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Linear Algebra : Matrix-Vector Product

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Example Questions

Example Question #21 : Matrix Vector Product

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

$A = \begin{bmatrix} 0 & 8 & 3 & 6\\ 9 & 0 & 8 & -4 \end{bmatrix}$

$\bf{x} = \begin{bmatrix} 5 \\ 8 \\ -2 \\ 7\end{bmatrix}$

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

Possible Answers:

$A\bf{x} = \begin{bmatrix} 100 & 1 \end{bmatrix}$

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

$A\bf{x} = \begin{bmatrix} 10 \\ 10 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 100 & 99 \end{bmatrix}$

$A\bf{x} = \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

Correct answer:

$A\bf{x} = \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

Explanation:

Matrix has 4 columns and vector $\bf{x}$ has 4 rows. The dimensions match and the product exists.

$A\bf{x} = \begin{bmatrix} 0 & 8 & 3 & 6\\ 9 & 0 & 8 & -4 \end{bmatrix} \cdot \begin{bmatrix} 5 \\ 8 \\ -2 \\ 7\end{bmatrix} = \begin{bmatrix} (0)(5)+(8)(8)+(3)(-2)+(6)(7) \\ (9)(5)+(0)(8)+(8)(-2)+(-4)(7)\end{bmatrix}$

$A\bf{x}= \begin{bmatrix} 100 \\ 1 \end{bmatrix}$

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Example Question #122 : Matrices

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:

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

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

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

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

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

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 #124 : Matrices

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:

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

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

is not defined.

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

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

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 #811 : Linear Algebra

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

Possible Answers:

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

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

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

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

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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Linear Algebra : Matrix-Vector Product

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #21 : Matrix Vector Product

Example Question #122 : Matrices

Example Question #123 : Matrices

Example Question #124 : Matrices

Example Question #811 : Linear Algebra

All Linear Algebra Resources

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