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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 #145 : Matrices

Let $\bf{a}$ and $\bf{b}$ be vectors defined by

$\bf{a} = [ 9, 0, 9, 5], \bf{b} = [4, 3, 0, 9]$ .

Find the dot product $\bf{a}\cdot\bf{b}$ .

Possible Answers:

$\bf{a}\cdot\bf{b} = \begin{bmatrix}9 & 0 & 9 & 5 \\ 4 & 3 & 0 & 9 \end{bmatrix}$

The dimensions do not match and the dot product does not exist.

$\bf{a}\cdot\bf{b} = 108$

$\bf{a}\cdot\bf{b} = 81$

$\bf{a}\cdot\bf{b} = 45$

Correct answer:

$\bf{a}\cdot\bf{b} = 81$

Explanation:

Vectors $\bf{a}$ and $\bf{b}$ are both of length 4. The dimensions match and the dot product exists.

$\bf{a}\cdot\bf{b} = [ 9, 0, 9, 5]\cdot [4, 3, 0, 9] = (9)(4)+(0)(3)+(9)(0)+(5)(9)=81$

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Example Question #146 : Matrices

Let $\bf{a}$ and $\bf{b}$ be vectors defined by

$\bf{a} = [8, -1, 4], \bf{b} = [7, 4, 3]$ .

Find the cross product $\bf{a}\times\bf{b}$ .

Possible Answers:

$\bf{a}\times\bf{b} &= -20i-4j+39k$

The cross product does not exist.

$\bf{a}\times\bf{b} &= -20i+4j+32k$

$\bf{a}\times\bf{b} &= -19i-4j+16k$

$\bf{a}\times\bf{b} &= -19i+4j+39k$

Correct answer:

$\bf{a}\times\bf{b} &= -19i+4j+39k$

Explanation:

We can find the cross product by calculating the determinant of the following matrix

$\bf{a}\times\bf{b} = \det \begin{bmatrix} i& j & k\\ 8 & -1 & 4\\ 7&4&3 \end{bmatrix}$

$\begin{align*} \bf{a}\times\bf{b} &= i ((-1)(3)-(4)(4))-j((8)(3)-(4)(7))+k((8)(4)-(-1)(7)) \\ \bf{a}\times\bf{b} &= -19i+4j+39k \end{align*}$

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Example Question #147 : Matrices

Let $\bf{u}$ and $\bf{v}$ be vectors defined by

$\bf{u} = [4,-3, 5], \bf{v}=[8, -2, 4]$ .

Find the cross product $\bf{u}\times\bf{v}$ .

Possible Answers:

The cross product does not exist.

$\bf{u}\times \bf{v} &= -2i - 24j +24k$

$\bf{u}\times \bf{v} &= -12i + 24j -16k$

$\bf{u}\times \bf{v} &= -2i + 24j +16k$

$\bf{u}\times \bf{v} &= -16i + 2j +8k$

Correct answer:

$\bf{u}\times \bf{v} &= -2i + 24j +16k$

Explanation:

We find the cross product by finding the determinant of the following matrix

$\bf{u}\times \bf{v}=\det \begin{bmatrix} i & j & k\\ 4 & -3 & 5\\ 8 & -2 & 4 \end{bmatrix}$

$\begin{align*} \bf{u}\times \bf{v} &= i ((-3)(4)-(5)(-2)) - j((4)(4)-(5)(8)) + k((4)(-2)-(-3)(8)) \\ \bf{u}\times \bf{v} &= -2i + 24j +16k \end{align*}$

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Example Question #148 : Matrices

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

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

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

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

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

The degree of a term of a polynomial is the sum of the exponents of its variables. Each term in this polynomial has exponent sum 5, so each term has degree 5. The degree of the polynomial is the greatest of the degrees, so the polynomial has degree 5.

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Example Question #21 : Vector Vector Product

$\textbf{u} = (1, x, x^{2}, x ^{3}, x^{4}, x^{5}, x^{6})$

$(x+ 1)^{6} = \textbf{u} \cdot \textbf{v}$ , where $\textbf{v}$ is which vector?

