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Linear Algebra : Linear Algebra

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Example Questions

Example Question #52 : Matrices

Markov 2

It is recommended that you use a calculator with matrix arithmetic capability for this problem.

The above diagram shows a board for a game of chance. A player moves according to the flip of a fair coin, depending on his current location. For example, if he is on the green square, he will move to the orange square if the coin comes up heads, and to the pink square if it comes up tails.

If, after eight moves, a player ends up on the green square, he wins. Which square should he start out on in order to maximize his probability of winning?

Possible Answers:

Blue

Orange

Pink

Green

Correct answer:

Pink

Explanation:

Since the coin is fair, each outcome will come up with probability 0.5. Therefore, a player on orange end up on orange or pink with 0.5 probability each; a player on pink will end up on orange or blue with probability 0.5 each; a player on blue will end up on pink or green with probability 0.5 each; and a player on green will end up on orange or pink with probability 0.5 each. The stochastic matrix that models this is

$P = \begin{bmatrix} 0.5 & 0.5 & 0&0.5 \\ 0.5 &0 & 0.5 & 0.5\\ 0 & 0.5 &0 & 0 \\ 0 & 0 & 0.5 & 0 \end{bmatrix}$ .

The matrix that models the probabilities that the player will end up on a given square after eight moves, given that he starts on a particular square, is $P^{8}$ , which through calculation is

$P^{8} = \begin{bmatrix} 0.4180 & 0.4141&0.4180 &0.4180 \\ 0.3320 &0.3359 &0.3320 &0.3320 \\ 0.1680 & 0.1641 & 0.1680 &0.1680 \\ 0.0820 & 0.0859 &0.0820 & 0.0820 \end{bmatrix}$

Since he wants to land on green after eight turns, examine the fourth row. The greatest probability of ending up on green, albeit by very little, appears in the second row. Therefore, to maximize his chances of ending up on green, the player should start on pink.

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Example Question #53 : Matrices

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ .

Calculate $A ^{T}A$ .

Possible Answers:

$\begin{bmatrix}16 & -6-5i \\ -6-5i &25\end{bmatrix}$

$\begin{bmatrix}19+4i &3-3i& -13+2i \\ 3-3i & 8-4i & 4+6i \\ -13+2i &4+6i& 8\end{bmatrix}$

$A ^{T}A$ is undefined.

$\begin{bmatrix}10-2i & -4+i \\ -4+i & 25+2i \end{bmatrix}$

$\begin{bmatrix} 21&1-9i & -11+2i \\1+9i & 12& 2\\ -11-2i &2& 10\end{bmatrix}$

Correct answer:

$\begin{bmatrix}19+4i &3-3i& -13+2i \\ 3-3i & 8-4i & 4+6i \\ -13+2i &4+6i& 8\end{bmatrix}$

Explanation:

$A^{T}$ , the transpose of , is the result of switching the rows of with the columns.

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ ,

$A^{T} = \begin{bmatrix} 2+ i & -4 \\ 3- i & 1+ i \\i & 3 \end{bmatrix}$

Find the product $A ^{T}A$ by multiplying the rows of $A^{T}$ by the columns of ; that is, add the product of the terms in corresponding positions:

$A ^{T}A$

$= \begin{bmatrix} 2+ i & -4 \\ 3- i & 1+ i \\i & 3 \end{bmatrix} \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$

$= \begin{bmatrix} (2+ i) (2+ i) + (-4)(-4) &(2+ i) (3 - i) + (-4)(1+i) & (2+ i) ( i) + (-4)(3) \\ (3-i) (2+ i) + (1+i)(-4) & (3-i) (3 - i) +(1+i)(1+i) & (3-i) ( i) +(1+i)(3) \\ ( i) (2+ i) + (3)(-4) &( i) (3 - i) + (3)(1+i) & ( i) ( i) + (3)(3)\end{bmatrix}$

$= \begin{bmatrix}19+4i &3-3i& -13+2i \\ 3-3i & 8-4i & 4+6i \\ -13+2i &4+6i& 8\end{bmatrix}$

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Example Question #54 : Matrices

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ .

