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Linear Algebra : Matrix-Matrix Product

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Example Question #101 : Matrix Matrix Product

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

Calculate $(AA^{T})^{99}$ .

Possible Answers:

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

$(AA^{T})^{99}$ is not defined.

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

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

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

Correct answer:

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

Explanation:

$A^{T}$ , the transpose of , can be found by transposing rows with columns.

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

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

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

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

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

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

This is a square matrix, so it can be raised to a power. To raise a diagonal matrix to a power, simply raise each number in the main diagonal to that power:

$(AA^{T})^{99} = \begin{bmatrix} (-1)^{99} & 0 \\ 0 & (-1)^{99} \end{bmatrix}= \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$ .

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Example Question #102 : Matrix Matrix Product

$A=\begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}$ and $C = \begin{bmatrix} c & -1 \\ 1 & d \end{bmatrix}$ , where all four variables stand for real quantities.

Which must be true of and regardless of the values of the variables?

Possible Answers:

$AC= -(CA)^{T}$

AC= CA

None of the statements given in the other choices are correct.

AC=- (CA)

$AC= (CA)^{T}$

Correct answer:

None of the statements given in the other choices are correct.

Explanation:

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AC=\begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}\begin{bmatrix} c & -1 \\ 1 & d \end{bmatrix}$

$=\begin{bmatrix} a (c)+ 1(1)& a (-1)+ 1(d)\\ -1 (c)+ b(1)& -1 (-1)+ b(d)\end{bmatrix}$

$=\begin{bmatrix} ac+1&-a+d \\ b-c& bd+1\end{bmatrix}$

$CA=\begin{bmatrix} c & -1 \\ 1 & d \end{bmatrix}\begin{bmatrix} a & 1 \\ -1 & b \end{bmatrix}$

$=\begin{bmatrix} c (a)+ (-1)(-1)& c(1)+(-1)(b)\\ 1 (a)+ d(-1)& 1(1)+d(b)\end{bmatrix}$

$=\begin{bmatrix} ac+1& -b+c\\ a-d& bd+1\end{bmatrix}$

AC= CA is false in general.

$(CA)^{T}$ , the transpose of , is the result of transposing rows and columns:

$(CA)^{T} =\begin{bmatrix} ac+1& a-d \\ -b+c & bd+1\end{bmatrix}$ . so

$AC= (CA)^{T}$ is false in general.

$-(CA) =\begin{bmatrix} -ac-1& b-c\\ -a-d&- bd-1\end{bmatrix}$ , so

AC=- (CA) is false in general.

$-(CA)^{T} =\begin{bmatrix} -ac-1&-a-d \\ b-c &- bd-1\end{bmatrix}$ , so

$AC= -(CA)^{T}$ is false in general.

Thus, none of the four given statements need be true.

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Example Question #103 : Matrix Matrix Product

Let $A = \begin{bmatrix} \cos \frac{\pi }{5} \\ \sin \frac{\pi }{5} \end{bmatrix}$

Find $(AA^{T})^{2}$ .

Possible Answers:

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} - \sin^{2} \frac{\pi }{5}\end{bmatrix}$

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} & 1 \\ 1& \sin^{2} \frac{\pi }{5}\end{bmatrix}$

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

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} & \sin \frac{\pi }{5}\cos \frac{\pi }{5} \\ \sin \frac{\pi }{5}\cos \frac{\pi }{5} & \sin^{2} \frac{\pi }{5}\end{bmatrix}$

$(AA^{T})^{2}$ is undefined.

Correct answer:

$\begin{bmatrix} \cos ^{2}\frac{\pi }{5} & \sin \frac{\pi }{5}\cos \frac{\pi }{5} \\ \sin \frac{\pi }{5}\cos \frac{\pi }{5} & \sin^{2} \frac{\pi }{5}\end{bmatrix}$

Explanation:

is equal to the two-entry column matrix $\begin{bmatrix} \cos \frac{\pi }{5} \\ \sin \frac{\pi }{5} \end{bmatrix}$ , so $A^{T}$ , the transpose, is the row matrix

$A^{T} = \begin{bmatrix} \cos \frac{\pi }{5} & \sin \frac{\pi }{5} \end{bmatrix}$

The product of two matrices is calculated by multiplying rows by columns - adding the corresponding entries in the rows of the first matrix by the columns of the second - so

$AA^{T}= \begin{bmatrix} \cos \frac{\pi }{5} \\ \sin \frac{\pi }{5} \end{bmatrix}\begin{bmatrix} \cos \frac{\pi }{5} & \sin \frac{\pi }{5} \end{bmatrix}$

$= \begin{bmatrix} \cos \frac{\pi }{5}(\cos \frac{\pi }{5}) & \sin \frac{\pi }{5}(\cos \frac{\pi }{5}) \\ \cos \frac{\pi }{5}(\sin \frac{\pi }{5}) & \sin \frac{\pi }{5}(\sin \frac{\pi }{5}) \end{bmatrix}$

$= \begin{bmatrix} \cos ^{2}\frac{\pi }{5} & \sin \frac{\pi }{5}\cos \frac{\pi }{5} \\ \sin \frac{\pi }{5}\cos \frac{\pi }{5} & \sin^{2} \frac{\pi }{5}\end{bmatrix}$

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Linear Algebra : Matrix-Matrix Product

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #101 : Matrix Matrix Product

Example Question #102 : Matrix Matrix Product

Example Question #103 : Matrix Matrix Product

All Linear Algebra Resources

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