18.06 Problem Set 1 Solutions









§1.4 Matrix Equation Ax = b: Linear Combination (I)

b2 b3.. . Question: For what values of b1b2
week


18.06 Problem Set 1 Solutions

Feb 11 2010 If E21 subtracts row 1 from row. 2
pset s soln


Math 2331 – Linear Algebra - 1.4 The Matrix Equation Ax=b

1.4 The Matrix Equation Ax = b. Definition Theorem Span Rm. Matrix-Vector Multiplication: Examples. Example 1 −4. 3. 2.
sec


2.5 Inverse Matrices

Elimination solves Ax D b without explicitly using the matrix A. 1 . Note 2 Find the inverses (directly or from the 2 by 2 formula) of A;B;C:.
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Now let's show that V ar(aX + b) = a 2V ar(X). This is for a b

Same kind of idea works but just want to remember this. V ar(aX + b) = E((aX + b)2) − (E(aX + b))2. = 
prob


Matrix-Vector Products and the Matrix Equation Ax= b

Jan 31 2018 has a solution. 2. Indeed
Lecture


Chapter 2 - Matrices and Linear Algebra

Ax = b. In this way we see that with ci (A) denoting the ith column of A the system is expressible as x1c1 (A) + ··· + xncn (A) = b. From this equation it 
chapter


The Matrix Equation Ax = b Section 1.5: Solution Sets of Linear

This section is about solving the “matrix equation” Ax = b where A is an m Exercise 2 (1.7.1): Check if the following vectors are linearly independent:.
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Table of Integrals

u(x)v (x)dx = u(x)v(x) v(x)u (x)dx. RATIONAL FUNCTIONS. (5). 1 ax + b dx = 1 a ln(ax + b). (6). 1. (x + a)2 dx = 1 x + a. (7). (x + a)n dx = (x + a)n.
IntegralTable


Math 215 HW #4 Solutions

Execute the six steps following equation (6) to findthee column space and nullspace of A and the solution to Ax = b: A =.. 1 1 2 2. 2 5 7 6. 2 3 5 2.
hw solutions


212359 18.06 Problem Set 1 Solutions

18.06 Problem Set 1 Solutions

Due Thursday, 11 February 2010 at 4 pm in 2-106.

Total: 100 points

Section 1.2. Problem 23:The gure shows that cos() =v1=kvkand sin() = v

2=kvk. Similarly cos() isand sin() is. The angleis.

Substitute into the trigonometry formula cos()cos()+sin()sin() for cos() to nd cos() =vw=kvkkwk.Solution(4 points) First blank:w1=kwk. Second blank:w2=kwk. Substituting into the trigonome- try formula yields cos() = (w1=kwk)(v1=kvk) + (w2=kwk)(v2=kvk) =vw=kvkkwk: Section 1.2. Problem 28:Can three vectors in thexyplane haveuv <0 and vw <0 anduw <0?Solution(12 points)

Yes. For instance takeu= (1;0),v= (12

;p3 2 ),w= (12 ;p3 2 ). Notice uv=vw=uw=12 Section 1.3. Problem 4:Find a combinationx1w1+x2w2+x3w3that gives the zero vector: w 1=2 41
2 33
5 ;w2=2 44
5 63
5 ;w3=2 47
8 93
5 Those vectors are (independent)(dependent). The three vectors lie in a. The matrixWwith those columns isnot invertible.Solution(4 points) Observew12w2+w3= 0. The vectors aredependent. They lie in aplane. Section 1.3. Problem 13:The very last words say that the 5 by 5 centered dierence matrixis notinvertible. Write down the 5 equationsCx=b. Find a 1 combination of left sides that gives zero. What combination ofb1;b2;b3;b4;b5must be zero?Solution(12 points)

The 5 by 5 centered dierence matrix is

C=0 B

BBB@0 1 0 0 0

1 0 1 0 0

01 0 1 0

0 01 0 1

0 0 01 01

C CCCA:

The ve equationsCx=bare

x

2=b1;x1+x3=b2;x2+x4=b3;x3+x5=b4;x4=b5:

Observe that the sum of the rst, third, and fth equations is zero. Similarly, b

1+b3+b5= 0.

Section 2.1. Problem 29:Start with the vectoru0= (1;0). Multiply again and again by the same \Markov matrix"A= [:8:3;:2:7]. The next three vectors are u

1;u2;u3:

u

1=:8:3

:2:7 1 0 =:8 :2 u

2=Au1=u

3=Au2=:

What property do you notice for all four vectorsu0,u1,u2,u3.Solution(4 points)

Computing, we get

u 2=:7 :3 u 3=:65 :35

All four vectors have components that sum to one.