Calculate $AA ^{*}$ .

Possible Answers:

$\begin{bmatrix}19+4i &3-3i& -13+2i \\ 3-3i & 8-4i & 4+6i \\ -13+2i &4+6i& 8\end{bmatrix}$

$\begin{bmatrix} 21&1-9i & -11+2i \\1+9i & 12& 2\\ -11-2i &2& 10\end{bmatrix}$

$\begin{bmatrix}10-2i & -4+i \\ -4+i & 25+2i \end{bmatrix}$

$AA ^{*}$ is undefined.

$\begin{bmatrix}16 & -6-5i \\ -6-5i &27\end{bmatrix}$

Correct answer:

$\begin{bmatrix}16 & -6-5i \\ -6-5i &27\end{bmatrix}$

Explanation:

$A^{*}$ , the conjugate transpose of , can be found by first taking the transpose of :

$A = \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}$ ,

$A^{T} = \begin{bmatrix} 2+ i & -4 \\ 3- i & 1+ i \\i & 3 \end{bmatrix}$

then changing each element to its complex conjugate:

$A^{*} = \begin{bmatrix} 2- i & -4 \\ 3+i & 1- i \\-i & 3 \end{bmatrix}$

Find the product $AA^{*}$ by multiplying the rows of by the columns of $A^{*}$ ; that is, add the product of the terms in corresponding positions:

$AA^{*}$

$= \begin{bmatrix} 2+ i & 3 - i & i \\ -4 & 1+ i & 3 \end{bmatrix}\begin{bmatrix} 2- i & -4 \\ 3+i & 1- i \\-i & 3 \end{bmatrix}$

$= \begin{bmatrix}( 2+ i )( 2- i ) + (3 - i) (3 + i)+ i(-i) & ( 2+ i )( -4 ) + (3 - i) (1-i)+ i(3) \\ ( -4 )( 2- i ) + (1+i)(3 + i)+ 3(-i) & -4(-4) +(1+ i)(1-i) + 3(3) \end{bmatrix}$

$= \begin{bmatrix}16 & -6-5i \\ -6-5i &27\end{bmatrix}$

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Example Question #52 : Matrix Matrix Product

is a $2 \times 2$ nonsingular matrix; is a $3 \times 2$ matrix; is a $2 \times 2$ singular matrix.

Which of the following is defined?

Possible Answers:

ACB

$BA^{-1}C$

$CA^{-1}B^{-1}$

$A^{-1}B^{T}C$

$C^{-1}AB^{T}$

Correct answer:

$BA^{-1}C$

Explanation:

$CA^{-1}B^{-1}$ can be eliminated as a choice; is not a square matrix, so its inverse, $B^{-1}$ , is undefined.

$C^{-1}AB^{T}$ can be eliminated as a choice, since it is given that is singular - that is, $C^{-1}$ does not exist.

For the product of three matrices to be defined, they must be, in order, an $m \times n$ matrix, an $n \times p$ matrix, and a $p \times q$ matrix. We examine the three remaining choices.

$A^{-1}B^{T}C$ :

is a $2 \times 2$ matrix, so $A^{-1}$ is as well; is a $3 \times 2$ matrix, so its transpose $B^{T}$ is a $2 \times 3$ matrix; is a $2 \times 2$ matrix. These are incompatible, so $A^{-1}B^{T}C$ is undefined.

ACB :

is a $2 \times 2$ matrix; is a $2 \times 2$ matrix; is a $3 \times 2$ matrix; These are also incompatible, so ACB is undefined.

$BA^{-1}C$ :

is a $3 \times 2$ matrix; $A^{-1}$ is a $2 \times 2$ matrix, as stated before; is a $2 \times 2$ matrix. These are compatible, so $BA^{-1}C$ is defined. This is the correct choice.

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Example Question #55 : Matrices

Game

Some children are playing a game that employs the above board and a pair of fair six-sided dice.

According to the rules of the game, if you are on one of the six spaces of the hexagonal board at left:

1) If you roll a 6, you go to the gray "Time Out!" square.

2) If you roll any other number, you move that many squares clockwise.

