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Linear Algebra : Linear Algebra

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Example Questions

Example Question #154 : Matrices

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

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

If $\textbf{u} \cdot \textbf{v} = \ln a$ , then evaluate .

Possible Answers:

a = 64

$a = 4e^{3}$

a = 4e

$a = \frac{4}{e}$

a = 12

Correct answer:

$a = 4e^{3}$

Explanation:

The dot product $\textbf{u} \cdot \textbf{v}$ is equal to the sum of the products of the numbers in corresponding positions, so

$\textbf{u} \cdot \textbf{v} = (\ln 2) \cdot 2 + 1 \cdot 1 + (-1 )(-2)$

$\textbf{u} \cdot \textbf{v} = 2\ln 2+ 1 + 2$

$\textbf{u} \cdot \textbf{v} = 3+ 2\ln 2$

Applying the properties of logarithms:

$\textbf{u} \cdot \textbf{v} = 3+ \ln 2^{2}$

$\textbf{u} \cdot \textbf{v} = 3+ \ln 4$

$\textbf{u} \cdot \textbf{v} = \ln e^{3}+ \ln 4$

$\textbf{u} \cdot \textbf{v} = \ln 4 e^{3}$

Therefore, $a = 4e^{3}$ .

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Example Question #155 : Matrices

$\textbf{u} = (1, x, x^{2}, x ^{3}, x^{4}, x^{5})$

$\textbf{v} = (y^{5}, y^{3} , y, y^{4}, 1, y^{2})$

The expression $\textbf{u} \cdot \textbf{v}$ yields a polynomial of what degree?

Possible Answers:

None of the other choices gives a correct response.

Correct answer:

Explanation:

The dot product $\textbf{u} \cdot \textbf{v}$ is the sum of the products of entries in corresponding positions, so

$\textbf{u} \cdot \textbf{v} = 1 (y^{5}) + x (y^{3})+ x^{2}(y )+ x^{3}(y^{4})+ x^{4} (1)+ x^{5 }(y^{2})$

$\textbf{u} \cdot \textbf{v} = y^{5} + x y^{3}+ x^{2}y + x^{3}y^{4}+ x^{4} + x^{5 }y^{2}$

The degree of a term of a polynomial is the sum of the exponents of its variables; the individual terms have degrees 5, 4, 3, 7, 4, and 7, in that order. the degree of the polynomial is the highest of these, which is 7.

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Example Question #152 : Matrices

A triangle has two sides of length and ; their included angle has measure $\theta$ . The measure of the third side can be obtained from the expression

$\sqrt{\textbf{u} \cdot \textbf{v}}$ ,

where $\textbf{u}= (a , b, 2ab)$ and:

Possible Answers:

$\textbf{v} = (b, a, -\cos \theta)$

$\textbf{v} = (a, b, -\sin \theta)$

$\textbf{v} = (b, a, -\sin \theta)$

$\textbf{v} = (a, b, \cos \theta)$

$\textbf{v} = (a, b, -\cos \theta)$

Correct answer:

$\textbf{v} = (a, b, -\cos \theta)$

Explanation:

Given the lengths and of two sides of a triangle, and the measure of their included angle, $\theta$ , the length of the third side of a triangle can be calculated using the Law of Cosines, which states that

$c^{2} = a^{2}+ b^{2} - 2ab\cos \theta$ .

The dot product $\textbf{u} \cdot \textbf{v}$ is equal to the sum of the products of their corresponding entries, and since $c= \sqrt{\textbf{u} \cdot \textbf{v}}$ , we can substitute $\textbf{u} \cdot \textbf{v}$ for $c^{2}$ :

$\textbf{u} \cdot \textbf{v} = a^{2}+ b^{2} - 2ab\cos \theta$

$\textbf{u} \cdot \textbf{v} = a \cdot a + b \cdot b + 2ab \cdot ( - \cos \theta)$

$\textbf{u} = (a , b , 2ab )$ ; it follows that $\textbf{v} = (a, b, -\cos \theta)$ .

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Example Question #28 : Vector Vector Product

and are differentiable functions.

$A = \begin{bmatrix} f &g\end{bmatrix}$

$AB = \begin{bmatrix} \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{f}{g} \right )\end{bmatrix}$

Which value of makes this statement true?

