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Linear Algebra : Linear Algebra

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Example Questions

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 are Linearly Independent

The vectors aren't 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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Example Question #1 : Linear Independence And Rank

Determine if the following matrix is linearly independent or not.

$A=\begin{bmatrix} 5 & 3\\ 10 & 6 \end{bmatrix}$

Possible Answers:

Linearly Independent

Linearly Dependent

Correct answer:

Linearly Dependent

Explanation:

Since the matrix is $2\times2$ , we can simply take the determinant. If the determinant is not equal to zero, it's linearly independent. Otherwise it's linearly dependent.

$\det(A)=5(6)-(10)(3)=30-30=0$

Since the determinant is zero, the matrix is linearly dependent.

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Example Question #1 : Linear Independence And Rank

If matrix A is a 5x8 matrix with a two-dimensional null space, what is the rank of A?

Possible Answers:

rank (A) = 5

rank( A) = 6

rank (A) = 8

rank (A) = 3

None of the other answers.

Correct answer:

rank( A) = 6

Explanation:

Given that rank A + dimensional null space of A = total number of columns, we can determine rank A = total number of columns-dimensional null space of A. Using the information given in the question we can solve for rank A:

rank (A) = 8-2 = 6

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

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

Example Question #1 : Linear Independence And Rank

Example Question #1 : Linear Independence And Rank

All Linear Algebra Resources

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