§1.4 Matrix Equation Ax = b: Linear Combination (I)
b2 b3.. . Question: For what values of b1b2
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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.
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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. =
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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
Math 215 HW #4 Solutions
1.Problem 2.1.6. LetPbe the plane in 3-space with equationx+ 2y+z= 6. What is the
equation of the planeP0through the origin parallel toP? ArePandP0subspaces ofR3? Answer:For any real numberr, the planex+ 2y+z=ris parallel toP, since all such planes have a common normal vectori+2j+k=? ?1 2 1? . In particular, notice that the plane determined by the equation x+ 2y+z= 0 (?) is parallel toPand passes through the origin (since (x,y,z) = (0,0,0) is a solution of the above equation). Hence, this is the equation which determines the planeP0.Now, suppose?
?x 1 y 1 z 1? ?x 2 y 2 z 2? ?P0; i.e. the triples (x1,y1,z1) and (x2,y2,z2) both satisfy the equation (?). Then (x1+x2) + 2(y1+y2) + (z1+z2) = (x1+ 2y1+z1) + (x2+ 2y2+z2) = 0 + 0 = 0, so we have that ?x 1 y 1 z 1? ?x 2 y 2 z 2? ?x 1+x2 y 1+y2 z 1+z2? ?P0.Also, ifc?R, then
cx1+ 2(cy1) +cz1=c(x1+ 2y1+z1) =c(0) = 0,
so c? ?x 1 y 1 z 1? ?cx 1 cy 1 cz 1? ?P0.Therefore,P0is a subspace ofR3.
On the other hand,?
?6 0 0? and? ?0 3 0? are both inP, but ?6 0 0? ?0 3 0? ?6 3 0? is not inPsince6 + 2(3) + 0 = 12?= 6.
Therefore, we see thatPis not a subspace ofR3.1
2.Problem 2.1.12. The functionsf(x) =x2andg(x) = 5xare "vectors" in the vector spaceF
of all real functions. The combination 3f(x)-4g(x) is the functionh(x) =. Which rule is broken if multiplyingf(x) bycgives the functionf(cx)? Answer:The combination 3f(x)-4g(x) is the function h(x) = 3x2-20x. If we tried to define scalar multiplication ascf(x) =f(cx) we would run into problems. Note that f(5x) = (5x)2= 25x2, but f(2x) +f(3x) = (2x)2+ (3x)2= 4x2+ 9x2= 13x2. Hence, this attempted definition of scalar multiplication would not satisfy rule 8 in the defi- nition of a vector space.3.Problem 2.1.18. (a)The intersection of two planes through (0,0,0) is probably abut it could be a . It can"t be the zero vectorZ! Answer:The intersection of two planes through the origin inR3is probably a line, butit could be a plane (if the two planes coincide).(b)The intersection of a plane through (0,0,0) with a line through (0,0,0) is probably abut it could be a.
Answer:The intersection of a plane through the origin with a line through the origin inR3is probably just the single point (0,0,0), but it could be a whole line (if the linelies in the plane).(c)IfSandTare subspaces ofR5, their intersectionS∩T(vectors in both subspaces) is a
subspace ofR5.Check the requirements onx+yandcx. Answer:To see thatS∩Tis a subspace, supposex,y?S∩Tand thatc?R. Then, sincexandyare both inSand sinceSis a subspace (meaning that it is closed under addition), we have that x+y?S. Likewise, sincexandyare both elements ofTand sinceTis a subspace, we have that x+y?T. Therefore, sincex+yis in bothSandT, we have that x+y?S∩T. Likewise, sincex?SandSis a subspace (meaning thatSis closed under scalar multiplication), we have thatcx?S; similarly,cx?T. Therefore, cx?S∩T. Since our choices ofx,y, andcwere completely arbitrary, we see thatS∩Tis a subspace ofR5.24.Problem 2.1.22. For which right-hand sides (find a condition onb1,b2,b3) are these systems
solvable? (a) ?1 4 2 2 8 4 -1-4-2? ?x 1 x 2 x 3? ?b 1 b 2 b 3? .(b)? ?1 4 2 9 -1-4? ??x1 x 2? ?b 1 b 2 b 3? .(a)Answer:Form the augmented matrix ?1 4 2b12 8 4b2
-1-4-2b3? The goal is to use elimination to get this into reduced echelon form. Subtract twice row1 from row 2 and add row 1 to row 3 to get:
?1 4 2b10 0 0b2-2b1
0 0 0b3+b1?
