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Linear Algebra : Linear Algebra

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Example Questions

Example Question #22 : Matrix Vector Product

Let be a matrix and $\bf{x}$ be a vector defined by

$A = \begin{bmatrix} -7 & 3 & 0 \\ -9 & 0 & 2 \\ 0 & 5 & -9 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} -5 \\ 4 \\ 8 \\ 3 \\ 8\end{bmatrix}$

Find the product $A\bf{x}$ .

Possible Answers:

$A\bf{x} = \begin{bmatrix} 47 \\ 61 \\ -52\end{bmatrix}$

The product does not exist because the dimensions do not match.

$A\bf{x} = \begin{bmatrix} 47 & 61 & -52\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 35 \\ 0 \\ -72\end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 42 \\ 9 \\ -7\end{bmatrix}$

Correct answer:

The product does not exist because the dimensions do not match.

Explanation:

The matrix has 3 columns and the vector $\bf {x}$ has 5 rows. The dimensions do not match and the product does not exist.

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Example Question #123 : Matrices

Rewrite the system of equations:

x+2y+z=3

-x+5y+7z=10

2x+3y+6=2

into a matrix vector product:

Av=b

where is a 3x3 matrix and v,b are vectors in $\mathbb{R}^3$ .

Possible Answers:

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 10\\ 3\\2 \end{pmatrix}$

$\begin{pmatrix} -1&5 &7 \\ 1& 2 &3 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

$\begin{pmatrix} 1& -1 &2 \\ 2&5 &3 \\ 3&7 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

Correct answer:

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

Explanation:

To write

x+2y+z=3

-x+5y+7z=10

2x+3y+6=2

into matrix vector form, we recall that matrix multiplication with a vector is done such that the first element in the resulting vector is the dot product of the first row of with the vector v=(x,y,z)^T , the second element is the dot product of the second row with , and so on. The first row is thus (1,2,3) , the second row is (-1,5,7) , and the third row is (2,3,6) . So the left side of the equality is

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}$

The right side is the vector (3,10,2)^T , so the final answer is

$\begin{pmatrix} 1& 2 &3 \\ -1&5 &7 \\ 2&3 &6\end{pmatrix}\begin{pmatrix} x\\ y\\z \end{pmatrix}=\begin{pmatrix} 3\\ 10\\2 \end{pmatrix}$

which is equivalent to

x+2y+z=3

-x+5y+7z=10

2x+3y+6=2

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Example Question #812 : Linear Algebra

Let $A = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \\ - \sqrt{2} & 2\sqrt{3} \end{bmatrix}$ and $B = \begin{bmatrix} \sqrt{2} \\ \sqrt{3} \end{bmatrix}$ .

Find .

Possible Answers:

$\begin{bmatrix}7 \\ 4 \end{bmatrix}$

$\begin{bmatrix} 4 & \sqrt{6} \\ - \sqrt{6} &6\end{bmatrix}$

is not defined.

$\begin{bmatrix} 4 - \sqrt{6} \\ 6+ \sqrt{6} \end{bmatrix}$

$\begin{bmatrix} 4 & 3 \\ -2 & 6 \end{bmatrix}$

Correct answer:

$\begin{bmatrix}7 \\ 4 \end{bmatrix}$

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AB = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \\ - \sqrt{2} & 2\sqrt{3} \end{bmatrix} \begin{bmatrix} \sqrt{2} \\ \sqrt{3} \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{2} (\sqrt{2})+ \sqrt{3} (\sqrt{3} )\\ - \sqrt{2} (\sqrt{2})+ 2\sqrt{3}(\sqrt{3} ) \end{bmatrix}$

$= \begin{bmatrix}7 \\ 4 \end{bmatrix}$

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Example Question #23 : Matrix Vector Product

Multiply $\begin{bmatrix} 5 &3 &-1 \\ 0&6 &3 \\1 &-2 &1 \end{bmatrix} \cdot \begin{bmatrix} 1\\5 \\-2 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 7\\29 \\16 \end{bmatrix}$

$\begin{bmatrix} 22\\24 \\-11 \end{bmatrix}$

$\begin{bmatrix} 3\\ 37\\12 \end{bmatrix}$

$\begin{bmatrix} 18\\ 36\\ -8\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 22\\24 \\-11 \end{bmatrix}$

Explanation:

To multiply, add:

$1 \cdot \begin{bmatrix} 5\\ 0\\ 1\end{bmatrix} + 5 \cdot \begin{bmatrix} 3\\ 6\\ -2\end{bmatrix} + -2 \cdot \begin{bmatrix} -1\\3 \\1 \end{bmatrix}$

$\begin{bmatrix} 5\\ 0\\ 1\end{bmatrix} + \begin{bmatrix} 15\\ 30\\ -10\end{bmatrix} + \begin{bmatrix} 2\\ -6\\ -2\end{bmatrix} = \begin{bmatrix} 22\\ 24\\ -11\end{bmatrix}$

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Example Question #1 : Vector Vector Product

Compute $A\cdot B$ , where

$A=\begin{bmatrix} 2\\ 3\\ 4\\ \end{bmatrix}$

$B=\begin{bmatrix} 1 & 3 & 5 \end{bmatrix}$

Possible Answers:

$A\cdot B =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

Not possible

$A\cdot B =\begin{bmatrix} 2 & 9& 10\\ 3& 6& 15 \\ 4& 20& 12 \end{bmatrix}$

Correct answer:

$A\cdot B =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

Explanation:

Before we compute the product of , and , we need to check if it is possible to take the product. We will check the dimensions. is $3\times1$ , and is $1\times3$ , so the dimensions of the resulting matrix will be $3\times3$ . Now let's compute it.