If you are on the gray "Time Out!" square:

1) If you roll anything but doubles, you stay on the gray square.

2) If you roll doubles, you re-enter the hexagonal board on the space indicated by the particular double - for example, if you roll 1-1, you re-enter the board on the red space; 2-2, the blue space; and so forth.

Construct a stochastic matrix to model the probability of ending up on each space on any given turn. Let the first row/column represent the red space, the second row/column represent the blue space, and so forth clockwise, and let the seventh row/column represent the gray square.

Possible Answers:

$\begin{bmatrix}0 &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{36} \\ \\ \frac{1}{6} &0& \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} \\ \\ \frac{1}{6} & \frac{1}{6} & 0&\frac{1}{6}&\frac{1}{6}& \frac{1}{6}&\frac{1}{36} \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}&0&\frac{1}{6} &\frac{1}{6}&\frac{1}{36} \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &0 & \frac{1}{6} &\frac{1}{36}\\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & 0 & \frac{1}{36} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{5}{6}\end{bmatrix}$

$\begin{bmatrix} \frac{1}{36} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{36} \\ \\ \frac{1}{6} & \frac{1}{36} & \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{36} &\frac{1}{6}&\frac{1}{6}& \frac{1}{6}&\frac{1}{36} \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}& \frac{1}{36} &\frac{1}{6} &\frac{1}{6}&\frac{1}{36} \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} & \frac{1}{6} &\frac{1}{36}\\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} & \frac{1}{36} \\ \\ \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36}& \frac{5}{6}\end{bmatrix}$

$\begin{bmatrix} \frac{5}{36} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{36} \\ \\ \frac{1}{6} & \frac{5}{36} & \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{5}{36} &\frac{1}{6}&\frac{1}{6}& \frac{1}{6}&\frac{1}{36} \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}& \frac{5}{36} &\frac{1}{6} &\frac{1}{6}&\frac{1}{36} \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{5}{36} & \frac{1}{6} &\frac{1}{36}\\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{5}{36} & \frac{1}{36} \\ \\ \frac{1}{36} & \frac{1}{36} & \frac{1}{36} & \frac{1}{36} & \frac{1}{36} & \frac{1}{36}& \frac{5}{6}\end{bmatrix}$

$\begin{bmatrix} \frac{1}{36} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} \\ \\ \frac{1}{6} & \frac{1}{36} & \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{6} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{36} &\frac{1}{6}&\frac{1}{6}& \frac{1}{6}&\frac{1}{6} \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}& \frac{1}{36} &\frac{1}{6} &\frac{1}{6}&\frac{1}{6} \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} & \frac{1}{6} &\frac{1}{6}\\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} & \frac{1}{6} \\ \\ \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36}& \frac{1}{6}\end{bmatrix}$

$\begin{bmatrix}0 &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} \\ \\ \frac{1}{6} &0& \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{6} \\ \\ \frac{1}{6} & \frac{1}{6} & 0&\frac{1}{6}&\frac{1}{6}& \frac{1}{6}&\frac{1}{6} \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}&0&\frac{1}{6} &\frac{1}{6}&\frac{1}{6} \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &0 & \frac{1}{6} &\frac{1}{6}\\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & 0 & \frac{1}{6} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & 0\end{bmatrix}$

Correct answer:

Explanation:

If a player is on the red space, he will end up on each of the other spaces with the indicated rolls:

Red: 12; 1 roll out of 36 - probability $\frac{1}{36}$ .

Blue: 7; 6 rolls out of 36 possible - probability $\frac{1}{6}$ .

Green: 2 (1 roll) or 8 (5 rolls); 6 rolls out of 36 - probability $\frac{1}{6}$ .

Orange: 3 (2 rolls) or 9 (4 rolls); 6 rolls out of 36 - probability $\frac{1}{6}$ .

Lavender: 4 (3 rolls) or 10 (3 rolls); 6 rolls out of 36 - probability $\frac{1}{6}$ .

Brown: 5 (4 rolls) or 11 (2 rolls); 6 rolls out of 36 - probability $\frac{1}{6}$ .

Gray ("Time out"): 6; 5 rolls out of 36 - probability $\frac{5}{36}$ .