Hence, the given equation is solvable only if
b2-2b1= 0 andb3+b1= 0.
In other words, the right-hand side of the equation must be a vector of the form ?b 1 2b1 -b1? =b1? ?1 2 -1? for any real numberb1. In other words, the column space of the given matrix is the line containing the vector? ?1 2 -1? .(b)Answer:Form the augmented matrix ?1 4b1 2 9b2 -1-4b3? Then the goal is to get this into reduced echelon form. To do so, subtract twice row 1 from row 2 and add row 1 to row 3, yielding: ?1 4b10 1b2-2b1
0 0b3+b1?
The given equation is solvable only if
b3+b1= 0,3
or, equivalently, ifb3=-b1. Hence, the possible right-hand sides are vectors of the form ?b 1 b 2 -b1? =b1? ?1 0Math 215 HW #4 Solutions
1.Problem 2.1.6. LetPbe the plane in 3-space with equationx+ 2y+z= 6. What is the
equation of the planeP0through the origin parallel toP? ArePandP0subspaces ofR3? Answer:For any real numberr, the planex+ 2y+z=ris parallel toP, since all such planes have a common normal vectori+2j+k=? ?1 2 1? . In particular, notice that the plane determined by the equation x+ 2y+z= 0 (?) is parallel toPand passes through the origin (since (x,y,z) = (0,0,0) is a solution of the above equation). Hence, this is the equation which determines the planeP0.Now, suppose?
?x 1 y 1 z 1? ?x 2 y 2 z 2? ?P0; i.e. the triples (x1,y1,z1) and (x2,y2,z2) both satisfy the equation (?). Then (x1+x2) + 2(y1+y2) + (z1+z2) = (x1+ 2y1+z1) + (x2+ 2y2+z2) = 0 + 0 = 0, so we have that ?x 1 y 1 z 1? ?x 2 y 2 z 2? ?x 1+x2 y 1+y2 z 1+z2? ?P0.Also, ifc?R, then
cx1+ 2(cy1) +cz1=c(x1+ 2y1+z1) =c(0) = 0,
so c? ?x 1 y 1 z 1? ?cx 1 cy 1 cz 1? ?P0.Therefore,P0is a subspace ofR3.
On the other hand,?
?6 0 0? and? ?0 3 0? are both inP, but ?6 0 0? ?0 3 0? ?6 3 0? is not inPsince6 + 2(3) + 0 = 12?= 6.
Therefore, we see thatPis not a subspace ofR3.1
2.Problem 2.1.12. The functionsf(x) =x2andg(x) = 5xare "vectors" in the vector spaceF
of all real functions. The combination 3f(x)-4g(x) is the functionh(x) =. Which rule is broken if multiplyingf(x) bycgives the functionf(cx)? Answer:The combination 3f(x)-4g(x) is the function h(x) = 3x2-20x. If we tried to define scalar multiplication ascf(x) =f(cx) we would run into problems. Note that f(5x) = (5x)2= 25x2, but f(2x) +f(3x) = (2x)2+ (3x)2= 4x2+ 9x2= 13x2. Hence, this attempted definition of scalar multiplication would not satisfy rule 8 in the defi- nition of a vector space.3.Problem 2.1.18. (a)The intersection of two planes through (0,0,0) is probably abut it could be a . It can"t be the zero vectorZ! Answer:The intersection of two planes through the origin inR3is probably a line, butit could be a plane (if the two planes coincide).(b)The intersection of a plane through (0,0,0) with a line through (0,0,0) is probably abut it could be a.
Answer:The intersection of a plane through the origin with a line through the origin inR3is probably just the single point (0,0,0), but it could be a whole line (if the linelies in the plane).(c)IfSandTare subspaces ofR5, their intersectionS∩T(vectors in both subspaces) is a
subspace ofR5.Check the requirements onx+yandcx. Answer:To see thatS∩Tis a subspace, supposex,y?S∩Tand thatc?R. Then, sincexandyare both inSand sinceSis a subspace (meaning that it is closed under addition), we have that x+y?S. Likewise, sincexandyare both elements ofTand sinceTis a subspace, we have that x+y?T. Therefore, sincex+yis in bothSandT, we have that x+y?S∩T. Likewise, sincex?SandSis a subspace (meaning thatSis closed under scalar multiplication), we have thatcx?S; similarly,cx?T. Therefore, cx?S∩T. Since our choices ofx,y, andcwere completely arbitrary, we see thatS∩Tis a subspace ofR5.2