Section 2.1. Problem 30:Continue Problem 29 fromu0= (1;0) tou7, and also fromv0= (0;1) tov7. What do you notice aboutu7andv7? Here are two MATLAB codes, with while and for. They plotu0tou7andv0tov7. Theu's and thev's are approaching a steady state vectors. Guess that vector and check thatAs=s. If you start withs, then you stay withs. 2

Solution(12 points)

Here is a screenshot of the code entered into matlab.In this graph, we see that the sequenceu1;u2;:::;u7is approaching (:6;:4).

3 In this graph, we see that the sequencev1;v2;:::;v7is approaching (:6;:4).4 From the graphs, we guess thats= (:6;:4) is a steady state vector. We verify this with the computation

As=:8:3

:2:7 :6 :4 =:6 :4 Section 2.2. Problem 20:Three planes can fail to have an intersection point,

even if no planes are parallel. The system is singular if row 3 ofAis aof the rst two rows. Find a third equation that can't be solved together with

x+y+z= 0 andx2yz= 1.Solution(4 points) The system is singular if row 3 ofAis alinear combinationof the rst two rows. There are many possible choices of a third equation that cannot be solved together with the ones given. An example is 2x+ 5y+ 4z= 1. Note that the left hand side of the third equation is the three times the left hand side of thte rst minus the left hand side of the second. However, the right hand side does not satisfy this relation. Section 2.2. Problem 32:Start with 100 equationsAx= 0 for 100 unknowns x= (x1;:::;x100). Suppose elimination reduces the 100th equation to 0 = 0, so the system is \singular". (a) Elimination takes linear combinations of the rows. So this singular system has the singular property: Some linear combination of the 100rowsis. (b) Singular systemsAx= 0 have innitely many solutions. This means that some linear combination of the 100columnsis. (c) Invent a 100 by 100 singular matrix with no zero entries. (d) For your matrix, describe in words the row picture and the column picture of Ax= 0. Not necessary to draw 100-dimensional space.Solution(12 points) (a) Zero. (b) Zero. (c) There are many possible answers. For instance, the matrix for which every row is (1 2 3100). (d) The row picture is 100 copies of the hyperplane in 100-space dened by the equation x

1+ 2x2+ 3x3++ 100x100= 0:

The column picture is the 100 vectors proportional to (1 1 11) of lengths

10;20;:::;1000.

5 Section 2.3. Problem 22:The entries ofAandxareaijandxj. So the rst component ofAxisPa1jxj=a11x1++a1nxn. IfE21subtracts row 1 from row

2, write a formula for

(a) the third component ofAx (b) the (2;1) entry ofE21A (c) the (2;1) entry ofE21(E21A) (d) the rst component ofE21Ax.Solution(4 points) (a)Pa3jxj. (b)a21a11. (c)a212a11. (d)Pa1jxj. Section 2.3. Problem 29:Find the triangular matrixEthat reduces \Pascal's matrix" to a smaller Pascal: E 2 6

641 0 0 0

1 1 0 0

1 2 1 0

1 3 3 13

7 75=2
6

641 0 0 0

0 1 0 0

0 1 1 0

0 1 2 13

7 75:
Which matrixM(multiplying severalE's) reduces Pascal all the way toI?Solution(12 points) E=2 6

641 0 0 0

1 1 0 0

18.06 Problem Set 1 Solutions

Due Thursday, 11 February 2010 at 4 pm in 2-106.

Total: 100 points

Section 1.2. Problem 23:The gure shows that cos() =v1=kvkand sin() = v

2=kvk. Similarly cos() isand sin() is. The angleis.

Substitute into the trigonometry formula cos()cos()+sin()sin() for cos() to nd cos() =vw=kvkkwk.Solution(4 points) First blank:w1=kwk. Second blank:w2=kwk. Substituting into the trigonome- try formula yields cos() = (w1=kwk)(v1=kvk) + (w2=kwk)(v2=kvk) =vw=kvkkwk: Section 1.2. Problem 28:Can three vectors in thexyplane haveuv <0 and vw <0 anduw <0?Solution(12 points)

Yes. For instance takeu= (1;0),v= (12

;p3 2 ),w= (12 ;p3 2 ). Notice uv=vw=uw=12 Section 1.3. Problem 4:Find a combinationx1w1+x2w2+x3w3that gives the zero vector: w 1=2 41
2 33
5 ;w2=2 44
5 63
5 ;w3=2 47
8 93
5 Those vectors are (independent)(dependent). The three vectors lie in a. The matrixWwith those columns isnot invertible.Solution(4 points) Observew12w2+w3= 0. The vectors aredependent. They lie in aplane. Section 1.3. Problem 13:The very last words say that the 5 by 5 centered dierence matrixis notinvertible. Write down the 5 equationsCx=b. Find a 1 combination of left sides that gives zero. What combination ofb1;b2;b3;b4;b5must be zero?Solution(12 points)