The $7 \times 7$ stochastic matrix with the first column only filled in is

$\begin{bmatrix} \frac{1}{36} & \times& \times& \times& \times& \times& \times \\ \\ \frac{1}{6} & \times& \times& \times& \times& \times& \times \\ \\ \frac{1}{6} & \times& \times& \times& \times& \times& \times \\ \\ \frac{1}{6} & \times& \times& \times& \times& \times& \times \\ \\ \frac{1}{6} & \times& \times& \times& \times& \times& \times \\ \\ \frac{1}{6} & \times& \times& \times& \times& \times& \times \\ \\ \frac{5}{36} & \times& \times& \times& \times& \times& \times\end{bmatrix}$

The rules of the rolls are analogous for each of the other spaces, keeping the probabilities, relatively speaking, the same - the probability of staying on the same space is $\frac{1}{36}$ , that of landing on a given other space on the hexagon is $\frac{1}{6}$ , and the probability of ending up on the gray square is $\frac{5}{36}$ . The next five columns can be filled in as follows:

$\begin{bmatrix} \frac{1}{36} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \times \\ \\ \frac{1}{6} & \frac{1}{36} & \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \times \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{36} &\frac{1}{6}&\frac{1}{6}& \frac{1}{6}& \times \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}& \frac{1}{36} &\frac{1}{6} &\frac{1}{6}& \times \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} & \frac{1}{6} &\times\\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \frac{1}{36} & \times \\ \\ \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36} & \frac{5}{36}& \times\end{bmatrix}$

A player on the gray square can only end up any given space on the hexagon by rolling one specific pair of doubles, making each of these probabilities equal to $\frac{1}{36}$ . The other 30 rolls will keep him on the gray square, so there is a $\frac{5}{6}$ probability he will stay on the gray square on that turn. Therefore, the final column can be filled in as follows:

This matrix is the correct choice.

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Example Question #56 : Matrices

Game

Some children are playing a game that employs the above board and a pair of fair six-sided dice.

According to the rules of the game:

If you are on one of the six spaces of the hexagonal board at left, you roll one die. If roll a 6, you go to the gray square marked "Time Out!"; otherwise, you move clockwise the number of squares indicated.

If you are on the gray square marked "Time Out!", you roll both dice. If you roll anything but doubles, you stay on the gray square. If you roll doubles, you re-enter the hexagonal board on the space indicated by the particular double - for example, if you roll 1-1, you re-enter the board on the red space; 2-2, the blue space; and so forth.

Possible Answers:

Correct answer:

Explanation:

If a player is on one of the spaces on the hexagon, he will end up on each of the other spaces with probability $\frac{1}{6}$ ; he cannot stay on his current space, since it would require a roll of 6 - this, by the rules, takes him to the gray square. Therefore, the first six columns of the stochastic matrix for this game can be filled in as follows:

$\begin{bmatrix}0 &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \times \\ \\ \frac{1}{6} &0& \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & \times \\ \\ \frac{1}{6} & \frac{1}{6} & 0&\frac{1}{6}&\frac{1}{6}& \frac{1}{6}& \times \\ \\ \frac{1}{6}& \frac{1}{6} &\frac{1}{6}&0&\frac{1}{6} &\frac{1}{6}& \times \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &0 & \frac{1}{6} & \times \\ \\ \frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} &\frac{1}{6} & 0 & \times \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \times\end{bmatrix}$

This is the correct choice.

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Example Question #61 : Matrix Matrix Product

$A = \begin{bmatrix} \cos \frac{4\pi }{13}\\ \\i \sin \frac{4\pi }{13}\end{bmatrix}$

If $A^{T}A = \begin{bmatrix} a \end{bmatrix}$ , then which expression is equal to ?

