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Linear Algebra : Linear Algebra

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Example Questions

Example Question #24 : Matrices

Give the values of , , and so that the matrix

$\begin{bmatrix} 0.4& 0.5&0.6 \\ 0.8& 0.4 & 0.1 \\ a & b& c \end{bmatrix}$

is an example of a stochastic matrix (Markov chain).

Possible Answers:

a= - 0.2, b= 0.1, c = 0.3

a= 0.4, b= 0.1, c = 0.5

a= 1.2, b= 0.9, c = 0.7

a= 0.2, b= 0.1, c = 0.3

The matrix cannot be made into a stochastic matrix regardless of the values of the variables.

Correct answer:

The matrix cannot be made into a stochastic matrix regardless of the values of the variables.

Explanation:

A stochastic matrix is a matrix of probabilities in which the entry $p_{m,n}$ is the probability that, given the fact that a given system is in a state , the system will be in state next. As such, the elements in each column of , being the probabilities that the system will change from a given state to each other state, respectively, must add up to 1.

Therefore,

a + 0.4 + 0.8 = 1

a+ 1.2 = 1

a = -0.2

However, since a stochastic matrix is, by definition, a matrix of probabilities, each element of the matrix must fall in the interval [0,1] . This rule has been violated, so no values of the variables can make the matrix a stochastic matrix.

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Example Question #25 : Matrices

Markov 1

Mary and Marty are playing a game whose board comprises two spaces, marked "X' and "Y", as shown above. For each turn, one player rolls a fair six-sided die, then moves to the space indicated based on where the player is already and what the player rolls. For example, if the player is on space "X" and rolls a "2", he moves to space "Y", since one of the numbers on the arrow going from "X" to 'Y" is "2"; similarly, if he is on space "X" and rolls a "6", he will stay where he is.

Give the stochastic matrix which represents the probabilities that, given that a player is on a particular space, he or she will be on each given space two turns later.

Possible Answers:

$\begin{bmatrix} \frac{4}{9} & \frac{1}{6} \\ \\ \frac{5}{9} &\frac{5}{6} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{3}& \frac{1}{3} \\ \\ \frac{2}{3} &\frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} \frac{4}{9} &\frac{4}{9} \\ \\ \frac{5}{9} &\ \frac{5}{9} \end{bmatrix}$

$\begin{bmatrix} \frac{1}{6} & \frac{1}{3} \\ \\ \frac{5}{6} &\frac{2}{3} \end{bmatrix}$

$\begin{bmatrix} \frac{4}{9} & \frac{1}{3} \\ \\ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \frac{4}{9} & \frac{1}{3} \\ \\ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$

Explanation:

A stochastic matrix is a matrix of probabilities in which the entry $p_{m,n}$ is the probability that, given the fact that a given system is in a state , the system will be in state next.

We will let State 1 be that the player is on Space "X" and State 2 be that the player is on Space "Y". If the player is on Space "X", the probability that he will still be on that space after one turn is $\frac{1}{6}$ , since one roll out of six will allow him to stay there; similarly, the probability that he will be on Space "Y" is $\frac{5}{6}$ . If a player is on Space "B", the same probabilities are, respectively, $\frac{1}{2}$ and $\frac{1}{2}$ . The stochastic matrix representing these probabilities is

$P = \begin{bmatrix} \frac{1}{6}& \frac{1}{2}\\ \\ \frac{5}{6}& \frac{1}{2} \end{bmatrix}$ .

The stochastic matrix representing the probabilities of being on a particular space after two terms is the square of this matrix $P^{2}$ , which is

$P ^{2}= \begin{bmatrix} \frac{1}{6}& \frac{1}{2}\\ \\ \frac{5}{6}& \frac{1}{2} \end{bmatrix}\begin{bmatrix} \frac{1}{6}& \frac{1}{2}\\ \\ \frac{5}{6}& \frac{1}{2} \end{bmatrix}$

The entry in column , row of the product $PP= P^{2}$ is the product of row of and column of - the sum of the products of the numbers that appear in the corresponding positions of the row and the column. Therefore,

$P ^{2}= \begin{bmatrix} \frac{1}{6} \cdot \frac{1}{6}+ \frac{1}{2} \cdot \frac{5}{6} & \frac{1}{6} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \\ \\ \frac{5}{6} \cdot \frac{1}{6}+ \frac{1}{2}\cdot \frac{5}{6} &\frac{5}{6} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{36}+ \frac{5}{12} & \frac{1}{12} + \frac{1}{4} \\ \\ \frac{5}{36}+ \frac{5}{12} &\frac{5}{12} + \frac{1}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{4}{9} & \frac{1}{3} \\ \\ \frac{5}{9} &\frac{2}{3} \end{bmatrix}$ ,

the correct choice.

