(完整版)哈工大选修课LINEARALGEBRA试卷及答案,推荐文档
哈工大选修课 LINEAR ALGEBRA 试卷及答案
LINEAR ALGEBRAANDITS APPLICATIONS 姓名：易学号：成绩：1. Definitions(1) Pivot position in a matrix; (2) Echelon Form; (3) Elementary operations;(4) Onto mapping and one-to-one mapping; (5) Linearly independence.2. Describe the row reduction algorithm which produces a matrix in reduced echelon form.3. Find the 33⨯ matrix that corresponds to the composite transformation of a scaling by 0.3, a rotation of 90︒, and finally a translation that adds (-0.5, 2) to each point of a figure.4. Find a basis for the null space of the matrix361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦5. Find a basis for Col A of the matrix1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦6. Let a and b be positive numbers. Find the area of the region bounded by the ellipse whose equation is22221x y ab+=7. Provide twenty statements for the invertible matrix theorem. 8. Show and prove the Gram-Schmidt process. 9. Show and prove the diagonalization theorem.10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent.Answers:1. Definitions(1) Pivot position in a matrix:A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position.(2) Echelon Form:A rectangular matrix is in echelon form (or row echelon form) if it has the following three properties:1.All nonzero rows are above any rows of all zeros.2.Each leading entry of a row is in a column to the right of the leading entry of the row above it.3.All entries in a column below a leading entry are zeros.If a matrix in a echelon form satisfies the following additional conditions, then it is in reduced echelon form (or reduced row echelon form):4.The leading entry in each nonzero row is 1.5.Each leading 1 is the only nonzero entry in its column.(3)Elementary operations:Elementary operations can refer to elementary row operations or elementary column operations.There are three types of elementary matrices, which correspond to three types of row operations (respectively, column operations):1.(Replacement) Replace one row by the sum of itself anda multiple of another row.2.(Interchange) Interchange two rows.3.(scaling) Multiply all entries in a row by a nonzero constant.(4)Onto mapping and one-to-one mapping:A mapping T : n →m is said to be onto m if each b in m is the image of at least one x in n.A mapping T : n →m is said to be one-to-one if each b in m is the image of at most one x in n.(5)Linearly independence:An indexed set of vectors {V1, . . . ,V p} in n is said to be linearly independent if the vector equationx 1v 1+x 2v 2+ . . . +x p v p = 0Has only the trivial solution. The set {V 1, . . . ,V p } is said to be linearly dependent if there exist weights c 1, . . . ,c p , not all zero, such that c 1v 1+c 2v 2+ . . . +c p v p = 02. Describe the row reduction algorithm which produces a matrix in reduced echelon form. Solution: Step 1:Begin with the leftmost nonzero column. This is a pivot column. The pivot position is at the top. Step 2:Select a nonzero entry in the pivot column as a pivot. If necessary, interchange rows to move this entry into the pivot position. Step 3:Use row replacement operations to create zeros in all positions below the pivot. Step 4:Cover (or ignore) the row containing the pivot position and cover all rows, if any, above it. Apply steps 1-3 to the submatrix that remains. Repeat the process until there all no more nonzero rows to modify. Step 5:Beginning with the rightmost pivot and working upward and to the left, create zeros above each pivot. If a pivot is not 1, make it 1 by scaling operation.3. Find the 33⨯ matrix that corresponds to the composite transformation of a scaling by 0.3, a rotation of 90︒, and finally a translation that adds (-0.5, 2) to each point of a figure. Solution:If ψ=π/2, then sin ψ=1 and cos ψ=0. Then we have ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡110003.00003.01y x y x scale⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−→−110003.00003.0100001010y x R o t a t e⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−−→−110003.00003.0100001010125.0010001y x T r a n s l a t eThe matrix for the composite transformation is ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-10003.00003.0100001010125.0010001⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=10003.00003.0125.0001010⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=125.0003.003.004. Find a basis for the null space of the matrix 361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦Solution:First, write the solution of A X=0 in parametric vector form: A ~ ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---00002302101000201, x 1-2x 2 -x 4+3x 5=0 x 3+2x 4-2x 5=0 0=0The general solution is x 1=2x 2+x 4-3x 5, x 3=-2x 4+2x 5, with x 2, x 4, and x 5 free. ⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡+--+=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡10203012010001222325425454254254321x x x x x x x x x x x x x x x xu v w=x 2u+x 4v+x 5w (1)Equation (1) shows that Nul A coincides with the set of all linear conbinations of u, v and w . That is, {u, v, w}generates Nul A. In fact, this construction of u, v and w automatically makes them linearly independent, because (1) shows that 0=x 2u+x 4v+x 5w only if the weights x 2, x 4, and x 5 are all zero.So {u, v , w} is a basis for Nul A.5. Find a basis for Col A of the matrix 1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦Solution: A ~ ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---07490012002300130001, so the rank of A is 3. Then we have a basis for Col A of the matrix: U = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0001, v = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0013and w = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--07496. Let a and b be positive numbers. Find the area of the region bounded by the ellipse whose equation is22221x y ab+=Solution:We claim that E is the image of the unit disk D under the linear transformation Tdetermined by the matrix A=⎥⎦⎤⎢⎣⎡b a 00, because if u= ⎥⎦⎤⎢⎣⎡21u u , x=⎥⎦⎤⎢⎣⎡21x x , and x = Au, then u 1 =ax 1 and u 2 =bx 2It follows that u is in the unit disk, with 12221≤+u u , if and only if x is in E , with1)()(2221≤+b x a x . Then we have{area of ellipse} = {area of T (D )} = |det A| {area of D} = ab π(1)2= πab7. Provide twenty statements for the invertible matrix theorem.Let A be a square n n ⨯ matrix. Then the following statements are equivalent. That is, for a given A, the statements are either all true or false. a. A is an invertible matrix.b. A is row equivalent to the n n ⨯ identity matrix.c. A has n pivot positions.d. The equation Ax = 0 has only the trivial solution.e. The columns of A form a linearly independent set.f. The linear transformation x → Ax is one-to-one.g. The equation Ax = b has at least one solution for each b in n.h. The columns of A spann.i. The linear transformation x → Ax maps nonton.j. There is an n n ⨯ matrix C such that CA = I. k. There is an n n ⨯ matrix D such that AD = I. l. A T is an invertible matrix. m. If 0A ≠, then ()()T11T A A --=n. If A, B are all invertible, then (AB)* = B *A *o. T**T )(A )(A =p. If 0A ≠, then ()()*11*A A --=q. ()*1n *A 1)(A --=-r. If 0A ≠, then ()()L11L A A --= ( L is a natural number )s. ()*1n *A K)(KA --=-t. If 0A ≠, then *1A A1A =-8. Show and prove the Gram-Schmidt process.Solution:The Gram-Schmidt process:Given a basis {x 1, . . . , x p } for a subspace W of n, define11x v = 1112222v v v v x x v ⋅⋅-=222231111333v v v v x v v v v x x v ⋅⋅-⋅⋅-=. ..1p 1p 1p 1p p 2222p 1111p p p v v v v x v v v v x v v v v x x v ----⋅-⋅⋅⋅-⋅⋅-⋅⋅-=Then {v 1, . . . , v p } is an orthogonal basis for W. In additionSpan {v 1, . . . , v p } = {x 1, . . . , x p } for p k ≤≤1 PROOFFor p k ≤≤1, let W k = Span {v 1, . . . , v p }. Set 11x v =, so that Span {v 1} = Span {x 1}.Suppose, for some k < p, we have constructed v 1, . . . , v k so that {v 1, . . . , v k } is an orthogonal basis for W k . Define1k w1k 1k x p r o j x v k+++-= By the Orthogonal Decomposition Theorem, v k+1 is orthogonal to W k . Note that proj Wk x k+1 is in W k and hence also in W k+1. Since x k+1 is in W k+1, so is v k+1 (because W k+1 is a subspace and is closed under subtraction). Furthermore, 0v 1k ≠+ because x k+1 is not in W k = Span {x 1, . . . , x p }. Hence {v 1, . . . , v k } is an orthogonal set of nonzero vectors in the (k+1)-dismensional space W k+1. By the Basis Theorem, this set is an