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Linear Algebra : The Transpose

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Example Questions

Example Question #51 : Operations And Properties

Let , , and be real numbers such that

$A = \begin{bmatrix} a& b& c\\ 1& 2& 3\\ -4&-5 &-6 \end{bmatrix}$ , $B = \begin{bmatrix} a&1 &-4 \\ b& 2& -5\\ c& 3& -6 \end{bmatrix}$ ,

and the determinant of is 8.

True or false: The determinant of is 8.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is the transpose of - the matrix formed by interchanging the rows of with its columns. The determinant of a matrix and that of its transpose are equal, so, since has determinant 8, so does .

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Example Question #131 : Linear Algebra

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

Find $A^{*}$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

$A^{*}$ , the conjugate transpose of , is the result of transposing the matrix - interchanging rows with columns - and changing each entry to its complex conjugate. First, find transpose $A^{T}$ :

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

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

Change each entry to its complex conjugate:

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

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Example Question #132 : Linear Algebra

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

Find $A^{T}$ .

Possible Answers:

None of the other choices gives the correct response.

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #31 : The Transpose

and are skew-symmetric matrices.

Which of the following is true of $A^{T} B^{T}$ ?

Possible Answers:

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

$A^{T} B^{T} =- AB$

Correct answer:

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

Explanation:

By definition, the transpose $A^{T}$ of a skew-symmetric matrix is equal to its additive inverse . It follows that

$A^{T} B^{T} =( -A)(-B) =( -1) A (-1)B = (-1)(-1)AB = AB$

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Example Question #134 : Linear Algebra

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

Find $A^{*}$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

$A^{*}$ , the conjugate transpose of , is the result of transposing the matrix - interchanging rows with columns - and changing each entry to its complex conjugate. First, find transpose $A^{T}$ :

$A = \begin{bmatrix} 1 & 1 + i &0 \\ 1 & 1 - i & 0 \\ 0 & 1 & i \end{bmatrix}$ ,

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

Change each entry to its complex conjugate:

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

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Example Question #56 : Operations And Properties

$A = \begin{bmatrix} -2 & 1 & 3 \\ 5 & 4 &- 1 \\ 4 & -2 & 1 \end{bmatrix}$

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

Possible Answers:

$\begin{bmatrix} -2 & 5& 4 \\ 1 & 4 &- 2 \\ 3& -1&1 \end{bmatrix}$

$\begin{bmatrix} -2i & 5i & 4i \\ 1i & 4i &- 2i \\ 3i & -i & i \end{bmatrix}$

None of the other responses gives the correct answer.

$\begin{bmatrix} -2 & 5i & 4i \\ 1i & 4 &- 2i \\ 3i & -i & 1 \end{bmatrix}$

$\begin{bmatrix} -2i & 5& 4 \\ 1 & 4i &- 2 \\ 3& -i &1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -2 & 5& 4 \\ 1 & 4 &- 2 \\ 3& -1&1 \end{bmatrix}$

Explanation:

$A^{*}$ , the conjugate transpose of , is the result of transposing the matrix - interchanging rows with columns - and changing each entry to its complex conjugate. First, find transpose $A^{T}$ :

$A = \begin{bmatrix} -2 & 1 & 3 \\ 5 & 4 &- 1 \\ 4 & -2 & 1 \end{bmatrix}$

$A^{T}= \begin{bmatrix} -2 & 5& 4 \\ 1 & 4 &- 2 \\ 3& -1&1 \end{bmatrix}$

Each entry of $A^{*}$ is equal to the complex conjugate of the corresponding entry of $A^{T}$ . However, each entry in $A^{T}$ is real, so each entry is equal to its own complex conjugate, and

$A ^{*} = A^{T}= \begin{bmatrix} -2 & 5& 4 \\ 1 & 4 &- 2 \\ 3& -1&1 \end{bmatrix}$

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Example Question #135 : Linear Algebra

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

Find $A^{T}$ .

Possible Answers:

None of the other choices gives the correct response.

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

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

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

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

Correct answer:

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

Explanation:

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

$A = \begin{bmatrix} 1 & 1 + i &0 \\ 1 & 1 - i & 0 \\ 0 & 1 & i \end{bmatrix}$ ,

$A^{T}= \begin{bmatrix} 1 & 1 &0 \\ 1+ i & 1 - i & 1 \\ 0 & 0 & i \end{bmatrix}$ .

