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Linear Algebra : Linear Algebra

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Example Questions

Example Question #14 : Norms

The taxicab norm on $\mathbb{R}^n$ for a vector (x_1,...,x_n)^T is defined as

||x||_1=|x_1|+...+|x_n|

Given v=(1,2,-3)^T , find ||v||_1 .

Possible Answers:

$||v||_1=\sqrt{6}$

||v||_1=1

||v||_1=0

||v||_1=6

Correct answer:

||v||_1=6

Explanation:

To find ||v||_1 given v=(1,2,-3)^T , we simply do what the taxicab norm formula tells us:

||v||_1=|1|+|2|+|-3|=1+2+3=6

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Example Question #231 : Linear Algebra

Find the euclidean norm of the vector $(\pi,e)$

Possible Answers:

$\pi^2+e^2$

$\sqrt{\pi^2+e^2}$

$\pi+e$

$\sqrt{\pi+e}$

Correct answer:

$\sqrt{\pi^2+e^2}$

Explanation:

To find the euclidean norm of $(\pi,e)$ , we take the sum of the entries squared and take the square root:

$\sqrt{\pi^2+e^2}$

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Example Question #232 : Linear Algebra

Find the euclidean norm of (10,2,5) .

Possible Answers:

$\sqrt{129}$

8.5

$\sqrt{17}$

Correct answer:

$\sqrt{129}$

Explanation:

To find the euclidean norm of (10,2,5) , we take the sum of the entries squared and take the square root:

$\sqrt{10^2+2^2+5^2}=\sqrt{100+4+25}=\sqrt{129}$

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Example Question #233 : Linear Algebra

Find the euclidean norm of (1,1,1,1) .

Possible Answers:

$\sqrt{2}$

1.5

Correct answer:

Explanation:

To find the euclidean norm of (1,1,1,1) , we take the sum of the entries squared and take the square root:

$\sqrt{1^2+1^2+1^2+1^2}=\sqrt{1+1+1=1}=\sqrt{4}=2$

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Example Question #23 : Norms

$\textbf{u} = \left ( \frac{1}{4}, \frac{1}{3} , \ln a \right )$ .

Evaluate to make $\textbf{u}$ a unit vector.

Possible Answers:

0.8188 or 1.2214

0.6592 or 1.5169

$\textbf{u}$ cannot be a unit vector regardless of the value of .

0.8406 or 1.1896

0.4029 or 2.4818

Correct answer:

0.4029 or 2.4818

Explanation:

$\textbf{u}$ is a unit vector if and only if $\left \| \textbf{u} \right \| = 1$

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left \| \textbf{u} \right \| =\sqrt{\left ( \frac{1}{4} \right ) ^{2} +\left ( \frac{1}{3} \right ) ^{2} +( \ln a) ^{2}}$

$\left \| \textbf{u} \right \| =\sqrt{ \frac{1}{16} + \frac{1}{9} +( \ln a) ^{2}}$

$\left \| \textbf{u} \right \| =\sqrt{ \frac{25}{144} +( \ln a) ^{2}}$

Set this expression equal to 1:

$\sqrt{ \frac{25}{144} +( \ln a) ^{2}} = 1$

$\frac{25}{144} +( \ln a) ^{2} = 1$

$( \ln a) ^{2} = \frac{119}{144}$

$\ln a = \pm \frac{\sqrt{119 }}{12 } \approx \pm 0.9091$

$a \approx e^{-0.9091} \approx 0.4029$

$a \approx e^{ 0.9091} \approx 2.4818$

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Example Question #234 : Linear Algebra

$\textbf{u} = \left (\cos \theta , \frac{1}{2} , \frac{2}{3} \right )$

Evaluate $\theta \in [0, 2 \pi]$ (nearest hundredth of a radian) to make $\textbf{u}$ a unit vector.

Possible Answers:

0.59

0.17

1.40

0.99

$\textbf{u}$ cannot be a unit vector regardless of the value of $\theta$ .