Possible Answers:

$\textbf{v} = (6, 5, 4, 3, 2, 1 )$

$\textbf{v} = (1, 6, 15, 20, 15, 6, 1)$

$\textbf{v} = (1, 6, 36, 216, 1296, 7776 , 46656 )$

$\textbf{v} = ( 46656, 7776, 1296, 216, 36, 6, 1 )$

$\textbf{v} = (1, 2, 3, 4, 5,6 )$

Correct answer:

$\textbf{v} = (1, 6, 15, 20, 15, 6, 1)$

Explanation:

Let $\textbf{v}= (v_{0}, v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6})$

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

$\textbf{u} \cdot \textbf{v} = v_{0} + v_{1}x+ v_{2}x^{2} + v_{3}x^{3}+ v_{4} x^{4}+ v_{5}x^{5}+ v_{6}x^{6}$

Therefore, $\textbf{v}$ is the vector of coefficients of the powers of of $(x+ 1)^{6}$ , in ascending order of exponent.

By the Binomial Theorem,

$(x+ 1)^{6} = \sum_{k = 0}^{6} C(6, k)x^{k}$ .

Therefore, $\textbf{v}$ has as its entries the binomial coefficients for 6, which are:

C(6,0) = c(6,6) = 1

C(6,1) = c(6,5) = 6

C(6,2) = c(6,4) = 15

C(6,3) = 20

It follows that $\textbf{v} = (1, 6, 15, 20, 15, 6, 1)$ .

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Example Question #22 : Vector Vector Product

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

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

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

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

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

The degree of a term of a polynomial is the sum of the exponents of its variables; the individual terms have degrees 0, 2, 4, 6, 8, 10, in that order. the degree of the polynomial is the highest of these, which is 10.

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Example Question #23 : Vector Vector Product

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

$\textbf{u} \cdot \textbf{v}$ is equal to the fifth-degree Maclaurin series for $e^{2x}$ for:

Possible Answers:

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

$\textbf{v}=\left ( 1, 2, 1, \frac{1}{3}, \frac{1}{12}, \frac{1}{60} \right )$

None of the other choices gives the correct response.

$\textbf{v}=\left ( \frac{1}{60} , \frac{1}{12} , \frac{1}{3}, 1, 2, 1\right )$

$\textbf{v}=\left ( \frac{4}{15} , \frac{2}{3} , \frac{4}{3} , 2, 2, 1 \right )$

Correct answer:

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

Explanation:

The th-degree Maclaurin series for a function f(x) is the polynomial

$\frac{f(0)}{0!} + \frac{f'(0)}{1!}x+ \frac{f''(0)}{2!}x^{2}+ ... + \frac{f^{(n)}(0)}{n!}x^{n}$

If $\textbf{v} = (v_{0}, v_{1}, v_{2}, v_{3}, v_{4}, v_{5})$ ,

then

$\textbf{u} \cdot \textbf{v} = v_{0}+ v_{1}x+ v_{2}x^{2}+ ...+v_{5}x^{5}$ .

Therefore, we want $\textbf{v}$ to be the vector of Maclaurin coefficients by ascending order of degree.

The fifth-degree Maclaurin series for $e^{x}$ is

$e^{x}=1+x+ \frac{1}{2}x^{2}+ \frac{1}{6}x^{3}+ \frac{1}{24}x^{4}+ \frac{1}{120}x^{5}$

The Maclaurin series for $e^{2x}$ can be derived from this by replacing with :

$e^{2x}=1+2x+ \frac{1}{2}(2x)^{2}+ \frac{1}{6}(2x)^{3}+ \frac{1}{24}(2x)^{4}+ \frac{1}{120}(2x)^{5}$

$e^{2x}=1+2x+ \frac{1}{2}(4x^{2})+ \frac{1}{6}(8x^{3})+ \frac{1}{24}(16x^{4})+ \frac{1}{120}(32x^{5})$

$e^{2x}=1+2x+ 2 x^{2}+ \frac{4}{3}x^{3}+ \frac{2}{3}x^{4}+ \frac{4}{15} x^{5}$

Therefore,

$\textbf{v}=\left ( 1, 2, 2 , \frac{4}{3}, \frac{2}{3}, \frac{4}{15} \right )$

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Example Question #21 : Vector Vector Product

$\textbf{u} = (-3, 2, 4 ,1 )$

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

Which of the following is undefined, $\textbf{u} \cdot \textbf{v}$ or $\textbf{u} \times \textbf{v}$ ?