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Example Question #31 : The Transpose

$A^{*} = \begin{bmatrix} 3 + i & 5 - 2i \\3 + 2i & 3i \end{bmatrix}$

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

Possible Answers:

$A^{T} = \begin{bmatrix} 3 + i &3 - 2i \\ 5 + 2i & 3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 - i & 5 + 2i \\3 - 2i & 3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 - i & 5 + 2i \\3 - 2i & -3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 + i & 5 + 2i \\3 - 2i & 3i \end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 - i &3 - 2i \\ 5 + 2i & -3i \end{bmatrix}$

Correct answer:

$A^{T} = \begin{bmatrix} 3 - i & 5 + 2i \\3 - 2i & -3i \end{bmatrix}$

Explanation:

$A^{T}$ is the transpose of - the result of interchanging the rows of with its columns. $A^{*}$ is the conjugate transpose of - the result of changing each entry of $A^{T}$ to its complex conjugate. Therefore, if

$A^{*} = \begin{bmatrix} 3 + i & 5 - 2i \\3 + 2i & 3i \end{bmatrix}$ ,

we can find $A^{T}$ by simply changing each entry in $A^{*}$ to its complex conjugate:

$A^{T} = \begin{bmatrix} \overline{3 + i }& \overline{5 - 2i} \\\overline{3 + 2i} & \overline{3i }\end{bmatrix}$

$A^{T} = \begin{bmatrix} 3 - i & 5 + 2i \\3 - 2i & -3i \end{bmatrix}$

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Example Question #61 : Operations And Properties

$A^{*} = \begin{bmatrix} 3+ 5i & 0 & 0& 0 \\ 2 - 5i & 3+ 6i & 0 & 0 \\1 + 6i &2 - 8 i & 4 + 8 i & 0 \\ 1-2i & 3 - 7i & 2 - 7i & 3- 5i \end{bmatrix}$

True or false: is an upper triangular matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

$A^{*}$ is the result of interchanging rows of with columns, then changing each entry to its complex conjugate. Also, is equal to $(A^{*} )^{*} = A$ , so perform the same process on $A^{*}$ :

$A^{*} = \begin{bmatrix} 3+ 5i & 0 & 0& 0 \\ 2 - 5i & 3+ 6i & 0 & 0 \\1 + 6i &2 - 8 i & 4 + 8 i & 0 \\ 1-2i & 3 - 7i & 2 - 7i & 3- 5i \end{bmatrix}$

$(A^{*})^{T} = \begin{bmatrix} 3+ 5i & 2 - 5i & 1 + 6i & 1-2i \\ 0& 3+ 6i & 2 - 8 i & 3 - 7i \\ 0& 0& 4 + 8 i & 2 - 7i \\ 0& 0 & 0 & 3- 5i \end{bmatrix}$

$A= (A^{*})^{*} = \begin{bmatrix} 3- 5i & 2 + 5i & 1 - 6i & 1+2i \\ {\color{Red} \textbf{0} }& 3-6i & 2 + 8 i & 3+ 7i \\ {\color{Red} \textbf{0} } & {\color{Red} \textbf{0} } & 4 - 8 i & 2 +7i \\ {\color{Red} \textbf{0} }& {\color{Red} \textbf{0} } & {\color{Red} \textbf{0} } & 3+ 5i \end{bmatrix}$

A matrix is upper triangular if all elements below its main (upper-left corner to lower right corner) diagonal are equal to 0. These elements in are displayed in red above. Since all of the lower-triangular elements of are zeroes, is upper triangular.

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Example Question #32 : The Transpose

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

Determine $A+A^{T}$ .

Possible Answers:

$\begin{bmatrix} -6 & 3 & 4 \\ 3 & -2& 3 \\ 4 & 3 & 0 \end{bmatrix}$

$\begin{bmatrix} -6 & 3 &0 \\ 3 & -2& 0 \\ 0 & 0 & 0 \end{bmatrix}$

$A+A^{T}$ is undefined.

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

Correct answer:

$A+A^{T}$ is undefined.

Explanation:

is a two-by-three matrix. It follows that its transpose, $A^{T}$ , the result of switching rows with columns, is a three-by-two matrix. Since and $A^{T}$ have different dimensions, $A+A^{T}$ is an undefined expression.

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Linear Algebra : The Transpose

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #51 : Operations And Properties

Example Question #131 : Linear Algebra

Example Question #132 : Linear Algebra

Example Question #31 : The Transpose

Example Question #134 : Linear Algebra

Example Question #56 : Operations And Properties

Example Question #135 : Linear Algebra

Example Question #31 : The Transpose

Example Question #61 : Operations And Properties

Example Question #32 : The Transpose

All Linear Algebra Resources

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