Possible Answers:

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\- \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \end{bmatrix}$

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \end{bmatrix}$

Correct answer:

$B = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$

Explanation:

Recall the quotient rule of differentiation:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}}$

This can be rewritten as

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x}+ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g }{g^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =\frac{-f \cdot \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}}+\frac{ \frac{\mathrm{d}f }{\mathrm{d} x} \cdot g }{g^{2}}$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =f \left (- \frac{ \frac{\mathrm{d}g }{\mathrm{d} x} }{g^{2}} \right )+g\left ( \frac{ \frac{\mathrm{d}f }{\mathrm{d} x} }{g^{2}} \right )$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{f}{g} \right ) =f \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x} +g\left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x}$

If $A = \begin{bmatrix} f &g\end{bmatrix}$ and $B = \begin{bmatrix} a \\ b \end{bmatrix}$ ,

then multiply corresponding elements and add the products to get the sole element in :

$AB = \begin{bmatrix} f \cdot a + g \cdot b \end{bmatrix}$

Since we want

$AB = \begin{bmatrix} \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{f}{g} \right )\end{bmatrix}=\begin{bmatrix} f \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x} +g\left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x} \end{bmatrix}$ ,

It follows that, of the given choices, $a= \left (- \frac{1 }{g^{2}} \right ) \frac{\mathrm{d}g }{\mathrm{d} x}$ and $b= \left ( \frac{1}{g^{2}} \right ) \frac{\mathrm{d}f }{\mathrm{d} x}$ , and

$B = \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} - \frac{1}{g^{2}} \cdot \frac{\mathrm{d} g}{\mathrm{d} x} \\ \\ \frac{1}{g^{2}} \cdot \frac{\mathrm{d} f}{\mathrm{d} x} \end{bmatrix}$ .

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Example Question #31 : Vector Vector Product

$\textbf{u} = (- 3, 4, 1, 0)$

$\textbf{v} = (0, 1, -2, 4)$

Calculate the angle (nearest degree) between $\textbf{u}$ and $\textbf{v}$ .

Possible Answers:

$85^{\circ }$

$175^{\circ }$

$105^{\circ }$

$5^{\circ }$

The angle is undefined, since the vectors are in $\mathbb{R}^{4}$ .

Correct answer:

$85^{\circ }$

Explanation:

$\textbf{u} = (- 3, 4, 1, 0)$

$\textbf{v} = (0, 1, -2, 4)$

The angle $\; \theta$ between vectors $\textbf{u}$ and $\textbf{v}$ can be calculated using the formula

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

$\textbf{u} \cdot \textbf{v}$ , the dot product, is the sum of the products of corresponding entries:

$\textbf{u} \cdot \textbf{v} = -3(0)+4(1)+1(-2)+0(4) = 0 + 4 + (-2)+ 0 = 2$

$\left \| \textbf{u} \right \|$ , the norm of $\textbf{u}$ , is the square root of the sum of the squares of its entries; $\left \| \textbf{v} \right \|$ is defined similarly:

$\left \| \textbf{u} \right \|= \sqrt{ (- 3)^{2}+ 4^{2} + 1^{2} +0^{2}} = \sqrt{9+16+1+0} = \sqrt{26}$

$\left \| \textbf{v} \right \|= \sqrt{ 0^{2}+ 1^{2} +(-2)^{2} +4^{2}} = \sqrt{0+1+4+16} = \sqrt{21}$

$\cos \theta = \frac{2}{ \sqrt{26} \cdot \sqrt{21}} \approx 0.0856$

$\theta \approx \cos ^{-1}0.0856 \approx 85^{\circ }$

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Example Question #153 : Matrices

, , and give the length, width, and height of a rectangular prism.

$\textbf{u} = (L, W, H )$ and $\textbf{v} = (W, H, L )$ .

True or false: $\textbf{u} \cdot \textbf{v}$ gives the surface area of the prism.

Possible Answers:

False

True

Correct answer:

False

Explanation:

The dot product $\textbf{u} \cdot \textbf{v}$ can be calculated by adding the products of the elements in corresponding locations, so

$\textbf{u} \cdot \textbf{v} = LW + WH + LH$ .

The surface area of the prism, , can be found by using the formula:

$A = 2 (LW + WH + LH) = 2( \textbf{u} \cdot \textbf{v})$

Equivalently, $\textbf{u} \cdot \textbf{v}$ gives half the surface area of the prism. The statement is false.