The 5 by 5 centered dierence matrix is

C=0 B

BBB@0 1 0 0 0

1 0 1 0 0

01 0 1 0

0 01 0 1

0 0 01 01

C CCCA:

The ve equationsCx=bare

x

2=b1;x1+x3=b2;x2+x4=b3;x3+x5=b4;x4=b5:

Observe that the sum of the rst, third, and fth equations is zero. Similarly, b

1+b3+b5= 0.

Section 2.1. Problem 29:Start with the vectoru0= (1;0). Multiply again and again by the same \Markov matrix"A= [:8:3;:2:7]. The next three vectors are u

1;u2;u3:

u

1=:8:3

:2:7 1 0 =:8 :2 u

2=Au1=u

3=Au2=:

What property do you notice for all four vectorsu0,u1,u2,u3.Solution(4 points)

Computing, we get

u 2=:7 :3 u 3=:65 :35

All four vectors have components that sum to one.

Section 2.1. Problem 30:Continue Problem 29 fromu0= (1;0) tou7, and also fromv0= (0;1) tov7. What do you notice aboutu7andv7? Here are two MATLAB codes, with while and for. They plotu0tou7andv0tov7. Theu's and thev's are approaching a steady state vectors. Guess that vector and check thatAs=s. If you start withs, then you stay withs. 2

Solution(12 points)

Here is a screenshot of the code entered into matlab.In this graph, we see that the sequenceu1;u2;:::;u7is approaching (:6;:4).

3 In this graph, we see that the sequencev1;v2;:::;v7is approaching (:6;:4).4 From the graphs, we guess thats= (:6;:4) is a steady state vector. We verify this with the computation

As=:8:3

:2:7 :6 :4 =:6 :4 Section 2.2. Problem 20:Three planes can fail to have an intersection point,

even if no planes are parallel. The system is singular if row 3 ofAis aof the rst two rows. Find a third equation that can't be solved together with

x+y+z= 0 andx2yz= 1.Solution(4 points) The system is singular if row 3 ofAis alinear combinationof the rst two rows. There are many possible choices of a third equation that cannot be solved together with the ones given. An example is 2x+ 5y+ 4z= 1. Note that the left hand side of the third equation is the three times the left hand side of thte rst minus the left hand side of the second. However, the right hand side does not satisfy this relation. Section 2.2. Problem 32:Start with 100 equationsAx= 0 for 100 unknowns x= (x1;:::;x100). Suppose elimination reduces the 100th equation to 0 = 0, so the system is \singular". (a) Elimination takes linear combinations of the rows. So this singular system has the singular property: Some linear combination of the 100rowsis. (b) Singular systemsAx= 0 have innitely many solutions. This means that some linear combination of the 100columnsis. (c) Invent a 100 by 100 singular matrix with no zero entries. (d) For your matrix, describe in words the row picture and the column picture of Ax= 0. Not necessary to draw 100-dimensional space.Solution(12 points) (a) Zero. (b) Zero. (c) There are many possible answers. For instance, the matrix for which every row is (1 2 3100). (d) The row picture is 100 copies of the hyperplane in 100-space dened by the equation x

1+ 2x2+ 3x3++ 100x100= 0:

The column picture is the 100 vectors proportional to (1 1 11) of lengths

10;20;:::;1000.

5 Section 2.3. Problem 22:The entries ofAandxareaijandxj. So the rst component ofAxisPa1jxj=a11x1++a1nxn. IfE21subtracts row 1 from row

2, write a formula for

(a) the third component ofAx (b) the (2;1) entry ofE21A (c) the (2;1) entry ofE21(E21A) (d) the rst component ofE21Ax.Solution(4 points) (a)Pa3jxj. (b)a21a11. (c)a212a11. (d)Pa1jxj. Section 2.3. Problem 29:Find the triangular matrixEthat reduces \Pascal's matrix" to a smaller Pascal: E 2 6

641 0 0 0

1 1 0 0

1 2 1 0

1 3 3 13

7 75=2
6

641 0 0 0

0 1 0 0

0 1 1 0

0 1 2 13

7 75:
Which matrixM(multiplying severalE's) reduces Pascal all the way toI?Solution(12 points) E=2 6

641 0 0 0

1 1 0 0


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