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Linear Algebra : Matrices

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Example Questions

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:

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

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

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

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

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 #61 : Matrices

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

Which expression is defined?

Possible Answers:

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

$AB + (BA)^{T}$

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

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

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

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:

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

a = 1

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

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

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

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 #62 : Matrices

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

Find $A^{6}$ .

Possible Answers:

None of the other choices gives the correct response.

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

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

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

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

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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Example Question #751 : Linear Algebra

$A = \begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix}$

Which of the following is equal to $A^{4}$ ?

Possible Answers:

$\begin{bmatrix}-50 & 42 \\ -63 & -29\end{bmatrix}$

None of the other choices gives the correct response.

$\begin{bmatrix}-50 & -63 \\ 42& -29\end{bmatrix}$

$\begin{bmatrix}-50 & -42 \\ 63 & -29\end{bmatrix}$

$\begin{bmatrix}-50 & 63 \\ -42& -29\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-50 & -63 \\ 42& -29\end{bmatrix}$

Explanation:

An easy way to find this is to note that $A^{4} = (A^{2} )^{2}$ ; therefore, we can find $A^{4}$ by squaring and squaring the result.

Matrix multiplication is worked row by column - each row in the former matrix is multiplied by each column in the latter by adding the products of elements in corresponding positions, as follows:

$A^{2} = \begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -2 & 1 \end{bmatrix}$

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

$= \begin{bmatrix}-2 & 9 \\-6& -5\end{bmatrix}$

Now square this:

$A^{4} = (A^{2} )^{2}= \begin{bmatrix}-2 & 9 \\-6& -5\end{bmatrix} \begin{bmatrix}-2 & 9 \\-6& -5\end{bmatrix}$

$= \begin{bmatrix}-2(-2) + 9(-6) & -2(9) + 9(-5) \\ -6(-2) + (-5)(-6) & -6(9) + (-5)(-5) \end{bmatrix}$

$= \begin{bmatrix}-50 & -63 \\ 42& -29\end{bmatrix}$

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Example Question #63 : Matrices

$A = \begin{bmatrix} \cos \frac{ \pi}{7} - i \sin \frac{ \pi}{7} & 0 \\ 0 & \cos \frac{ \pi}{5} - i \sin \frac{ \pi}{5} \end{bmatrix}$

Evaluate $A^{35}$ .

Possible Answers:

None of the other choices gives the correct response.

$\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$

$\begin{bmatrix}-1 & 0 \\ 0 & -1 \end{bmatrix}$

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

$\begin{bmatrix}-1 & 0 \\ 0 & 1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix}-1 & 0 \\ 0 & -1 \end{bmatrix}$

Explanation:

is a diagonal matrix, so it can be raised to a power by raising the individual entries to that power:

$A^{35} = \begin{bmatrix} \left (\cos \frac{ \pi}{7} - i \sin \frac{ \pi}{7} \right ) ^{35}& 0 \\ 0 & \left (\cos \frac{ \pi}{5} - i \sin \frac{ \pi}{5} \right ) ^{35}\end{bmatrix}$

By DeMoivre's Theorem,

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

We will need to rewrite the entries in the matrix as sums, not differences. We can do this by noting that the cosine and sine functions are even and odd, respectively, so $A^{35}$ can be rewritten as

$A^{35} = \begin{bmatrix} \left (\cos\left ( -\frac{ \pi}{7} \right )+ i \sin\left ( -\frac{ \pi}{7} \right )\right ) ^{35}& 0 \\ 0 & \left (\cos\left ( -\frac{ \pi}{5} \right )+ i \sin\left ( -\frac{ \pi}{5} \right )\right ) ^{35}\end{bmatrix}$

Applying DeMoivre's Theorem,

$A^{35} = \begin{bmatrix} \cos \left (35\cdot \left ( -\frac{ \pi}{7} \right ) \right )+ i \sin \left (35\cdot \left ( -\frac{ \pi}{7} \right ) \right ) & 0 \\ 0 & \cos \left (35\cdot \left ( -\frac{ \pi}{5} \right ) \right )+ i \sin \left (35\cdot \left ( -\frac{ \pi}{5} \right ) \right ) \end{bmatrix}$

$A^{35} = \begin{bmatrix} \cos \left (-5 \pi \right )+ i \sin \left (-5 \pi \right ) & 0 \\ 0 &\cos \left (-7\pi \right )+ i \sin \left (-7 \pi \right ) \end{bmatrix}$

The coterminal angle for both $-5 \pi$ and $-7\pi$ is $\pi$ , so

$A^{35} = \begin{bmatrix} \cos \pi + i \sin \pi & 0 \\ 0 & \cos \pi + i \sin \pi \end{bmatrix}$

$A^{35} = \begin{bmatrix}-1 + i (0)& 0 \\ 0 & -1 + i (0) \end{bmatrix}$

$A^{35} = \begin{bmatrix}-1 & 0 \\ 0 & -1 \end{bmatrix}$

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Example Question #753 : Linear Algebra

is a column matrix with seven entries. Which of the following is true of $A^{T}A$ ?