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Example Question #841 : Linear Algebra

$\textbf{u},\textbf{v} , \textbf{w}, \textbf{x} \in \mathbb{R}^{3}$

Which of the following applies to $(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x})$ , where " $\cdot$ " and " $\times$ " refer to the dot product and the cross product of two vectors?

Possible Answers:

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}$

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{3}$

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x})$ is an undefined expression.

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in M_{2 \times 2}$

Correct answer:

$(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{3}$

Explanation:

The cross product of two vectors in $\mathbb{R}^{3}$ is also a vector in $\mathbb{R}^{3}$ . It follows that $\textbf{u} \times \textbf{v} \in \mathbb{R}^{3}$ and $\textbf{w} \times \textbf{x} \in \mathbb{R}^{3}$ ; it further follows that $(\textbf{u} \times \textbf{v}) \times( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{3}$ .

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Example Question #34 : Vector Vector Product

$\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ ; $\textbf{w}, \textbf{x} \in \mathbb{R}^{4}$

Which of the following applies to $(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x})$ , where " $\times$ " refers to the cross product of two vectors, and "" refers to either scalar or vector addition, as applicable?

Possible Answers:

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x}) \in \mathbb{R}$

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{3}$

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x}) \in M_{2 \times 3}$

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x})$ is an undefined expression.

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x}) \in \mathbb{R}^{2}$

Correct answer:

$(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x})$ is an undefined expression.

Explanation:

The cross product of two vectors is defined only if both vectors are in $\mathbb{R}^{3}$ . $\textbf{w}$ and $\textbf{x}$ are vectors in $\mathbb{R}^{4}$ , so $\textbf{w} \times \textbf{x}$ is undefined; consequently, so is $(\textbf{u} \times \textbf{v}) + ( \textbf{w} \times \textbf{x})$ .

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Example Question #842 : Linear Algebra

$\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ ; $\textbf{w}, \textbf{x} \in \mathbb{R}^{4}$

Which of the following applies to $(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x})$ , where " $\cdot$ " refers to the dot product of two vectors, and "" refers to either scalar or vector addition, as applicable?

Possible Answers:

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}^{3}$

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}^{4}$

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x})$ is an undefined expression.

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}$

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in M_{3 \times 4 }$

Correct answer:

$(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}$

Explanation:

The dot product of two vectors in the same vector space is a scalar quantity. $\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ , so $\textbf{u}$ and $\textbf{v}$ are in the same vector space; their dot product is defined, and $\textbf{u} \cdot \textbf{v} \in \mathbb{R}$ . For similar reasons, $\textbf{w} \cdot \textbf{x} \in \mathbb{R}$ . Therefore, their sum is defined, and $(\textbf{u} \cdot \textbf{v}) + ( \textbf{w} \cdot \textbf{x}) \in \mathbb{R}$ .

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Example Question #35 : Vector Vector Product

$\textbf{u},\textbf{v} \in \mathbb{R}^{3}$ .

Which of the following applies to $\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v}$ , where " $\cdot$ " and " $\times$ " refer to the dot product and the cross product of two vectors, and "" refers to either scalar or vector addition, as applicable?

Possible Answers:

$\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v} \in M_{1 \times 3 }$

$\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v} \in \mathbb{R} ^{3}$

$\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v}$ is an undefined expression.

$\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v} \in M_{3 \times 1 }$

$\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v} \in \mathbb{R}$

Correct answer:

$\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v}$ is an undefined expression.

Explanation:

The cross product $\textbf{u} \times \textbf{v}$ of two vectors in $\mathbb{R}^{3}$ is also a vector in $\mathbb{R}^{3}$ . The dot product $\textbf{u} \cdot \textbf{v}$ of two such vectors is a scalar. Since a vector and a scalar cannot be added, $\textbf{u} \times \textbf{v} + \textbf{u} \cdot \textbf{v}$ is an undefined expression.

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #154 : Matrices

Example Question #155 : Matrices

Example Question #152 : Matrices

Example Question #28 : Vector Vector Product

Example Question #31 : Vector Vector Product

Example Question #153 : Matrices

Example Question #841 : Linear Algebra

Example Question #34 : Vector Vector Product

Example Question #842 : Linear Algebra

Example Question #35 : Vector Vector Product

All Linear Algebra Resources

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