Possible Answers:

$a= \frac{1}{2}\sin \frac{8 \pi}{13}$

$a= \cos \frac{8 \pi}{13}$

$a= \frac{1}{2}\cos \frac{8 \pi}{13}$

$a= \sin \frac{8 \pi}{13}$

a = 1

Correct answer:

$a= \cos \frac{8 \pi}{13}$

Explanation:

The transpose of a column matrix is the row matrix with the same entries, so

$A = \begin{bmatrix} \cos \frac{4\pi }{13}\\ \\i \sin \frac{4\pi }{13}\end{bmatrix}$ , so $A^{T} = \begin{bmatrix} \cos \frac{4\pi }{13} & i \sin \frac{4\pi }{13}\end{bmatrix}$

Find $A^{T}A$ by simply adding the products of corresponding entries:

$A^{T}A= \begin{bmatrix} \cos \frac{4\pi }{13} & i \sin \frac{4\pi }{13}\end{bmatrix} \begin{bmatrix} \cos \frac{4\pi }{13}\\ \\i \sin \frac{4\pi }{13}\end{bmatrix}$

$\begin{bmatrix} \cos \frac{4\pi }{13} \cdot \cos \frac{4\pi }{13} + i \sin \frac{4\pi }{13} \cdot i \sin \frac{4\pi }{13} \end{bmatrix}$

$a= \cos \frac{4\pi }{13} \cdot \cos \frac{4\pi }{13} + i \sin \frac{4\pi }{13} \cdot i \sin \frac{4\pi }{13}$

$= \cos ^{2}\frac{4\pi }{13} + i^{2} \sin^{2} \frac{4\pi }{13}$

$= \cos ^{2}\frac{4\pi }{13} + (-1) \sin^{2} \frac{4\pi }{13}$

$= \cos ^{2}\frac{4\pi }{13} -\sin^{2} \frac{4\pi }{13}$

Apply the trigonometric identity

$\cos ^{2} \theta -\sin^{2} \theta = \cos 2\theta$ ,

setting $\theta =\frac{4\pi }{13}$ :

$a= \cos \frac{8 \pi}{13}$

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Example Question #62 : Matrix Matrix Product

is a $2 \times 3$ matrix. is a $3 \times 2$ matrix. and are both nonsingular.

Which expression is defined?

Possible Answers:

$(AB )^{T}+ (BA)^{-1}$

$(AB )^{-1}+BA$

$AB + B^{T}A^{T}$

$AB + B^{-1}A^{-1}$

$AB + (BA)^{T}$

Correct answer:

$AB + B^{T}A^{T}$

Explanation:

and , being nonsquare matrices, cannot have inverses, so $AB + B^{-1}A^{-1}$ can be eliminated as a choice.

The product of a $m \times n$ matrix and a $n \times p$ matrix is a $m \times p$ matrix. Therefore:

is a $2 \times 2$ matrix; $(AB)^{T}$ and $(AB)^{-1}$ are also $2 \times 2$ matrices.

Similarly,

, $(BA)^{T}$ , and $(BA)^{-1}$ are $3 \times 3$ matrices.

Matrices of different dimensions cannot be added, so $(AB )^{-1}+BA$ , $AB + (BA)^{T}$ , and $(AB )^{T}+ (BA)^{-1}$ can be eliminated.

$B^{T}$ and $A^{T}$ are $3 \times 2$ and $2 \times 3$ matrices, respectively. Therefore, $B^{T}A^{T}$ is a $3 \times 3$ matrix; $AB + B^{T}A^{T}$ , the sum of $3 \times 3$ matrices, is a defined expression.

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Example Question #63 : Matrix Matrix Product

$A = \begin{bmatrix} \sqrt{2 }\cos \frac{6\pi }{11}\\ \\i \end{bmatrix}$

If $A^{T}A = \begin{bmatrix} a \end{bmatrix}$ , then which expression is equal to ?