Possible Answers:

$A^{T}A$ is a scalar.

$A^{T}A$ is a column matrix with seven entries.

$A^{T}A$ is a row matrix with seven entries.

$A^{T}A$ is a $7 \times 7$ matrix.

$A^{T}A$ is a matrix with a single entry.

Correct answer:

$A^{T}A$ is a matrix with a single entry.

Explanation:

, a column matrix with ten entries, is a $7 \times 1$ matrix; $A^{T}$ , its transpose, is a $1 \times 7$ matrix.

If and are $m \times n$ and $n \times p$ matrices, respectively, then the product is a $m \times p$ matrix. Therefore, $A^{T}A$ is a $1 \times 1$ matrix - a matrix with a single entry.

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Example Question #754 : Linear Algebra

is a column matrix with ten entries. Which of the following is true of $AA^{T}$ ?

Possible Answers:

$AA^{T}$ is a column matrix with ten elements.

$AA^{T}$ is a matrix with a single entry.

$AA^{T}$ is a row matrix with ten elements.

$AA^{T}$ is a $10 \times 10$ matrix.

$AA^{T}$ is a scalar.

Correct answer:

$AA^{T}$ is a $10 \times 10$ matrix.

Explanation:

, a column matrix with ten entries, is a $10 \times 1$ matrix; $A^{T}$ , its transpose, is a $1 \times 10$ matrix.

If and are $m \times n$ and $n \times p$ matrices, respectively, then the product is a $m \times p$ matrix. Therefore, $AA^{T}$ is a $10 \times 10$ matrix.

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Example Question #755 : Linear Algebra

$A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

$B= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 &-2 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

$C = \begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Which of the following is equal to ABC ?

Possible Answers:

$\begin{bmatrix} 1 & -1 & -1 & -1 \\ 2 & -9 & 3 & -3 \\ 1 & 7 & -2 & 2 \\ 1 & -8 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & -1 & -1 \\ -6 & -3 & -9 & 9 \\ 13 & 5 &16 & -16 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} 1 & -9 & -1 & -1 \\ 2 & 15 & 3 & -3 \\ 1 &-9 & -2 & 2 \\ 1 & 0 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & -1 & -1 \\ -6 & -3 & -9 & 9 \\ -3 & -3 & -8 &8 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & -1 & -1 \\ -6 & -3 & -9 &9 \\ -2 &3 & 4 & -4 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1 & 1 & -1 & -1 \\ -6 & -3 & -9 & 9 \\ -3 & -3 & -8 &8 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Explanation:

and are both elementary matrices, in that each can be formed from the (four-by-four) identity matrix by a single row operation.

Since differs from in that the entry is in Row 2, Column 2, the row operation is $-3R2 \rightarrow R2$ . Since differs from in that entry is in Row 3, Column 2, the row operation is $-2R2+R3 \rightarrow R3$ .

Premultiplying a matrix by an elementary matrix has the effect of performing that row operation on the matrix. Looking at ABC as A(BC) :

$C = \begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Premultiply by by performing the operation $-2R2+R3 \rightarrow R3$ :

$BC = \begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ -3 & -3 & -8 &8 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Premultiply by by performing the operation $-3R2 \rightarrow R2$ :

$ABC = A (BC )= \begin{bmatrix} 1 & 1 & -1 & -1 \\ -6 & -3 & -9 & 9 \\ -3 & -3 & -8 &8 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

This is the correct product.

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Example Question #756 : Linear Algebra

refers to the $2 \times 2$ identity matrix.

$A^{7} = I$ . Which of the following matrices could be equal to ?

Possible Answers:

$A = \begin{bmatrix} \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7} & 0 \\ 0 & \cos \frac{4\pi}{7} + i \sin \frac{4\pi}{7} \end{bmatrix}$

$A = \begin{bmatrix} \cos \frac{ \pi}{7} + i \sin \frac{ \pi}{7} & 0 \\ 0 & \cos \frac{6\pi}{7} + i \sin \frac{6\pi}{7} \end{bmatrix}$

None of the other responses gives a correct answer.

$A = \begin{bmatrix} \cos \frac{5\pi}{14} + i \sin \frac{5 \pi}{14} & 0 \\ 0 & \cos \frac{9\pi}{14} + i \sin \frac{9 \pi }{14} \end{bmatrix}$

$A = \begin{bmatrix} \cos \frac{\pi}{14} + i \sin \frac{ \pi}{14} & 0 \\ 0 & \cos \frac{13\pi}{14} + i \sin \frac{13 \pi }{14} \end{bmatrix}$

Correct answer:

$A = \begin{bmatrix} \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7} & 0 \\ 0 & \cos \frac{4\pi}{7} + i \sin \frac{4\pi}{7} \end{bmatrix}$

Explanation:

All of the matrices are diagonal, so the seventh power of each can be determined by simply taking the seventh power of the individual entries in the main diagonal. Also, note that each entry in each choice is of the form

$\cos \theta + i \sin \theta$ .

By DeMoivre's Theorem, for any real ,

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

Combining these ideas, we can take the seventh power of each matrix and determine which exponentiation yields the identity.

$A = \begin{bmatrix} \cos \frac{5\pi}{14} + i \sin \frac{5 \pi}{14} & 0 \\ 0 & \cos \frac{9\pi}{14} + i \sin \frac{9 \pi }{14} \end{bmatrix}$ ,

then

$A^{7} = \begin{bmatrix}( \cos \frac{5\pi}{14} + i \sin \frac{5 \pi}{14} )^{7}& 0 \\ 0 & (\cos \frac{9\pi}{14} + i \sin \frac{9 \pi }{14} )^{7}\end{bmatrix}$

$= \begin{bmatrix}\cos \frac{5\pi}{2} + i \sin \frac{5 \pi}{2} & 0 \\ 0 & \cos \frac{9\pi}{2} + i \sin \frac{9 \pi }{2} \end{bmatrix}$

$= \begin{bmatrix}\cos \frac{\pi}{2} + i \sin \frac{ \pi}{2} & 0 \\ 0 & \cos \frac{\pi}{2} + i \sin \frac{ \pi }{2} \end{bmatrix}$

$= \begin{bmatrix}0 + i (1)& 0 \\ 0 & 0 + i (1) \end{bmatrix}$

$= \begin{bmatrix}i & 0 \\ 0 & i \end{bmatrix}$

$\ne I$

$A = \begin{bmatrix} \cos \frac{\pi}{14} + i \sin \frac{ \pi}{14} & 0 \\ 0 & \cos \frac{13\pi}{14} + i \sin \frac{13 \pi }{14} \end{bmatrix}$ ,

then

$A^{7} = \begin{bmatrix}( \cos \frac{\pi}{14} + i \sin \frac{ \pi}{14} )^{7}& 0 \\ 0 &( \cos \frac{13\pi}{14} + i \sin \frac{13 \pi }{14} )^{7}\end{bmatrix}$

$= \begin{bmatrix} \cos \frac{\pi}{2} + i \sin \frac{ \pi}{2} & 0 \\ 0 & \cos \frac{13\pi}{2} + i \sin \frac{13 \pi }{2} \end{bmatrix}$

$= \begin{bmatrix} \cos \frac{\pi}{2} + i \sin \frac{ \pi}{2} & 0 \\ 0 & \cos \frac{ \pi}{2} + i \sin \frac{ \pi }{2} \end{bmatrix}$

$= \begin{bmatrix}0 + i (1)& 0 \\ 0 & 0 + i (1) \end{bmatrix}$

$= \begin{bmatrix}i & 0 \\ 0 & i \end{bmatrix}$

$\ne I$

$A = \begin{bmatrix} \cos \frac{ \pi}{7} + i \sin \frac{ \pi}{7} & 0 \\ 0 & \cos \frac{6\pi}{7} + i \sin \frac{6\pi}{7} \end{bmatrix}$ ,

then

$A^{7} = \begin{bmatrix} (\cos \frac{ \pi}{7} + i \sin \frac{ \pi}{7} )^{7} & 0 \\ 0 &( \cos \frac{6\pi}{7} + i \sin \frac{6\pi}{7} )^{7} \end{bmatrix}$

$= \begin{bmatrix} \cos \pi + i \sin \pi & 0 \\ 0 & \cos 6 \pi + i \sin 6 \pi \end{bmatrix}$

$= \begin{bmatrix} \cos \pi + i \sin \pi & 0 \\ 0 & \cos 0 + i \sin 0 \end{bmatrix}$

$= \begin{bmatrix} -1 + i (0)& 0 \\ 0 & 1 + i (0) \end{bmatrix}$

$= \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

$\ne I$

$A = \begin{bmatrix} \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7} & 0 \\ 0 & \cos \frac{4\pi}{7} + i \sin \frac{4\pi}{7} \end{bmatrix}$ ,

then

$A^{7} = \begin{bmatrix} (\cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7})^{7} & 0 \\ 0 &( \cos \frac{4\pi}{7} + i \sin \frac{4\pi}{7} )^{7} \end{bmatrix}$

$A^{7} = \begin{bmatrix} \cos 2 \pi + i \sin 2 \pi & 0 \\ 0 & \cos 4\pi + i \sin4\pi \end{bmatrix}$

$= \begin{bmatrix} \cos 0 + i \sin 0 & 0 \\ 0 & \cos 0 + i \sin 0 \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

= I

$\begin{bmatrix} \cos \frac{2\pi}{7} + i \sin \frac{2\pi}{7} & 0 \\ 0 & \cos \frac{4\pi}{7} + i \sin \frac{4\pi}{7} \end{bmatrix}$ is the only possible matrix value of among the choices.

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Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #61 : Matrix Matrix Product

Example Question #61 : Matrices

Example Question #63 : Matrix Matrix Product

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Example Question #63 : Matrices

Example Question #753 : Linear Algebra

Example Question #754 : Linear Algebra

Example Question #755 : Linear Algebra

Example Question #756 : Linear Algebra

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