Call Now to Set Up Tutoring:

(888) 888-0446

Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

Create An Account

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Matrices

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

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 = 1

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

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

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

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

Report an Error

Example Question #62 : Matrices

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

Which expression is defined?

Possible Answers:

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

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

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

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

$AB + (BA)^{T}$

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.

Report an Error

Example Question #63 : Matrices

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

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

Possible Answers:

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

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

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

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

a = 1

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

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

Report an Error

Example Question #64 : 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.

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

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

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

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

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.

Report an Error

Example Question #65 : Matrices

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

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

Possible Answers:

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

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

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

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

None of the other choices gives the correct response.

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:

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

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

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

Report an Error

Example Question #66 : 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:

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

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

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

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

None of the other choices gives the correct response.

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

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

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

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

Report an Error

Example Question #67 : Matrices

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

Possible Answers:

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

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

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

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

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

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.

Report an Error

Example Question #68 : Matrices

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

Possible Answers:

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

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

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

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

$AA^{T}$ 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.

Report an Error

Example Question #69 : Matrix Matrix Product

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

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

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

Which of the following is equal to ABC ?

Possible Answers:

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

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

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

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

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

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.

Report an Error

Example Question #70 : Matrix Matrix Product

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

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

Possible Answers:

None of the other responses gives a correct answer.

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

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

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

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

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.

Report an Error

View Linear Algebra Tutors

Chelsea
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Physics. The University of Texas at Dallas, Master of Science, Bioinf...

View Linear Algebra Tutors

Nima
Certified Tutor

Sahand University of Technology, Master of Science, Electrical Engineering. Institut national de la recherche scientifique, D...

View Linear Algebra Tutors

Liban
Certified Tutor

Emory University, Bachelor of Science, Mathematics/Economics.

All Linear Algebra Resources

4 Diagnostic Tests 108 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Reading Tutors in Boston, Statistics Tutors in San Francisco-Bay Area, Calculus Tutors in Dallas Fort Worth, French Tutors in Dallas Fort Worth, Calculus Tutors in Boston, LSAT Tutors in Los Angeles, GMAT Tutors in Atlanta, French Tutors in Phoenix, SAT Tutors in Denver, SSAT Tutors in Los Angeles

Popular Courses & Classes

GRE Courses & Classes in Atlanta, ISEE Courses & Classes in New York City, LSAT Courses & Classes in Phoenix, GMAT Courses & Classes in Atlanta, LSAT Courses & Classes in Atlanta, Spanish Courses & Classes in Chicago, ACT Courses & Classes in New York City, ACT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in New York City, LSAT Courses & Classes in Philadelphia

Popular Test Prep

GMAT Test Prep in Washington DC, GRE Test Prep in San Francisco-Bay Area, GRE Test Prep in Atlanta, SAT Test Prep in Miami, SAT Test Prep in Los Angeles, GRE Test Prep in Dallas Fort Worth, ISEE Test Prep in Atlanta, ISEE Test Prep in Chicago, ACT Test Prep in New York City, MCAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #61 : Matrices

Example Question #62 : Matrices

Example Question #63 : Matrices

Example Question #64 : Matrices

Example Question #65 : Matrices

Example Question #66 : Matrices

Example Question #67 : Matrices

Example Question #68 : Matrices

Example Question #69 : Matrix Matrix Product

Example Question #70 : Matrix Matrix Product

All Linear Algebra Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: