\sqrt{1}$ $\displaystyle =\sqrt{1}$

$\displaystyle =1$

$\textbf{v}$ $\displaystyle \textbf{v}$ is a unit vector.

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Example Question #15 : Norms

Which of these functions could be that of a Euclidean norm operator? You may assume each function is onto.

Possible Answers:

All of the other answers are norm operators

$h: \mathbb{N}^3 \rightarrow \mathbb{Q}$ $\displaystyle h: \mathbb{N}^3 \rightarrow \mathbb{Q}$

$f:\mathbb{R}^7 \rightarrow \mathbb{R}$ $\displaystyle f:\mathbb{R}^7 \rightarrow \mathbb{R}$

$g: \mathbb{Q}^2 \rightarrow \mathbb{N}$ $\displaystyle g: \mathbb{Q}^2 \rightarrow \mathbb{N}$

Correct answer:

$f:\mathbb{R}^7 \rightarrow \mathbb{R}$ $\displaystyle f:\mathbb{R}^7 \rightarrow \mathbb{R}$

Explanation:

This function's range is $\mathbb{R}$ $\displaystyle \mathbb{R}$, the set of all real numbers. In short, this is set of all possible "distances between two given numbers" in elementary linear algebra. $\displaystyle h$ would not be a norm. For example, $h(1,1,1) = \sqrt{1+1+1} =\sqrt{3}$ $\displaystyle h(1,1,1) = \sqrt{1+1+1} =\sqrt{3}$, which is not a rational number (part of $\mathbb{Q}$ $\displaystyle \mathbb{Q}$). Similarly, $\displaystyle g$ is also not a norm. We have $g(1,1) = \sqrt{1+1} = \sqrt{2}$ $\displaystyle g(1,1) = \sqrt{1+1} = \sqrt{2}$, which is not a natural number.

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : Norms

Example Question #1 : Norms

Example Question #221 : Linear Algebra

Example Question #222 : Linear Algebra

Example Question #223 : Linear Algebra

Example Question #144 : Operations And Properties

Example Question #145 : Operations And Properties

Example Question #146 : Operations And Properties

Example Question #14 : Norms

Example Question #15 : Norms

All Linear Algebra Resources

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