Possible Answers:

Both

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

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

Neither

Correct answer:

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

Explanation:

$\textbf{u} \cdot \textbf{v}$ , the dot product of the vectors, is a defined quantity if and only if both vectors are elements of the same vector space. Each has four entries, so both are in $\mathbb{R}^{4}$ . Consequently, $\textbf{u} \cdot \textbf{v}$ is defined.

$\textbf{u} \times \textbf{v}$ , the cross product of the vectors, is a defined vector if and only if both vectors are elements in $\mathbb{R}^{3}$ . As previously mentioned, they are in $\mathbb{R}^{4}$ , so $\textbf{u} \times \textbf{v}$ is undefined.

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Example Question #25 : Vector Vector Product

$\textbf{u} = (3, 7, -2)$

$\textbf{v} = (1, -5 )$

Which of the following is undefined, $\textbf{u} \cdot \textbf{v}$ or $\textbf{u} \times \textbf{v}$ ?

Possible Answers:

Neither

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

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

Both

Correct answer:

Both

Explanation:

$\textbf{u} \cdot \textbf{v}$ , the dot product of the vectors, is a defined quantity if and only if both vectors are elements of the same vector space. $\textbf{u}$ has three entries, so $\textbf{u} \in \mathbb{R}^{3}$ ; $\textbf{v}$ has two entries, so $\textbf{v} \in \mathbb{R}^{2}$ . The two are in different vector spaces, so $\textbf{u} \cdot \textbf{v}$ is undefined.

$\textbf{u} \times \textbf{v}$ , the cross product of the vectors, is a defined vector if and only if both vectors are elements in $\mathbb{R}^{3}$ . As previously mentioned, $\textbf{v} \in \mathbb{R}^{2}$ , so $\textbf{u} \times \textbf{v}$ is undefined.

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Example Question #831 : Linear Algebra

$\textbf{u} = (3, 7, -2)$

$\textbf{v} = (1, -5, 1 )$

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

Possible Answers:

(-3, 5, -22)

(-3, -5, -22)

(3, 5,22)

( 17, 5, 8)

(-17, 5, -8)

Correct answer:

(-3, -5, -22)

Explanation:

One way to determine the cross-product of two vectors is to set up a matrix with the first row $\textbf{i}, \textbf{j}, \textbf{k}$ , where these are the unit vectors (1,0,0), (0,1,0),(0,0,1) , respectively, and with the entries of the vectors as the other two rows:

$\begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 3 & 7 & -2 \\ 1 & -5 & 1 \end{vmatrix}$

We can evaluate this as we would evaluate a determinant of a matrix with real entries. Take the products of the upper-left-to-lower-right diagonals, and subtract the products of the lower-left-to-upper-right diagonals:

Cross product

$\textbf{u} \times \textbf{v} = 7\textbf{i}+ ( - 2\textbf{j})+ (-15\textbf{k}) - 7\textbf{k}- 10\textbf{i} - 3 \textbf{j}$

$\textbf{u} \times \textbf{v} = -3\textbf{i}- 5\textbf{j}-22\textbf{k}$

$\textbf{u} \times \textbf{v} = -3(1,0,0 )- 5(0,1,0)-22 (0,0,1)$

$\textbf{u} \times \textbf{v} =(-3, -5, -22)$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #145 : Matrices

Example Question #146 : Matrices

Example Question #147 : Matrices

Example Question #148 : Matrices

Example Question #21 : Vector Vector Product

Example Question #22 : Vector Vector Product

Example Question #23 : Vector Vector Product

Example Question #21 : Vector Vector Product

Example Question #25 : Vector Vector Product

Example Question #831 : Linear Algebra

All Linear Algebra Resources

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