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Linear Algebra : Matrix-Matrix Product

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Example Questions

Example Question #81 : Matrices

is an idempotent matrix.

True or false: It follows that $A^{T}$ is an idempotent matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an idempotent matrix if $A^{2}= A$ .

For any A , B for which is defined, $(AB)^{T} = B^{T}A^{T}$ . Setting B = A , if is idempotent, this becomes

$(AA)^{T} = A^{T}A^{T}$

$(A^{2})^{T} =( A^{T} )^{2}$

$A^{T} =( A^{T} )^{2}$

It follows that $A^{T}$ is idempotent.

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Example Question #82 : Matrices

is an involutory matrix.

True or false: It follows that $A^{T}$ is an involutory matrix.

Possible Answers:

False

True

Correct answer:

True

Explanation:

is an involutory matrix if $A^{2}= I$ .

For any A , B for which is defined, $(AB)^{T} = B^{T}A^{T}$ . Setting B = A , if is involutory, this becomes

$(AA)^{T} = A^{T}A^{T}$

$(A^{2})^{T} = (A^{T}) ^{2}$

$I^{T} = (A^{T}) ^{2}$

$I = (A^{T}) ^{2}$

Thus, $A^{T}$ is involutory, making the statement true.

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Example Question #83 : Matrices

is a nilpotent matrix.

True or false: It follows that $A^{T}$ is a nilpotent matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

A matrix is nilpotent if there exists $k \in \mathbb{N}$ so that $A^{k} = O$ , the zero matrix.

Suppose is nilpotent.

Case 1: If A = O , which is trivially nilpotent, then $A^{T} = O$ , which is trivially nilpotent.

Case 2: Suppose $A^{2} = O$ . For any A , B for which is defined, $(AB)^{T} = B^{T}A^{T}$ . Setting B = A , if is nilpotent, this becomes

$(AA)^{T} = A^{T}A^{T}$

$O ^{T} =( A^{T} )^{2}$

$O =( A^{T} )^{2}$

It follows that $A^{T}$ is nilpotent. This reasoning can easily be extended to the case k >2 , since $(A_{1}A_{2}...A_{k})^{T} = A_{n}^{T}...A_{2}^{T}A_{1}^{T}$ .

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Example Question #84 : Matrices

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

A five year study of political affiliation reveals that most voters stay loyal to their political party. However, some changes of affiliation are noted over the course of a typical year, as follows:

3% of Republicans become Democrats; 5% of Republicans become independents; 1% of Republicans become Libertarians.

4% of Democrats become Republicans; 3% of Democrats become independents; 1% of Democrats become Libertarians.

17% of Independents become Republicans; 19% of independents become Democrats; 8% become Libertarians.

1% of Libertarians become Republicans; 1% of Libertarians become Democrats; 1% of Libertarians become independents.

Which group of voters has the same affiliation after the end of a forty-year period with greatest probability: Democrats Republicans, Independents, or Libertarians?

Possible Answers:

Libertarians

Republicans

Democrats

Independents

Correct answer:

Libertarians

Explanation:

Form a stochastic (probability) matrix for this system, where the columns represent current political affiliation (Republicans, Democrats, independents, and Libertarians, in that order), and the rows represent affiliation after one year (same order). This matrix

$P=\begin{bmatrix} 0.91 & 0.04 & 0.17 &0.01 \\ 0.03 & 0.92 & 0.19 &0.01 \\ 0.05 & 0.03 &0.56&0.01 \\ 0.01 & 0.01 & 0.08 & 0.97 \end{bmatrix}$

Raise this matrix to the power of 40 to obtain the stochastic matrix with the probabilities that political affiliations will change over 40 years. Observe the entries in the main diagonal, which represent voters who have the same party affiliation at the beginning and the end of the forty-year period:

$P^{40} = \begin{bmatrix}\textbf{0.3147} & 0.3120 & 0.3014 & 0.2395 \\ 0.3199 & \textbf{0.3290} & 0.3123 & 0.2478\\ 0.0648 & 0.0650 & \textbf{0.0631} & 0.0538\\ 0.3006 & 0.2940 & 0.3232 & \textbf{{\color{Red} 0.4589}} \end{bmatrix}$

The greatest number appears in the fourth entry - the voters who were Libertarians at the beginning and end of the forty-year period. The correct choice is Libertarians.

