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Linear Algebra : Linear Algebra

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Example Questions

Example Question #84 : Matrix Matrix Product

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

Possible Answers:

False

True

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 : Matrix Matrix Product

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

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 #86 : Matrix Matrix Product

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:

Independent

Libertarian

Republican

Democrat

Correct answer:

Libertarian

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 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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Example Question #91 : Matrix Matrix Product

is a $4 \times 3$ matrix.

is a $5 \times 6$ matrix.

is a $3 \times 8$ matrix.

is a $6 \times 4$ matrix.

is an $8 \times 5$ matrix.

Which of the following is a square matrix?

Possible Answers:

EBDAC

$A^{T} D^{T}B^{T}E^{T}C^{T}$

$B^{T}E^{T}C^{T}A^{T}D^{T}$

All four of the expressions given in the other choices are defined.

ACEBD

Correct answer:

All four of the expressions given in the other choices are defined.

Explanation:

Two matrices can be multiplied if and only if the number of columns in the first is equal to the number of columns in the second. The number of rows in the first matrix and the number of columns in the second are the number of rows and columns in the product, respectively.

For example, the following matrices can be put together in order:

is a $4 \times 3$ matrix; is a $3 \times 8$ matrix; is defined and is a $4 \times 8$ matrix.

is a $3 \times 8$ matrix; is an $8 \times 5$ matrix; is defined and is a $3 \times 5$ matrix.

By extension, matrices can be linked in a product of three or more such that, if two matrices appear together, this same relation must hold. For example, since is a $4 \times 3$ matrix, is a $3 \times 8$ matrix, and is an $8 \times 5$ matrix, ACE is defined and is a $4 \times 5$ matrix.

It follows by further extension that ACEBD is defined and is a $4 \times 4$ matrix, and that EBDAC is defined and is an $8 \times 8$ matrix.

DACEB is defined to be a $6 \times 6$ matrix. The transpose of the product of matrices is equal to the product of transposes in reverse, so

$B^{T}E^{T}C^{T}A^{T}D^{T} = (DACEB)^{T}$ , a $6 \times 6$ matrix.

Similarly, $A^{T} D^{T}B^{T}E^{T}C^{T} = (CEBDA)^{T}$ , a $3 \times 3$ matrix.

The correct response is that all four given matrices are square.

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Example Question #91 : Matrices

is a singular $4 \times 4$ matrix.

is a nonsingular $5 \times 5$ matrix.

is a $4 \times 5$ matrix.

All of the following are undefined except:

Possible Answers:

$B^{-1}C^{T}A$

$ABC^{T}$

$BC^{-1}A$

$A^{-1} C B$

None of the expressions given in the other choices are defined.

Correct answer:

$B^{-1}C^{T}A$

Explanation:

$BC^{-1}A$ can be eliminated as a choice; is not a square matrix, so the inverse of , $C^{-1}$ , does not exist.

$A^{-1} C B$ can also be eliminated; is a singular matrix, so, by definition, $A^{-1}$ does not exist.

Two matrices can be multiplied if and only if the number of columns in the first is equal to the number of columns in the second. $ABC^{T}$ can be eliminated, since has four columns and has five rows.

The remaining choice is $B^{-1}C^{T}A$ . is nonsingular, so $B^{-1}$ is defined. $B^{-1}$ , like , is a $5 \times 5$ matrix; is a $4 \times 5$ matrix, so its transpose $C^{T}$ is $5 \times 4$ ; thus, $B^{-1}C^{T}$ is $5 \times 4$ . is $4 \times 4$ , so $B^{-1}C^{T}A = (B^{-1}C^{T})A$ is $5 \times 4$ . $B^{-1}C^{T}A$ is defined and is the correct choice.

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Example Question #91 : Matrix Matrix Product

Let $A = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \end{bmatrix}$ and $B = \begin{bmatrix} \sqrt{2} & \sqrt{6} \\-2 \sqrt{6} & \sqrt{3} \end{bmatrix}$ .

Find .

Possible Answers:

is undefined.

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

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

$\begin{bmatrix} 7 & 6 \sqrt{2}+ 4\sqrt{3} \end{bmatrix}$

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

Correct answer:

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

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

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,

$A B = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \end{bmatrix} \begin{bmatrix} \sqrt{2} & \sqrt{6} \\-2 \sqrt{6} & \sqrt{3} \end{bmatrix}$

$= \begin{bmatrix} 2\sqrt{2} (\sqrt{2} )+ \sqrt{3}(-2 \sqrt{6} ) & 2\sqrt{2} (\sqrt{6} )+ \sqrt{3}( \sqrt{3} )\end{bmatrix}$

$= \begin{bmatrix} 4-6 \sqrt{2} & 3+ 4 \sqrt{3}\end{bmatrix}$

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Example Question #771 : Linear Algebra

Let $A = \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix}$ and $B = \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \\ \\ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ .

Find .

Possible Answers:

$\begin{bmatrix} \frac{63}{80} \\ \\ \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{53}{80} & \frac{1}{4} \end{bmatrix}$

is undefined.

$\begin{bmatrix} \frac{53}{80} \\ \\ \frac{1}{4} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

Explanation:

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

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,

$AB= \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix} \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \\ \\ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ .

$= \begin{bmatrix} \frac{2}{5}\left ( \frac{3}{2} \right ) + \frac{1}{4} \left ( \frac{3}{4} \right )& \frac{2}{5}\left ( \frac{1}{4} \right ) + \frac{1}{4} \left ( -\frac{1}{5} \right ) \end{bmatrix}$

$= \begin{bmatrix} \frac{3}{5} + \frac{3}{16} & \frac{1}{10} - \frac{1}{20} \end{bmatrix}$

$= \begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

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Example Questions

Example Question #84 : Matrix Matrix Product

Example Question #85 : Matrix Matrix Product

Example Question #86 : Matrices

Example Question #88 : Matrix Matrix Product

Example Question #86 : Matrix Matrix Product

Example Question #88 : Matrices

Example Question #91 : Matrix Matrix Product

Example Question #91 : Matrices

Example Question #91 : Matrix Matrix Product

Example Question #771 : Linear Algebra

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