orthogonal basis for W k+1. Hence W k+1 = Span {v 1, . . . , v k+1}. When k + 1 = p, the process stops.9. Show and prove the diagonalization theorem. Solution:diagonalization theorem:If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal. PROOFLet v 1 and v 2 be eigenvectors that correspond to distinct eigenvalues, say, 1λand 2λ. T o show that 0v v 21=⋅, compute2T 12T 11211v )(A v v )v (λv v λ==⋅ Since v 1 is an eigenvector ()()2T12T T1Avv v A v ==)(221v v Tλ=2122T12v v λv v λ⋅==Hence ()0v v λλ2121=⋅-, but ()0λλ21≠-, so 0v v 21=⋅10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent. Solution:If v 1, . . . , v r are eigenvectors that correspond to distinct eignvalues λ1, . . . , λr of an n n ⨯ matrix A.Suppose {v 1, . . . , v r } is linearly dependent. Since v 1 is nonzero, Theorem, Characterization of Linearly Dependent Sets, says that one of the vectors in the set is linear combination of the preceding vectors. Let p be the least index such that v p +1 is a linear combination of he preceding (linearly independent) vectors. Then there exist scalars c 1, . . . ,c p such that 1p p p 11v v c v c +=+⋅⋅⋅+ (1) Multiplying both sides of (1) by A and using the fact that Av k = λk v k for each k, we obtain 111+=+⋅⋅⋅+p p p Av Av c Av c11111++=+⋅⋅⋅+p p p p p v v c v c λλλ (2) Multiplying both sides of (1) by 1+p λ and subtracting the result from (2), we have0)()(11111=-+⋅⋅⋅+-++p p p p c v c λλλλ (3) Since {v 1, . . . , v p } is linearly independent, the weights in (3) are all zero. But none of the factors 1+-p i λλ are zero, because the eigenvalues are distinct. Hence 0=i c for i = 1, . . . ,p. But when (1) says that 01=+p v , which is impossible. Hence {v 1, . . . , v r } cannot be linearly dependent and therefore must be linearly independent.。
(完整版)哈工大选修课LINEARALGEBRA试卷及答案,推荐文档
LINEAR ALGEBRAANDITS APPLICATIONS 姓名：易学号：成绩：1. Definitions(1) Pivot position in a matrix;(2) Echelon Form;(3) Elementary operations;(4) Onto mapping and one-to-one mapping;(5) Linearly independence.2. Describe the row reduction algorithm which produces a matrix in reduced echelon form.3. Find the matrix that corresponds to the composite transformation of a scaling by 0.3, 33⨯a rotation of , and finally a translation that adds (-0.5, 2) to each point of a figure.90︒4. Find a basis for the null space of the matrix 361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦5. Find a basis for Col of the matrixA 1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦6. Let and be positive numbers. Find the area of the region bounded by the ellipse a b whose equation is22221x y a b +=7. Provide twenty statements for the invertible matrix theorem.8. Show and prove the Gram-Schmidt process.9. Show and prove the diagonalization theorem.10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent.Answers:1. Definitions(1) Pivot position in a matrix:A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position.(2) Echelon Form:A rectangular matrix is in echelon form (or row echelon form) if it has the following three properties:1. All nonzero rows are above any rows of all zeros.2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.3. All entries in a column below a leading entry are zeros.If a matrix in a echelon form satisfies the following additional conditions, then it is in reduced echelon form (or reduced row echelon form):4. The leading entry in each nonzero row is 1.5. Each leading 1 is the only nonzero entry in its column.(3) Elementary operations:Elementary operations can refer to elementary row operations or elementary column operations.There are three types of elementary matrices, which correspond to three types of row operations (respectively, column operations):1. (Replacement) Replace one row by the sum of itself anda multiple of another row.2. (Interchange) Interchange two rows.3. (scaling) Multiply all entries in a row by a nonzero constant.(4) Onto mapping and one-to-one mapping:A mapping T : R n → R m is said to be onto R m if each b in R m is the image of at least one x in R n.A mapping T : R n → R m is said to be one-to-one if each b in R m is the image of at most one x in R n.