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Linear Algebra : Operations and Properties

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Example Questions

Example Question #43 : Norms

$\textbf{u} = (4, 2, -3)$

$\textbf{v} = (6,2,2)$

These two vectors form two sides of a triangle. In which range does the area of the triangle fall?

Possible Answers:

$A \in (10 , 15)$

$A \in (25,30)$

$A \in (20 , 25)$

$A \in (15,20)$

$A \in (5,10)$

Correct answer:

$A \in (10 , 15)$

Explanation:

The area of a triangle formed by two vectors in $\mathbb{R}^{3}$ is half the norm of their cross-product - that is

$A =\frac{1}{2} \left \| \textbf{u} \times \textbf{v} \right \|$

The cross-product is equal to the "determinant" of the matrix

$\begin{bmatrix} \textbf{i} & \textbf{j} & \textbf{k}\\ 4 & 2 & -3\\ 6 & 2& 2 \end{bmatrix}$ ,

where $\textbf{i} = (1,0,0), \textbf{j} = (0,1,0), \textbf{k} = (0,0,1)$ .

which can be calculated by adding the upper-left to lower-right products and subtracting the upper-right to lower-left products:

$\textbf{u} \times \textbf{v}$

$=\textbf{i} (2)(2)+ \textbf{j} (-3)(6)+ \textbf{k} (4)(2)- \textbf{k}(2)(6)- \textbf{i} (-3)(2) - \textbf{j} (4) (2)$

$=4 \textbf{i} -18 \textbf{j} +8 \textbf{k} - 12 \textbf{k}+6 \textbf{i} -8 \textbf{j}$

$=10 \textbf{i} -26\textbf{j} - 4 \textbf{k}$

= (10, -26, -4)

The norm of this vector is the square root of the sum of the squares of the entries:

$\left \| \textbf{u} \times \textbf{v} \right \| = \sqrt{10^{2}+ (-26)^{2}+ (-4)^{2}} \approx 28.1$

$A \approx \frac{1}{2} \cdot 28.1 \approx 14.1$

The area of the triangle falls in the range $A \in (10 , 15)$ .

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Example Question #42 : Norms

$\textbf{u} = (4, 2, -3)$

$\textbf{v} = (6,2,2)$

$\textbf{u}$ and $\textbf{v}$ form two sides of a triangle. Is the triangle acute, right, or obtuse?

Possible Answers:

Right

Acute

Obtuse

Correct answer:

Acute

Explanation:

First, find the angle $\theta$ between the vectors using the formula

$\cos \theta = \frac{\textbf{u} \cdot \textbf{v}}{\left \| \textbf{u} \right \| \left \| \textbf{v} \right \|}$

Find the dot product $\textbf{u} \cdot \textbf{v}$ by adding the products of corresponding entries:

$\textbf{u} \cdot \textbf{v}= 4(6)+ 2(2)+ (-3)(2) = 22$

Find the lengths, or norms, of $\textbf{u}$ and $\textbf{v}$ , by taking the square roots of the sums of the squares of their entries:

$\left \| \textbf{u} \right \|= \sqrt{4^{2}+2^{2}+ (-3)^{2} } = \sqrt{29}$

$\left \| \textbf{v} \right \| =\sqrt{ 6^{2}+ 2^{2}+ 2^{2} }=\sqrt{44}$

Therefore,

$\cos \theta = \frac{22}{ \sqrt{29} \sqrt{44}} \approx 0.6159$

$\theta \approx 52.0^{\circ }$

To find the measures of the other angles, it is necessary to find the length of the third side, which is equal to $\left \|\textbf{v}-\textbf{u}\right \|$ .

$\textbf{v}-\textbf{u} = (-2, 0, -5 )$ , so

$\left \|\textbf{v}-\textbf{u}\right \| = \sqrt{(-2)^{2}+ 0^{2}+ (-5 )^{2}} = \sqrt{29}$ .

It follows that the triangle is isosceles. This third side, which is one of the two congruent sides, is opposite the $52.0^{\circ }$ angle; by the Isosceles Triangle Theorem the angle opposite the other congruent side is also $52.0^{\circ }$ . The measure of the third angle is

$180^{\circ } -( 52.0 ^{\circ }+ 52.0^{\circ } ) =76^{\circ }$ .

All three angles are acute, so the triangle is an acute triangle.