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Example Question #23 : Matrix Matrix Product

A B and are both defined products. has four rows and three columns.

Which of the following is true about the number of rows and columns in ?

Possible Answers:

can have any number of rows but must have four columns.

must have three rows but can have any number of columns.

must either have three rows and four columns or four rows and three columns.

must have three rows and four columns.

must have four rows and three columns.

Correct answer:

must have three rows and four columns.

Explanation:

For a matrix product A B to be defined, it must hold that the number of rows in is equal to the number of columns in . has three columns, so must have three rows. For to be defined, by similar reasoning, must have four columns.

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Example Question #28 : Matrices

AB = C

where

$B = \begin{bmatrix} 2& 3& -4& 2\\ -1&4 & 1& 0\\ 3& -2& 5&1 \\ 1&1 &2 &3 \end{bmatrix}$

and

$C = \begin{bmatrix} 2& 3& -4& 2\\ 3&-12 & -3& 0\\ 3& -2& 5&1 \\ 1&1 &2 &3 \end{bmatrix}$

Evaluate .

(You may assume that both and have inverses. )

Possible Answers:

None of the other choices gives a correct response.

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

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

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

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

Correct answer:

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

Explanation:

While A B =C implies that $A= C B^{-1}$ , it is not necessary to do the calculations in order to find .

differs from only in its second row, in which each element is times the corresponding element in :

$-1 \cdot( -3 )= 3$

$4 \cdot( -3 )= -12$

$1 \cdot( -3 )=- 3$

$0 \cdot( -3 )= 0$

Therefore, is the result of a row operation on , namely,

$-3R 2 \rightarrow R2$

Therefore, is the product of an elementary matrix and ; the elementary matrix for the given row operation is the one in which the operation is performed on the (four-by-four) identity matrix, which is

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

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Example Question #22 : Matrices

AB = C

where

$B = \begin{bmatrix} 2& 3& -4& 2\\ -1&4 & 1& 0\\ 3& -2& 5&1 \\ 1&1 &2 &3 \end{bmatrix}$

and

$C = \begin{bmatrix} 3& -2& 5&1 \\ -1&4 & 1& 0 \\ 2& 3& -4& 2 \\ 1&1 &2 &3 \end{bmatrix}$

Evaluate .

(You may assume that both and have inverses. )

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

While A B =C implies that $A= C B^{-1}$ , it is not necessary to do the calculations in order to find .

differs from only in that its first and third rows have reversed positions. Therefore, is the result of a row operation on , namely,

$R1 \leftrightarrow R3$

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

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Example Question #28 : Matrices

Let $L = \begin{bmatrix} 1& 4\\ 0& 2 \end{bmatrix}$ , $U = \begin{bmatrix} 2&2 \\ 0& -3 \end{bmatrix}$ , and $A = \begin{bmatrix} 2&-10 \\ 0& -6 \end{bmatrix}$

True or false: is an example of a valid -factorization of .

Possible Answers:

False, because is not the right kind of matrix.

True

False, because is not a factorization of .

Correct answer:

False, because is not the right kind of matrix.

Explanation:

An -factorization is a way of expressing a matrix as a product of two matrices and . For the factorization to be valid:

1) must be a Lower triangular matrix - all elements above its main diagonal (upper left corner to lower right corner) must be "0".

2) must be an Upper triangular matrix - all elements below its main diagonal must be "0".

3) LU = A

can be seen to have a nonzero element above its main diagonal - the 4 in Row 1, Column 2. Consequently, is not a lower triangular matrix. This makes invalid as an -factorization.

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Example Question #31 : Matrix Matrix Product

Let $L = \begin{bmatrix} 1& 0\\ 4& 2 \end{bmatrix}$ , $U = \begin{bmatrix} 2&-5 \\ 0& 1 \end{bmatrix}$ , and $A = \begin{bmatrix} 2&-5\\ 8& -18 \end{bmatrix}$ .

True or false: is an example of a valid -factorization of .

Possible Answers:

False, because is not the right kind of matrix.