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Example Question #84 : Matrices

True or false: $A^ {3}= I$ if and only if A= I .

Possible Answers:

True

False

Correct answer:

False

Explanation:

One direction of the biconditional holds: if A = I , then $A^{3} = I$ . But the other does not, as proved by counterexample.

We give an example of a nonidentity matrix whose cube is equal to the identity. Let

$A = \begin{bmatrix} \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} & 0 \\ \\ 0 & \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \end{bmatrix}$

This matrix is diagonal having its only nonzero entries on its main diagonal; its cube can be taken by cubing each diagonal element:

$A^{3} = \begin{bmatrix}\left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right ) ^{3}& 0 \\ \\ 0 &\left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right ) ^{3} \end{bmatrix}$

By DeMoivre's Theorem,

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

setting $\theta = \frac{2 \pi}{3} , n = 3$ ,

$\left ( \cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3} \right ) ^{3}$

$= \cos \left ( 3 \cdot \frac{2 \pi}{3} \right )+ i \sin \left ( 3 \cdot \frac{2 \pi}{3} \right )$

$= \cos 2 \pi + i \sin 2 \pi$

and

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

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Example Question #85 : Matrices

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

Calculate $A ^{101}$ .

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

This task isn't as difficult as it seems if we search for a pattern.

First, find the product $A^{2}$ . This can be found by multiplying rows of by columns of - adding the products of corresponding entries, as follows:

$A^{2} = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

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

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

Thus,

$A ^{101} = A^{100}A =( A^{2 })^{50} A = I^{50}A = IA = A$

The correct response is $\begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$ .

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Example Question #86 : Matrices

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

Calculate $A^{8}$ .

Possible Answers:

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

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

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

$\begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

$\begin{bmatrix} 256& 0 & 0 & 256\\ 0 & 256& 256& 0 \\ 0 & 256& 256& 0 \\ 256& 0 & 0 & 256\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

Explanation:

The question is actually much easier than it looks if you note that

$A^{8} =\left [ (A^{2} )^{2} \right ]^{2}$ .

Calculate $A^{2}$ by multiplying rows by columns, adding the products of corresponding entries:

$A^{2}= \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix}$

$A^{2}= \begin{bmatrix} 1(1)+0(0)+0(0) + 1(1)& 1(0)+0(1)+0(1) + 1(0) & 1(0)+0(1)+0(1) + 1(0) & 1(1)+0(0)+0(0) + 1(1) \\ 0(1) + 1(0)+1(0)+0(1) & 0(0) + 1(1)+1(1)+0(0) & 0(0) + 1(1)+1(1)+0(0) & 0(1) + 1(0)+1(0)+0(1) \\ 0(1) + 1(0)+1(0)+0(1) & 0(0) + 1(1)+1(1)+0(0) & 0(0) + 1(1)+1(1)+0(0) & 0(1) + 1(0)+1(0)+0(1)\\ 1(1)+0(0)+0(0) + 1(1) & 1(0)+0(1)+0(1) + 1(0) & 1(0)+0(1)+0(1) + 1(0) & 1(1)+0(0)+0(0) + 1(1) \end{bmatrix}$

$A^{2}= \begin{bmatrix} 2 & 0 & 0 &2 \\ 0 & 2&2 & 0 \\ 0 & 2 & 2 & 0 \\2 & 0 & 0 &2 \end{bmatrix}$

= 2A

$A^{8} =\left [ (A^{2} )^{2} \right ]^{2}$

From which we get

$A^{8} = \left [ (2A )^{2} \right ]^{2}$

$=(4A^{2})^{2}$

$=[ 4 (2A) ] ^{2}$

$=(8A)^{2}$

$= 64A^{2}$

= 128A

$= \begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

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Example Question #88 : Matrix Matrix Product

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

Two married couples - Mr. and Mrs. Bernoulli and Mr. and Mrs. Pascal - are playing a game with a pair of fair six-sided dice. They call the game 'King and Queen" and agree to the following rules:

Mr. Bernoulli starts out as "King".