$A\cdot B=\begin{bmatrix} 2\cdot1 & 2\cdot3 & 2\cdot 5\\ 3 \cdot 1& 3\cdot3 & 3 \cdot 5 \\ 4 \cdot 1 & 4\cdot3 & 4\cdot 5 \end{bmatrix} =\begin{bmatrix} 2 & 6& 10\\ 3& 9& 15 \\ 4& 12& 20 \end{bmatrix}$

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Example Question #2 : Vector Vector Product

Find the vector-vector product of the following vectors.

$A=[1 \ 2]$

$B=\begin{bmatrix} 3\\ 5 \end{bmatrix}$

Possible Answers:

It's not possible to multiply these vectors

Correct answer:

Explanation:

$A\cdot B=1\cdot3+2\cdot5=3+10=13$

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Example Question #3 : Vector Vector Product

Calculate $\bf{A} \times \bf{B}$ , given

$\textbf{A}= 4\hat{i}-2\hat{j}$

$\textbf{B}= 1\hat{i}+3\hat{j}$

Possible Answers:

$\textbf{A} \times \textbf{B} = 14 \hat{j}$

$\textbf{A} \times \textbf{B} = 14$

$\textbf{A} \times \textbf{B} = 14 \hat{i}$

$\textbf{A} \times \textbf{B} = 14 \hat{k}$

Correct answer:

$\textbf{A} \times \textbf{B} = 14 \hat{k}$

Explanation:

By definition,

$\textbf{A} \times \textbf{B} = \begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 4& -2& 0\\ 1& 3& 0 \end{vmatrix} =\hat{i}(0)-\hat{j}(0)+\hat{k}[4(3)-1(-2)] = 14 \hat{k}$ .

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Example Question #131 : Matrices

What is the physical significance of the resultant vector $\bf{C}$ , if $\bf{C}= \bf{A} \times \bf{B}$ ?

Possible Answers:

$\bf{C}$ is a scalar.

$\bf{C}$ lies in the same plane that contains both $\bf{A}$ and $\bf{B}$ .

$\bf{C}$ is the projection of $\bf{A}$ onto $\bf{B}$ .

$\bf{C}$ is orthogonal to both $\bf{A}$ and $\bf{B}$ .

Correct answer:

$\bf{C}$ is orthogonal to both $\bf{A}$ and $\bf{B}$ .

Explanation:

By definition, the resultant cross product vector (in this case, $\bf{C}$ ) is orthogonal to the original vectors that were crossed (in this case, $\bf{A}$ and $\bf{B}$ ). In , this means that $\bf{C}$ is a vector that is normal to the plane containing $\bf{A}$ and $\bf{B}$ .

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Example Question #4 : Vector Vector Product

$\begin{align*}&\text{Find the product }A\times B \\&\text{Where }A=\begin{bmatrix}-18&12&-11&4&-19&13\end{bmatrix}\text{ and }B=\begin{bmatrix}-6\\-2\\-17\\-20\\-8\\10\end{bmatrix}\end{align*}$

Possible Answers:

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

519

473

556

Correct answer:

473

Explanation:

$\begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since A has dimensions: }1\times6\\&\text{and B has dimensions: }6\times1\\&\text{The two vectors can be multiplied:}\\&\begin{bmatrix}-18&12&-11&4&-19&13\end{bmatrix}\times\begin{bmatrix}-6\\-2\\-17\\-20\\-8\\10\end{bmatrix}\\&(-18)(-6)+(12)(-2)+(-11)(-17)+(4)(-20)+(-19)(-8)+(13)(10)\\&473\end{align*}$

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Example Question #4 : Vector Vector Product

$\begin{align*}&\text{Find the vector product }\\&\begin{bmatrix}-7&-6\end{bmatrix}\times\begin{bmatrix}11\\-9\end{bmatrix}\end{align*}$

Possible Answers:

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

Correct answer:

Explanation:

$\begin{align*}&\text{In order to multiply two vectors, }A\times b\text{, the respective dimensions of each}\\&\text{must be of the form }1\times n\text{ and }n\times 1\\&\text{Note that order does matter:}\\&(A\times b \neq b \times A)\\&\text{Since our vectors have dimensions: }1\times2\text{ and }2\times1\\&\text{The two can be multiplied to find a product:}\\&\begin{bmatrix}-7&-6\end{bmatrix}\times\begin{bmatrix}11\\-9\end{bmatrix}\\&(-7)(11)+(-6)(-9)\\&-23\end{align*}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #22 : Matrix Vector Product

Example Question #123 : Matrices

Example Question #812 : Linear Algebra

Example Question #23 : Matrix Vector Product

Example Question #1 : Vector Vector Product

Example Question #2 : Vector Vector Product

Example Question #3 : Vector Vector Product

Example Question #131 : Matrices

Example Question #4 : Vector Vector Product

Example Question #4 : Vector Vector Product

All Linear Algebra Resources

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