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Example Question #44 : Norms

$\textbf{u}= (-2, 4, 7)$

Find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left ( -\frac{2}{\sqrt{13}}, \frac{4}{\sqrt{13}}, \frac{7}{\sqrt{13}} \right )$

$\left ( -\frac{2}{\sqrt{69}}, \frac{4}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

$\left ( -\frac{2}{\sqrt{61}}, \frac{4}{\sqrt{61}}, \frac{7}{\sqrt{61}} \right )$

$\left ( -\frac{2}{\sqrt{53}}, \frac{4}{\sqrt{53}}, \frac{7}{\sqrt{53}} \right )$

None of the other choices gives the correct response.

Correct answer:

$\left ( -\frac{2}{\sqrt{69}}, \frac{4}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{(-2)^{2}+4^{2}+7^{2}} = \sqrt{4+16+49} = \sqrt{69}$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} = \frac{ (-2, 4, 7)}{\sqrt{69}} = \left ( -\frac{2}{\sqrt{69}}, \frac{4}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

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Example Question #181 : Operations And Properties

$\textbf{u}= (-4 , -2, 7)$

Find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\textbf{u}$ itself is a unit vector.

$\left ( -\frac{4}{\sqrt{69}},-\frac{2}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

$\left ( -\frac{4}{\sqrt{61}},-\frac{2}{\sqrt{61}}, \frac{7}{\sqrt{61}} \right )$

$\left ( -\frac{4}{\sqrt{53}},-\frac{2}{\sqrt{53}}, \frac{7}{\sqrt{53}} \right )$

$\left ( -\frac{4}{\sqrt{13}},-\frac{2}{\sqrt{13}}, \frac{7}{\sqrt{13}} \right )$

Correct answer:

$\left ( -\frac{4}{\sqrt{69}},-\frac{2}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{(-4)^{2}+ (-2)^{2} +7^{2}} = \sqrt{16+4+ 49} = \sqrt{69}$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} = \frac{ (-4 , -2, 7) }{\sqrt{69}}$

$= \left ( -\frac{4}{\sqrt{69}},-\frac{2}{\sqrt{69}}, \frac{7}{\sqrt{69}} \right )$

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Example Question #51 : Norms

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

In terms of , find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left (-\frac{3}{\sqrt{n^{2}-8}}, \frac{1}{\sqrt{n^{2}-8}} ,\frac{n}{\sqrt{n^{2}-8}} \right )$

$\left (-\frac{3}{\sqrt{3n}}, \frac{1}{\sqrt{3n}} ,\frac{n}{\sqrt{3n}} \right )$

$\left (-\frac{3}{\sqrt{n^{2}+ 10}}, \frac{1}{\sqrt{n^{2}+ 10}} ,\frac{n}{\sqrt{n^{2}+ 10}} \right )$

$\left (-\frac{3}{\sqrt{n-2}}, \frac{1}{\sqrt{n-2}} ,\frac{n}{\sqrt{n-2}} \right )$

$\left (-\frac{3}{\sqrt{n+4}}, \frac{1}{\sqrt{n+4}} ,\frac{n}{\sqrt{n+4}} \right )$

Correct answer:

$\left (-\frac{3}{\sqrt{n^{2}+ 10}}, \frac{1}{\sqrt{n^{2}+ 10}} ,\frac{n}{\sqrt{n^{2}+ 10}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{(-3)^{2}+ 1^{2}+ n^{2}} = \sqrt{9+1+n^{2}} = \sqrt{n^{2}+ 10}$ .

Therefore, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} =\frac{(-3, 1,n)}{\sqrt{n^{2}+ 10}}$

$= \left (-\frac{3}{\sqrt{n^{2}+ 10}}, \frac{1}{\sqrt{n^{2}+ 10}} ,\frac{n}{\sqrt{n^{2}+ 10}} \right )$ .

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Example Question #52 : Norms

$\textbf{u} = (2, -1 , 4, -6)$ .

Find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\textbf{u}$ is itself a unit vector

$\left ( \frac{2}{\sqrt{57}}, -\frac{1 }{\sqrt{57}}, \frac{4}{\sqrt{57}}, -\frac{6}{\sqrt{57}} \right )$

$\left ( \frac{2}{\sqrt{13}}, -\frac{1 }{\sqrt{13}}, \frac{4}{\sqrt{13}}, -\frac{6}{\sqrt{13}} \right )$

$\left ( \frac{2}{\sqrt{23}}, -\frac{1 }{\sqrt{23}}, \frac{4}{\sqrt{23}}, -\frac{6}{\sqrt{23}} \right )$

$\left ( \frac{2}{\sqrt{48}}, -\frac{1 }{\sqrt{48}}, \frac{4}{\sqrt{48}}, -\frac{6}{\sqrt{48}} \right )$

Correct answer:

$\left ( \frac{2}{\sqrt{57}}, -\frac{1 }{\sqrt{57}}, \frac{4}{\sqrt{57}}, -\frac{6}{\sqrt{57}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is $\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$ .