Correct answer:

0.99

Explanation:

$\textbf{u}$ is a unit vector if and only if $\left \| \textbf{u} \right \| = 1$

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left \| \textbf{u} \right \| = \sqrt{ \cos^{2} \theta+ \left ( \frac{1}{2} \right ) ^{2}+ \left ( \frac{2}{3} \right ) ^{2}}$

$\left \| \textbf{u} \right \| = \sqrt{ \cos^{2} \theta+ \frac{1}{4} + \frac{4}{9} }$

$\left \| \textbf{u} \right \| = \sqrt{ \cos^{2} \theta+ \frac{25}{36} }$

Set this value equal to 1:

$\sqrt{ \cos^{2} \theta+ \frac{25}{36} } = 1$

$\cos^{2} \theta+ \frac{25}{36} = 1$

$\cos^{2} \theta= \frac{11}{36}$

$\cos \theta=\sqrt{ \frac{11}{36}} \approx 0.5528$

We are looking for a value in radians $\theta \in [0, 2 \pi]$ , so

$\theta \approx \cos^{-1} 0.5528 \approx 0.99$ .

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Example Question #23 : Norms

$\textbf{u} = \left ( \cos \theta , \frac{1}{6}, \sin \theta \right )$ .

To the nearest hundredth (radian), which of the following values of $\theta$ would make $\textbf{u}$ a unit vector?

Possible Answers:

0.59

2.56

2.91

$\textbf{u}$ cannot be a unit vector regardless of the value of $\theta$ .

0.99

Correct answer:

$\textbf{u}$ cannot be a unit vector regardless of the value of $\theta$ .

Explanation:

$\textbf{u}$ is a unit vector if and only if $\left \| \textbf{u} \right \| = 1$

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left \| \textbf{u} \right \| =\sqrt{ \cos ^{2}\theta +\left ( \frac{1}{6} \right ) ^{2} + \sin ^{2}\theta}$

$\left \| \textbf{u} \right \| =\sqrt{ \sin ^{2}\theta+\cos ^{2}\theta + \frac{1}{36} }$

Since by a trigonometric identity,

$\sin ^{2}\theta+\cos ^{2}\theta =1$ for all $\theta$ ,

$\left \| \textbf{u} \right \| =\sqrt{ 1+ \frac{1}{36} }$

$\left \| \textbf{u} \right \| =\sqrt{ \frac{37}{36} } = \frac{\sqrt{ 37 } } {6}$ .

Therefore, for any value of $\theta$ , $\left \| \textbf{u} \right \| \ne 1$ . $\textbf{u}$ cannot be a unit vector.

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Example Question #23 : Norms

$\textbf{u} = \left (\frac{3}{5} \cos \theta ,\frac{4}{5} \sin \theta , -\frac{3}{5} \sin \theta, -\frac{4}{5} \cos \theta \right )$

True or false: $\textbf{u}$ is a unit vector regardless of the value of $\theta$ .

Possible Answers:

True

False

Correct answer:

True

Explanation:

$\textbf{u}$ is a unit vector if and only if $\left \| \textbf{u} \right \| = 1$

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ can be found by adding the squares of the entries and taking the square root of the sum:

$\left \| \textbf{u} \right \| =\sqrt{ \left (\frac{3}{5} \cos \theta \right ) ^{2}+ \left (\frac{4}{5} \sin \theta\right )^{2} + \left ( -\frac{3}{5} \sin \theta\right )^{2}+ \left ( -\frac{4}{5} \cos \theta \right )^{2}}$

$\left \| \textbf{u} \right \| =\sqrt{ \frac{9}{25} \cos^{2} \theta + \frac{16}{25} \sin^{2} \theta+ \frac{9}{25} \sin^{2} \theta + \frac{16}{25} \cos^{2} \theta}$

$\left \| \textbf{u} \right \| =\sqrt{ \sin^{2} \theta + \cos^{2} \theta}$

Applying a trigonometric identity:

$\left \| \textbf{u} \right \| =\sqrt{1} = 1$ .

