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Linear Algebra : Matrices

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Example Questions

Example Question #171 : Matrices

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

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

A parallelogram has these two vectors as sides. Find so that the parallelogram is a rectangle.

Possible Answers:

$a =- \frac{3}{2}$

$a = -\frac{9}{4}$

a = 6

a = 28

$a = 2\sqrt{7}$

Correct answer:

$a = -\frac{9}{4}$

Explanation:

For the parallelogram formed by $\textbf{u}$ and $\textbf{v}$ to be a rectangle, the vectors must be perpendicular - that is, orthogonal, This is true if and only if $\textbf{u} \cdot \textbf{v} = 0$ .

The dot product of the vectors can be found by adding the products of corresponding entries:

$\textbf{u} \cdot \textbf{v} = 4(2)+ (-1)(-1)+4(a) = 4a+9$

Set equal to 0 and solve for :

$\textbf{u} \cdot \textbf{v} = 0$

4a+9 = 0

4a = -9

$a = -\frac{9}{4}$ .

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Example Question #43 : Vector Vector Product

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

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

Find a value of (nearest tenth) so that $\textbf{u}$ and $\textbf{v}$ are the sides of a parallelogram of area 100.

Possible Answers:

22.2

22.6

23.8

26.4

25.9

Correct answer:

26.4

Explanation:

The area of a parallelogram with sides $\textbf{u}$ and $\textbf{v}$ in $\mathbb{R}^{3}$ is the norm of their cross-product:

$A = \left \|\textbf{u} \times \textbf{v} \right \|$

To find the cross-product, take the "determinant" of the matrix formed from the entries of the vectors, as follows

$\textbf{u} \times \textbf{v} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 4 & -1 & 4 \\ 2 & -1 & a \end{vmatrix}$

where $\textbf{i}= (1,0,0), \textbf{j} = (0,1,0), \textbf{k} = (0,0,1)$ .

Find this "determinant" as you would a numeric determinant - add the upper-left to lower-right products, and subtract the upper-right to lower-left products.

$\begin{vmatrix}{\color{Red} \textbf{i}} & {\color{DarkGreen} \textbf{j}} &{\color{Purple} \textbf{k}} \\ 4 & {\color{Red} \textbf{-1}} & {\color{DarkGreen} \textbf{4}}\\ 2 & -1 &{\color{Red} \textbf{a}}\end{vmatrix}\begin{matrix} \textbf{i} & \textbf{j} \\ {\color{Purple} \textbf{4}} & -1 \\ {\color{DarkGreen} \textbf{2}} & {\color{Purple} \textbf{-1}} \end{matrix}$ $\begin{vmatrix} \textbf{i} & \textbf{j} & {\color{Red} \textbf{k} } \\ 4 & {\color{Red}\textbf{-1}} & {\color{DarkGreen}\textbf{4}} \\ {\color{Red}\textbf{2}} & {\color{DarkGreen}\textbf{-1}} & {\color{Purple}\textbf{a}} \end{vmatrix}\begin{matrix} {\color{DarkGreen}\textbf{i} }& {\color{Purple} \textbf{j}} \\ {\color{Purple} \textbf{4}} & -1 \\ 2 & -1 \end{matrix}$

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

$= \textbf{i}(-1)(a)+ \textbf{j}(4)(2)+ \textbf{k}(4)(-1) - \textbf{k}(-1)(1)- \textbf{i}(4)(-1)- \textbf{j}(4)(a)$

$=-a \textbf{i}+8 \textbf{j} -4 \textbf{k} + \textbf{k} +4 \textbf{i} -4a \textbf{j}$

$=(-a+4) \textbf{i}+( -4a+8 ) \textbf{j} -3 \textbf{k}$

=(-a+4, -4a+ 8, -3 )

The norm is the result of adding the squares of the entries, and taking the square root. In terms of , this is

$\left \|\textbf{u} \times \textbf{v} \right \|$

$= \sqrt{(-a+4)^{2}+ ( -4a+ 8)^{2}+( -3 )^{2}}$

$= \sqrt{a^{2}-8a + 16 + 16a^{2} -64a + 64 + 9 }$

$= \sqrt{17a^{2}-72a +25}$

Set this equal to 100:

$\sqrt{17a^{2}-72a +25} = 100$

$17a^{2}-72a +25 = 10,000$

$17a^{2}-72a -9,975 = 0$

Through the quadratic formula:

$a=\frac{ 72 \pm \sqrt{72^{2}- 4(17)(-9,975)}}{2(17)}$

After calculation, we get solutions

$a \approx -22.2, a \approx 26.4$ .

All of the choices are positive, so select 26.4.

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Example Question #172 : Matrices

A parallelogram in $\mathbb{R}^{3}$ has as two of its sides the vectors

$\textbf{u} = (3, 5, 4 )$

and

$\textbf{v}= (5, 0, -5)$ .

Which statement is true of the parallelogram?

Possible Answers:

The parallelogram is neither a rectangle nor a rhombus.

The parallelogram is a rectangle, but not a rhombus.

The parallelogram is a rhombus, but not a rectangle.

The parallelogram is a square.

Correct answer:

The parallelogram is a rhombus, but not a rectangle.

Explanation:

The parallelogram is a rectangle if and only if the two vectors that form the adjacent sides are perpendicular - that is, orthogonal. This happens if the dot product - the sum of the products of elements in corresponding positions - is equal to 0. Test this:

$\textbf{u} \cdot \textbf{v} = 3(5)+5(0)+ 4(-5) = -5 \ne 0$

The parallelogram is not a rectangle.

The parallelogram is a rhombus if and only if the two vectors that form the adjacent sides are of equal lenght - that is, if their norms are equal. The norm of a vector is the square root of the sum of the squares of its elements, so:

$\left \| \textbf{u} \right \| = \sqrt{3^{2}+5^{2}+4^{2}} = \sqrt{9+ 25 + 16}= \sqrt{50}$

$\left \| \textbf{v} \right \| = \sqrt{5^{2}+0^{2}+(-5)^{2}} = \sqrt{ 25 + 0+25}= \sqrt{50}$

$\left \| \textbf{u} \right \| = \left \| \textbf{v} \right \|$ , so the parallelogram is a rhombus.

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Example Question #173 : Matrices

$\textbf{u} = (8, -1,- 3)$

$\textbf{v}= (8, 1, 3)$

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

Possible Answers:

$43^{\circ}$

$137^{\circ}$

$47^{\circ}$

$90^{\circ}$

$133^{\circ}$

Correct answer:

$43^{\circ}$

Explanation:

The angle between two vectors $\textbf{u}$ and $\textbf{v}$ is $\theta$ , where

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

The dot product of two vectors is the sum of the products of their corresponding entries:

$\textbf{u} \cdot \textbf{v} = 8(8)+ (-1)(1)+ (-3)3= 54$

The norm of a vector is the square root of the squares of its entries:

$\left \| \textbf{u} \right \| = \sqrt{8^{2}+(-1)^{2}+ (-3)^{2}} = \sqrt{74}$

$\left \| \textbf{v} \right \| = \sqrt{8^{2}+1^{2}+ 3^{2}} = \sqrt{74}$

$\cos \theta = \frac{54}{( \sqrt{74}) (\sqrt{74} )}$

$\cos \theta = \frac{54}{74}$

$\theta = \cos^{-1} \frac{54}{74} \approx 43^{\circ}$

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Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #171 : Matrices

Example Question #43 : Vector Vector Product

Example Question #172 : Matrices

Example Question #173 : Matrices

All Linear Algebra Resources

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