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{2^{2}+(-1)^{2}+4^{2}+(-6)^{2}}$

$= \sqrt{4+1+16+36}$

$=\sqrt{57}$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|} =\frac{ (2, -1 , 4, -6)}{\sqrt{57}}$

$= \left ( \frac{2}{\sqrt{57}}, -\frac{1 }{\sqrt{57}}, \frac{4}{\sqrt{57}}, -\frac{6}{\sqrt{57}} \right )$

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Example Question #53 : Norms

$\textbf{u}=\left ( \frac{1}{5}, -\frac{1}{5}, \frac{4}{5} \right )$

Find the unit vector in the same direction as $\textbf{u}$ .

Possible Answers:

$\left ( \frac{1}{5}, -\frac{1}{ 5}, -\frac{4}{ 5} \right )$

$\left ( \frac{1}{ 4}, -\frac{1}{4},1 \right )$

$\textbf{u}$ itself is a unit vector.

$\left ( \frac{1}{3 \sqrt{2}}, -\frac{1}{ 3\sqrt{2}}, \frac{4}{ 3\sqrt{2}} \right )$

$\left ( \frac{1}{2 \sqrt{3}}, -\frac{1}{ 2\sqrt{3}}, \frac{2}{ \sqrt{3}} \right )$

Correct answer:

$\left ( \frac{1}{3 \sqrt{2}}, -\frac{1}{ 3\sqrt{2}}, \frac{4}{ 3\sqrt{2}} \right )$

Explanation:

The unit vector in the same direction as $\textbf{u}$ is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}$

$\left \| \textbf{u} \right \|$ is the norm of $\textbf{u}$ , which can be calculated by finding the square root of the sum of the squares of its entries:

$\left \| \textbf{u} \right \|=\sqrt{\left ( \frac{1}{5} \right )^{2} + \left ( -\frac{1}{5} \right )^{2} +\left ( \frac{4}{5} \right )^{2} }$

$=\sqrt{\frac{1}{25}+ \frac{1}{25}+ \frac{16}{25} }$

$=\sqrt{\frac{18}{25} }$

$= \frac{\sqrt{18}}{5 }$

Thus, the unit vector is

$\frac{ \textbf{u} }{\left \| \textbf{u} \right \|}=\frac{\left ( \frac{1}{5}, -\frac{1}{5}, \frac{4}{5} \right ) }{ \frac{\sqrt{18}}{5 }}$

$=\left ( \frac{1}{ \sqrt{18}}, -\frac{1}{ \sqrt{18}}, \frac{4}{ \sqrt{18}} \right )$

$=\left ( \frac{1}{3 \sqrt{2}}, -\frac{1}{ 3\sqrt{2}}, \frac{4}{ 3\sqrt{2}} \right )$

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Example Question #1 : Linear Independence And Rank

Determine whether the following vectors in Matrix form are Linearly Independent.

$A=\begin{bmatrix} 2 & 4 & 10\\ 3 & -7 & 11\\ -1 & 4 & 10 \end{bmatrix}$

Possible Answers:

The vectors aren't Linearly Independent

The vectors are Linearly Independent

Correct answer:

The vectors are Linearly Independent

Explanation:

To figure out if the matrix is independent, we need to get the matrix into reduced echelon form. If we get the Identity Matrix, then the matrix is Linearly Independent.

$\begin{bmatrix} 2 & 4 & 10\\ 3 & -7 & 11\\ -1 & 4 & 10 \end{bmatrix}$

$\frac{2}{3}R_2-R_1\rightarrow R_2$

$\begin{bmatrix} 2 & 4 & 10\\ 0 & -\frac{26}{3} & -\frac{8}{3}\\ -1 & 4 & 10 \end{bmatrix}$

$-2 R_3+R_1\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\\ 0 & -\frac{26}{3} & -\frac{8}{3}\\ 0 & -4 & -10 \end{bmatrix}$

$\\-3R_2\rightarrow R_2 \\-R_3\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\\ 0 & 26 & 8\\ 0 & 4 & 10 \end{bmatrix}$