Possible Answers:

a = 1

$a = \sin \frac{12\pi}{11}$

$a = 2 \sin \frac{12\pi}{11}$

$a = \cos \frac{12\pi}{11}$

$a = 2 \cos \frac{12\pi}{11}$

Correct answer:

$a = \cos \frac{12\pi}{11}$

Explanation:

The transpose of a column matrix is the row matrix with the same entries, so if

$A = \begin{bmatrix} \sqrt{2 }\cos \frac{6\pi }{11}\\ \\i \end{bmatrix}$ ,

then

$A^{T} = \begin{bmatrix} \sqrt{2 }\cos\cos \frac{6\pi }{11} & i \end{bmatrix}$

Find $A^{T}A$ by simply adding the products of corresponding entries:

$A^{T} A = \begin{bmatrix} \sqrt{2 }\cos \frac{6\pi }{11} \cdot \sqrt{2 } \cos \frac{6\pi }{11} + i \cdot i \end{bmatrix}$

$a = \sqrt{2 }\cos \frac{6\pi }{11} \cdot \sqrt{2 } \cos \frac{6\pi }{11} + i \cdot i$

$=2 \cos ^{2 } \frac{6\pi }{11} + i ^{2}$

$=2 \cos ^{2 } \frac{6\pi }{11} -1$

Apply the trigonometric identity

$2 \cos^{2} \theta -1 = \cos 2\theta$

Setting $\theta = \frac{6\pi }{11}$ ,

$a = \cos \frac{12\pi}{11}$ .

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Example Question #61 : Matrix Matrix Product

$A= \begin{bmatrix} \cos \frac{\pi }{36} + i \sin \frac{\pi }{36} & 0 \\ 0 & \cos \frac{\pi }{18} + i \sin \frac{\pi }{18} \end{bmatrix}$

Find $A^{6}$ .

Possible Answers:

$\begin{bmatrix} \frac{ \sqrt{3}}{2}+\frac{1}{2} i& 0 \\ 0 &\frac{1}{2} -\frac{ \sqrt{3}}{2} i \end{bmatrix}$

None of the other choices gives the correct response.

$\begin{bmatrix} \frac{ \sqrt{3}}{2}-\frac{1}{2} i& 0 \\ 0 &\frac{1}{2} +\frac{ \sqrt{3}}{2} i \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2} +\frac{ \sqrt{3}}{2} i & 0 \\ 0 & \frac{ \sqrt{3}}{2}+\frac{1}{2} i \end{bmatrix}$

$\begin{bmatrix} \frac{1}{2} -\frac{ \sqrt{3}}{2} i & 0 \\ 0 & \frac{ \sqrt{3}}{2}-\frac{1}{2} i \end{bmatrix}$

Correct answer:

None of the other choices gives the correct response.

Explanation:

is a diagonal matrix. The product of two diagonal matrices can be found by multiplying elements in corresponding diagonal positions; the idea can be extended to powers of matrices, so

$A^{6}= \begin{bmatrix} \left (\cos \frac{\pi }{36} + i \sin \frac{\pi }{36} \right ) ^{6}& 0 \\ 0 & \left (\cos \frac{\pi }{18} + i \sin \frac{\pi }{18} \right )^{6} \end{bmatrix}$

By DeMoivre's Theorem,

$\left ( \cos \theta + i \sin \theta \right )^{n} = \cos n \theta + i \sin n \theta$

Set n = 6 and $\theta = \frac{\pi }{36 }$ in the upper left element:

$\left (\cos \frac{\pi }{36} + i \sin \frac{\pi }{36} \right ) ^{6}$

$= \cos \left ( 6 \cdot \frac{\pi}{36}\right )+ i \sin\left ( 6 \cdot \frac{\pi}{36}\right )$

$= \cos \frac{\pi}{6} + i \sin \frac{\pi}{6}$

$=\frac{ \sqrt{3}}{2}+\frac{1}{2} i$

Set n = 6 and $\theta = \frac{\pi }{18}$ in the lower right element:

$\left (\cos \frac{\pi }{18} + i \sin \frac{\pi }{18} \right ) ^{6}$

$= \cos \left ( 6 \cdot \frac{\pi}{18}\right )+ i \sin\left ( 6 \cdot \frac{\pi}{18}\right )$

$= \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$

$=\frac{1}{2} +\frac{ \sqrt{3}}{2} i$

Therefore,

$A^{6}= \begin{bmatrix} \frac{ \sqrt{3}}{2}+\frac{1}{2} i& 0 \\ 0 &\frac{1}{2} +\frac{ \sqrt{3}}{2} i \end{bmatrix}$ ,

which is not among the choices.

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