False, because is not a factorization of .

False, because is not the right kind of matrix.

True

Correct answer:

True

Explanation:

An -factorization is a way of expressing a matrix as a product of two matrices and . For the factorization to be valid:

1) must be a Lower triangular matrix - all elements above its main diagonal (upper left corner to lower right corner) must be "0".

2) must be an Upper triangular matrix - all elements below its main diagonal must be "0".

3) LU = A

The factorization can be seen to satisfy the first two criteria - is lower triangular in that there are no nonzero elements above its main diagonal, and is, analogously, upper triangular. It remains to be shown that LU = A .

Multiply each row in by each column in - add the products of each element in the former by the corresponding element in the latter - as follows:

$LU = \begin{bmatrix} 1& 0\\ 4& 2 \end{bmatrix}\begin{bmatrix} 2&-5 \\ 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 1(2)+ 0(0) & 1(-5)+ 0(1)\\ 4(2)+ 2(0) & 4(-5)+ 2(1) \end{bmatrix}$

$= \begin{bmatrix} 2&-5\\ 8& -18 \end{bmatrix}$

All three criteria are met, and gives a valid -factorization of .

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Example Question #31 : Matrix Matrix Product

Let $L = \begin{bmatrix} 1& 0\\ -3& 2 \end{bmatrix}$ , $U = \begin{bmatrix} 2&2 \\ 0& -3 \end{bmatrix}$ , and $A = \begin{bmatrix} 2& 0\\ 0& -6 \end{bmatrix}$

True or false: is an example of a valid -factorization of .

Possible Answers:

True

False, because is not the right kind of matrix.

False, because is not a factorization of .

False, because is not the right kind of matrix.

Correct answer:

False, because is not a factorization of .

Explanation:

An -factorization is a way of expressing a matrix as a product of two matrices and . For the factorization to be valid:

1) must be a Lower triangular matrix - all elements above its main diagonal (upper left corner to lower right corner) must be "0".

2) must be an Upper triangular matrix - all elements below its main diagonal must be "0".

3) LU = A

Multiply each row in by each column in - add the products of each element in the former by the corresponding element in the latter - as follows:

$LU = \begin{bmatrix} 1& 0\\ -3& 2 \end{bmatrix} \begin{bmatrix} 2&2 \\ 0& -3 \end{bmatrix}$

$= \begin{bmatrix} 1(2)+ 0(0) & 1(2)+0(-3)\\ (-3)(2)+2(0)& -3(2)+2(-3) \end{bmatrix}$

$= \begin{bmatrix} 2 &2\\-6&-12 \end{bmatrix}$

$LU \ne A$ , so does not give a valid -factorization of .

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Example Question #31 : Matrices

True or False: If is a square matrix, and A^2 = 0 , then is either or .

Possible Answers:

True

False

Correct answer:

False

Explanation:

For example, if $A = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$ , then $A^2 = \begin{bmatrix} 0& 1\\ 0& 0 \end{bmatrix}\begin{bmatrix} 0& 1\\ 0&0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0& 0 \end{bmatrix}$ , but itself is not or .

(If represented a single real number, then the question would be true, but since is a matrix, the question is not true anymore.)

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Example Question #33 : Matrix Matrix Product

What is dimension criteria to multiply two matrices $(A\times B)$ ?

Possible Answers:

$(A\times B)$ can only be multiplied if the number of rows in equals the number of rows in

$(A\times B)$ can only be multiplied if the number of columns in equals the number of columns in

$(A\times B)$ can only be multiplied if the number of columns in equals the number of rows in

$(A\times B)$ can only be multiplied if the number of rows in equals the number of columns in

Correct answer:

$(A\times B)$ can only be multiplied if the number of columns in equals the number of rows in

Explanation:

If is an $(m\times n)$ matrix and is an $(r\times s)$ matrix,

$(A\times B)$ can only be multiplied if n=r , or the number of columns in equals the number of rows in . Otherwise there is a mismatch, and the two matrices can not be multiplied.

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #24 : Matrices

Example Question #25 : Matrices

Example Question #23 : Matrix Matrix Product

Example Question #28 : Matrices

Example Question #22 : Matrices

Example Question #28 : Matrices

Example Question #31 : Matrix Matrix Product

Example Question #31 : Matrix Matrix Product

Example Question #31 : Matrices

Example Question #33 : Matrix Matrix Product

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