Whoever is "King" or "Queen" rolls the dice. If doubles are rolled, the "King"/"Queen" stays as such. If a 7 is rolled, the new "King"/"Queen" is the spouse of the old "King"/"Queen". If a 3, 4, 5, or 6, other than doubles, is rolled, the new "King"/"Queen" is the same-sex member of the other couple. If an 8, 9, 10, or 11, other than doubles, is rolled, the new "King"/"Queen" is the opposite-sex member of the other couple.

Play is to end after eight rolls. Who is the most likely to end up the "King" or "Queen"?

Possible Answers:

Mr. Pascal

Mr. Bernoulli

Mrs. Pascal

Mrs. Bernoulli

The probabilities are virtually equal.

Correct answer:

The probabilities are virtually equal.

Explanation:

If two fair six-sided dice are rolled, the probability of rolling doubles is $\frac{1}{6}$ , making this the probability that the current "King" or 'Queen" remain as such. The probability of rolling a 7 is also $\frac{1}{6}$ , making this the probability that the new "King" or "Queen" will be the spouse of the current one. The probability that a 3, 4, 5, or 6, other than doubles, will be rolled is $\frac{1}{3}$ , making the probability that the new "King" or "Queen" will be the same-sex member of the other couple. The probability that a 8, 9, 10, or 11, other than doubles, will be rolled is $\frac{1}{3}$ , making the probability that the new "King" or "Queen" will be the opposite-sex member of the other couple. These probabilities do not change for any player.

Construct a stochastic (probability) matrix, where columns represent the initial "King" or "Queen", and rows represent the "King" or "Queen" after the first turn. let the columns/rows represent Mr. Bernoulli, Mrs. Bernoulli, Mr. Pascal, and Mrs. Pascal, in that order:

$P= \begin{bmatrix} \frac{1}{6} & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} \\ \\ \frac{1}{6} & \frac{1}{6} & \frac{1}{3} & \frac{1}{3} \\ \\ \frac{1}{3} & \frac{1}{3} &\frac{1}{6} & \frac{1}{6} \\ \\ \frac{1}{3} & \frac{1}{3} &\frac{1}{6} & \frac{1}{6} \end{bmatrix}$

The stochastic matrix representing the probabilities after eight turns will be the eighth power of this. If we round to the nearest four decimal digits, the matrix is very close to:

$P^{8} = \begin{bmatrix} 0.2500 & 0.2500 & 0.2500 & 0.2500 \\ 0.2500 & 0.2500 & 0.2500 & 0.2500 \\ 0.2500 & 0.2500 & 0.2500 & 0.2500 \\ 0.2500 & 0.2500 & 0.2500 & 0.2500 \end{bmatrix}$

This makes each person's chances of ending up "King" or "Queen" virtually equiprobable no matter who actually starts that way.

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Example Question #87 : Matrices

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

A five year study of political affiliation reveals that most voters stay loyal to their political party. However, some changes of affiliation are noted over the course of a typical year, as follows:

3% of Republicans become Democrats; 5% of Republicans become independents; 1% of Republicans become Libertarians.

4% of Democrats become Republicans; 3% of Democrats become independents; 1% of Democrats become Libertarians.

17% of Independents become Republicans; 19% of independents become Democrats; 8% become Libertarians.

1% of Libertarians become Republicans; 1% of Libertarians become Democrats; 1% of Libertarians become independents.

Suppose a voter starts out as an independent. In forty years, which is he most likely to be - a Democrat, a Republican, an independent, or a Libertarian?