$\frac{13}{2} R_3-R_2\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\\ 0 & 26 & 8\\ 0 & 0 & 57 \end{bmatrix}$

$\frac{13}{2} R_3-R_2\rightarrow R_3$

$\begin{bmatrix} 2 & 4 & 10\\ 0 & 26 & 8\\ 0 & 0 & 57 \end{bmatrix}$

$\\\frac{1}{2}R_1\rightarrow R_1 \\ \\ \frac{1}{2}R_2\rightarrow R_2$

$\begin{bmatrix} 1 & 2 & 5\\ 0 & 13 & 4\\ 0 & 0 & 57 \end{bmatrix}$

$\frac{4}{57} R_3-R_2\rightarrow R_2$

$\begin{bmatrix} 1 & 2 & 5\\ 0 & -9 & 0\\ 0 & 0 & 57 \end{bmatrix}$

$\frac{5}{57} R_3-R_1\rightarrow R_1$

$\begin{bmatrix} 4 & 3 & 0\\ 0 & -9 & 0\\ 0 & 0 & 57 \end{bmatrix}$

$R_2+3R_1\rightarrow R_1$

$\begin{bmatrix} 12 & 0 & 0\\ 0 & -9 & 0\\ 0 & 0 & 57 \end{bmatrix}$

$\\\frac{1}{12}R_1\rightarrow R_1 \\ \\ -\frac{1}{9}R_2\rightarrow R_2 \\ \\ \frac{1}{57}R_3\rightarrow R_3$

$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1& 0\\ 0 & 0 & 1 \end{bmatrix}$

Since we got the Identity Matrix, we know that the matrix is Linearly Independent.

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Example Question #2 : Linear Independence And Rank

Find the rank of the following matrix.

$\begin{bmatrix} 10 & 2 \\ 4 & 5\\ 5& 10 \end{bmatrix}$

Possible Answers:

Rank=0

Rank=2

Rank=4

Rank=3

Rank=1

Correct answer:

Rank=2

Explanation:

We need to get the matrix into reduced echelon form, and then count all the non all zero rows.

$R_2-\frac{2}{5}R_1\rightarrow R_2$

$\begin{bmatrix} 10 & 2 \\ 0 & \frac{21}{5}\\ 5& 10 \end{bmatrix}$

$R_3-\frac{1}{2}R_1\rightarrow R_3$

$\begin{bmatrix} 10 & 2 \\ 0 & \frac{21}{5}\\ 0& 9 \end{bmatrix}$

$R_2 \leftrightarrow R_3$

$\begin{bmatrix} 10 & 2 \\ 0& 9 \\ 0 & \frac{21}{5}\end{bmatrix}$

$R_3-\frac{7}{15}R_2\rightarrow R_3$

$\begin{bmatrix} 10 & 2 \\ 0& 9 \\ 0 & 0\end{bmatrix}$

$\frac{1}{9}R_2\rightarrow R_2$

$\begin{bmatrix} 10 & 2 \\ 0& 1 \\ 0 & 0\end{bmatrix}$

$R_1-2R_2\rightarrow R_1$

$\begin{bmatrix} 10 & 0 \\ 0& 1 \\ 0 & 0\end{bmatrix}$

$\frac{1}{10}R_1\rightarrow R_1$

$\begin{bmatrix} 1 & 0 \\ 0& 1 \\ 0 & 0\end{bmatrix}$

The rank is 2, since there are 2 non all zero rows.

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Example Question #1 : Linear Independence And Rank

Calculate the Rank of the following matrix

$\begin{bmatrix} 1&2 &3 \\ 1& 2& 3\\ 1&2 &3 \end{bmatrix}$

Possible Answers:

Correct answer:

Explanation:

We need to put the matrix into reduced echelon form, and then count all the non-zero rows.

$\\R_1-R_2\rightarrow R_2 \\R_1-R_3\rightarrow R_3$

$\begin{bmatrix} 1&2 &3 \\ 0& 0& 0\\ 0&0 &0 \end{bmatrix}$

Since there is only 1 non-zero row, the Rank is 1.

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #43 : Norms

Example Question #42 : Norms

Example Question #44 : Norms

Example Question #181 : Operations And Properties

Example Question #51 : Norms

Example Question #52 : Norms

Example Question #53 : Norms

Example Question #1 : Linear Independence And Rank

Example Question #2 : Linear Independence And Rank

Example Question #1 : Linear Independence And Rank

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