Therefore, $\textbf{u}$ is a unit vector regardless of the value of $\theta$ .

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Example Question #25 : Norms

$\textbf{u} = (1, -3, 2 )$

$\textbf{v} = (0, 1, a)$

Express the distance between $\textbf{u}$ and $\textbf{v}$ in terms of .

Possible Answers:

None of the other choices gives the correct response.

$\sqrt{ a^{2}+4a-11}$

$\sqrt{ a^{2}+4a+21}$

$\sqrt{ a^{2}-4a+21}$

$\sqrt{ a^{2}-4a-11}$

Correct answer:

$\sqrt{ a^{2}-4a+21}$

Explanation:

The distance between the vectors $\textbf{u}$ and $\textbf{v}$ is $\left \| \textbf{u} - \textbf{v} \right \|$ , the norm of their difference.

First, find $\textbf{u} - \textbf{v}$ by elementwise subtraction:

$\textbf{u} - \textbf{v} = (1-0, -3-1, 2-a)$

$\textbf{u} - \textbf{v} = (1 , -4, 2-a)$

$\left \| \textbf{u} - \textbf{v} \right \|$ , the norm of this difference, can be found by adding the squares of the elements, then taking the square root:

$\left \| \textbf{u} - \textbf{v} \right \| = \sqrt{ 1^{2} + (-4)^{2}+ ( 2-a)^{2}}$

$\left \| \textbf{u} - \textbf{v} \right \|= \sqrt{1+16 + a^{2}-4a+4}$

$\left \| \textbf{u} - \textbf{v} \right \|= \sqrt{ a^{2}-4a+21}$ ,

the correct choice.

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Example Question #23 : Norms

$\textbf{u} = (3, -1, 6)$

$\textbf{v}= (5, 2, a)$ , an integer.

For which values of does it hold that $\left \| \textbf{u} \right \|< \left \| \textbf{v} \right \|$ ?

Possible Answers:

$a \in \left \{ -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \right \}$

$a \in \left \{ ... -6, -5,-4, 4, 5, 6, ...\right \}$

$a \in \left \{ -4, -3, -2, -1, 0, 1, 2, 3, 4 \right \}$

$a \in \left \{ 0, 1, 2, 3, 4 ,5,...\right \}$

$a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$

Correct answer:

$a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$

Explanation:

$\left \| \textbf{u} \right \|$ , the norm, or length, of $\textbf{u}$ , can be calculated by adding the squares of the numbers and taking the square root of the sum; $\left \| \textbf{v} \right \|$ can be calculated similarly.

We are seeking the real values of so that $\left \| \textbf{u} \right \|< \left \| \textbf{v} \right \|$ ; since both norms must be nonnegative, it suffices to find so that $\left \| \textbf{u} \right \|^{2} < \left \| \textbf{v} \right \| ^{2}$ .

$\left \| \textbf{u} \right \|^{2} = 3^{2}+ (-1)^{2}+ 6^{2} = 9 + 1 + 36 = 46$

$\left \| \textbf{v} \right \| ^{2} = 5^{2}+ 2^{2}+ a^{2} = 25 + 4 + a^{2} = a^{2}+ 29$

For $\left \| \textbf{u} \right \|^{2} < \left \| \textbf{v} \right \| ^{2}$ to hold, it must hold that

$46 < a^{2}+ 29$ , or

$a^{2}+ 29> 46$

This is true if

$a^{2} > 17$ ,

which in turn holds if

$|a| > \sqrt{17}$ .

Since it is specified that is an integer, it holds that $a \in \left \{ ...-7, -6, -5, 5, 6, 7, ...\right \}$ .

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #14 : Norms

Example Question #231 : Linear Algebra

Example Question #232 : Linear Algebra

Example Question #233 : Linear Algebra

Example Question #23 : Norms

Example Question #234 : Linear Algebra

Example Question #23 : Norms

Example Question #23 : Norms

Example Question #25 : Norms

Example Question #23 : Norms

All Linear Algebra Resources

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