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Linear Algebra : Matrices

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Example Questions

Example Question #11 : Vector Vector Product

$\begin{align*}&\text{Find the vector product }\\&\begin{bmatrix}-18&-4&-20\end{bmatrix}\times\begin{bmatrix}-9\\-11\\-11\end{bmatrix}\end{align*}$

Possible Answers:

426

428

522

$\text{The multiplication cannot be performed.}$

Correct answer:

426

Explanation:

$\begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times3\text{ and }3\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}-18&-4&-20\end{bmatrix}\times\begin{bmatrix}-9\\-11\\-11\end{bmatrix}\\&(-18)(-9)+(-4)(-11)+(-20)(-11)\\&426\end{align*}$

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Example Question #11 : Vector Vector Product

$\begin{align*}&\text{Find the vector product }\\&\begin{bmatrix}-8&-19&5&-9&2\end{bmatrix}\times\begin{bmatrix}-5\\-14\\0\\10\\12\end{bmatrix}\end{align*}$

Possible Answers:

$\text{The multiplication cannot be performed.}$

311

240

192

Correct answer:

240

Explanation:

$\begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times5\text{ and }5\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}-8&-19&5&-9&2\end{bmatrix}\times\begin{bmatrix}-5\\-14\\0\\10\\12\end{bmatrix}\\&(-8)(-5)+(-19)(-14)+(5)(0)+(-9)(10)+(2)(12)\\&240\end{align*}$

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Example Question #12 : Vector Vector Product

$\begin{align*}&\text{Find the vector product }\\&\begin{bmatrix}1&-7&8&9&-9\end{bmatrix}\times\begin{bmatrix}-20\\-8\\-7\\-2\\-20\end{bmatrix}\end{align*}$

Possible Answers:

142

165

$\text{The multiplication cannot be performed.}$

Correct answer:

142

Explanation:

$\begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times5\text{ and }5\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}1&-7&8&9&-9\end{bmatrix}\times\begin{bmatrix}-20\\-8\\-7\\-2\\-20\end{bmatrix}\\&(1)(-20)+(-7)(-8)+(8)(-7)+(9)(-2)+(-9)(-20)\\&142\end{align*}$

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Example Question #12 : Vector Vector Product

$\begin{align*}&\text{Find the vector product }\\&\begin{bmatrix}-3&-11&-9&-8&3\end{bmatrix}\times\begin{bmatrix}8\\-3\\4\\-20\\8\end{bmatrix}\end{align*}$

Possible Answers:

193

$\text{The multiplication cannot be performed.}$

157

Correct answer:

157

Explanation:

$\begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times5\text{ and }5\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}-3&-11&-9&-8&3\end{bmatrix}\times\begin{bmatrix}8\\-3\\4\\-20\\8\end{bmatrix}\\&(-3)(8)+(-11)(-3)+(-9)(4)+(-8)(-20)+(3)(8)\\&157\end{align*}$

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Example Question #13 : Vector Vector Product

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:

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

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

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

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

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

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 #13 : Vector Vector Product

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:

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

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

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

The cross product does not exist.

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

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 #831 : Linear Algebra

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

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

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 #832 : Linear Algebra

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

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

Possible Answers:

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

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

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

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

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

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 #21 : 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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Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : Vector Vector Product

Example Question #11 : Vector Vector Product

Example Question #12 : Vector Vector Product

Example Question #12 : Vector Vector Product

Example Question #13 : Vector Vector Product

Example Question #13 : Vector Vector Product

Example Question #147 : Matrices

Example Question #831 : Linear Algebra

Example Question #832 : Linear Algebra

Example Question #21 : Vector Vector Product

All Linear Algebra Resources

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