Possible Answers:

Libertarian

Democrat

Independent

Republican

Correct answer:

Libertarian

Explanation:

$P=\begin{bmatrix} 0.91 & 0.04 & 0.17 &0.01 \\ 0.03 & 0.92 & 0.19 &0.01 \\ 0.05 & 0.03 &0.56&0.01 \\ 0.01 & 0.01 & 0.08 & 0.97 \end{bmatrix}$

Raise this matrix to the power of 40 to obtain the stochastic matrix with the probabilities that political affiliations will change over 40 years. Observe the probabilities in the third (independent) column:

$P^{40} = \begin{bmatrix}0.3147 & 0.3120 & \textbf{0.3014} & 0.2395 \\ 0.3199 & 0.3290 & \textbf{0.3123} & 0.2478\\ 0.0648 & 0.0650 & \textbf{0.0631} & 0.0538\\ 0.3006 & 0.2940 & \textbf{{\color{Red} 0.3232}} & 0.4589 \end{bmatrix}$

The greatest value appears in the fourth - Libertarian - row. Libertarian is the correct choice.

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Example Question #88 : Matrices

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

Calculate $A^{8}$ .

Possible Answers:

$\begin{bmatrix} -128 & 0 & 0 & 128 \\ 0 & -128 & 128 & 0 \\ 0 & -128 & 128 & 0 \\ -128 & 0 & 0 & 128 \end{bmatrix}$

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

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

$\begin{bmatrix} 128 & 0 & 0 & 128 \\ 0 & 128 & 128 & 0 \\ 0 & 128 & 128 & 0 \\ 128 & 0 & 0 & 128 \end{bmatrix}$

$\begin{bmatrix} 128 & 0 & 0 & -128 \\ 0 & 128 & -128 & 0 \\ 0 & 128 & -128 & 0 \\ 128 & 0 & 0 & -128 \end{bmatrix}$

Correct answer:

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

Explanation:

The question is actually much easier than it looks if you note that

$A^{8} =\left [ (A^{2} )^{2} \right ]^{2}$ .

Calculate $A^{2}$ by multiplying rows by columns, adding the products of corresponding entries:

$A^{2}= \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & -1 \end{bmatrix}$ $= \begin{bmatrix} 1(1)+ 0(0) + 0(0)+ (-1)(1) & 1(0)+ 0(1) + 0(1)+ (-1)(0) &1(0)+ 0(-1) + 0(-1)+ (-1)(0) & 1(-1)+ 0(0) + 0(0)+ (-1)(-1) \\ 0(1)+ 1(0) + (-1)(0)+0(1) & 0(0)+ 1(1) + (-1)(1)+ 0(0) & 0(0)+ 1(-1) + (-1)(-1)+ 0(0)& 0(-1)+ 1(0) + (-1)(0)+0(-1) \\ 0(1)+ 1(0) + (-1)(0)+0(1) & 0(0)+ 1(1) + (-1)(1)+ 0(0) & 0(0)+ 1(-1) + (-1)(-1)+ 0(0)& 0(-1)+ 1(0) + (-1)(0)+0(-1) \\ 1(1)+ 0(0) + 0(0)+ (-1)(1) & 1(0)+ 0(1) + 0(1)+ (-1)(0) &1(0)+ 0(-1) + 0(-1)+ (-1)(0) & 1(-1)+ 0(0) + 0(0)+ (-1)(-1) \end{bmatrix}$

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

the zero matrix.

is nilpotent with index 2, so any higher power of is also equal to the zero matrix.

$A^{8} =\begin{bmatrix} 0 & 0& 0& 0 \\ 0 & 0& 0 & 0 \\ 0 & 0& 0 &0 \\ 0 & 0& 0 & 0 \end{bmatrix}$ .

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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 #81 : Matrices

Example Question #82 : Matrices

Example Question #83 : Matrices

Example Question #84 : Matrices

Example Question #84 : Matrices

Example Question #85 : Matrices

Example Question #86 : Matrices

Example Question #88 : Matrix Matrix Product

Example Question #87 : Matrices

Example Question #88 : Matrices

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