(5) Linearly independence:An indexed set of vectors {V1, . . . ,V p} in R n is said to be linearly independent if the vector equationx 1v 1+x 2v 2+ . . . +x p v p = 0Has only the trivial solution. The set {V 1, . . . ,V p } is said to be linearly dependent if there exist weights c 1, . . . ,c p , not all zero, such thatc 1v 1+c 2v 2+ . . . +c p v p = 02. Describe the row reduction algorithm which produces a matrix in reduced echelon form.Solution:Step 1:Begin with the leftmost nonzero column. This is a pivot column. The pivot position is at the top.Step 2:Select a nonzero entry in the pivot column as a pivot. If necessary, interchange rows to move this entry into the pivot position.Step 3:Use row replacement operations to create zeros in all positions below the pivot.Step 4:Cover (or ignore) the row containing the pivot position and cover all rows, if any, above it. Apply steps 1-3 to the submatrix that remains. Repeat the process until there all no more nonzero rows to modify.Step 5:Beginning with the rightmost pivot and working upward and to the left, create zeros above each pivot. If a pivot is not 1, make it 1 by scaling operation.3. Find the matrix that corresponds to the composite transformation of a scaling by 0.3, 33⨯a rotation of , and finally a translation that adds (-0.5, 2) to each point of a figure.90︒Solution:If ψ=π/2, then sin ψ=1 and cos ψ=0. Then we have⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡110003.00003.01y x y x scale ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−→−110003.00003.010*******y x Rotate⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−−→−110003.00003.0100001010125.0010001y x Translate The matrix for the composite transformation is⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-10003.00003.0100001010125.0010001⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=10003.00003.0125.0001010⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=125.0003.003.004. Find a basis for the null space of the matrix361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦Solution:First, write the solution of A X=0 in parametric vector form:A ~ , ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---000023021010002001x 1-2x 2 -x 4+3x 5=0x 3+2x 4-2x 5=00=0The general solution is x 1=2x 2+x 4-3x 5, x 3=-2x 4+2x 5, with x 2, x 4, and x 5 free.⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡+--+=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡10203012010001222325425454254254321x x x x x x x x x x x x x x x xu v w=x 2u+x 4v+x 5w (1)Equation (1) shows that Nul A coincides with the set of all linear conbinations of u, v and w. That is, {u, v, w}generates Nul A. In fact, this construction of u, v and w automatically makes them linearly independent, because (1) shows that 0=x 2u+x 4v+x 5w only if the weightsx 2, x 4, and x 5 are all zero.So {u, v, w} is a basis for Nul A.5. Find a basis for Col of the matrix A 1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦Solution:A ~ , so the rank of A is 3.⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---07490012002300130001Then we have a basis for Col of the matrix:A U = , v = and w = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0001⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0013⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--07496. Let and be positive numbers. Find the area of the region bounded by the ellipse a b whose equation is22221x y a b+=Solution:We claim that E is the image of the unit disk D under the linear transformation Tdetermined by the matrix A=, because if u= , x=, and x = Au, then ⎥⎦⎤⎢⎣⎡b a 00⎥⎦⎤⎢⎣⎡21u u ⎥⎦⎤⎢⎣⎡21x x u 1 =and u 2 = a x 1bx 2It follows that u is in the unit disk, with , if and only if x is in E , with 12221≤+u u1)()(2221≤+x a x . Then we have{area of ellipse} = {area of T (D )}= |det A| {area of D} = ab π(1)2 = πab7. Provide twenty statements for the invertible matrix theorem.Let A be a square matrix. Then the following statements are equivalent. That is, for a n n ⨯given A, the statements are either all true or false.a. A is an invertible matrix.b. A is row equivalent to the identity matrix.n n ⨯c. A has n pivot positions.d. The equation Ax = 0 has only the trivial solution.e. The columns of A form a linearly independent set.f. The linear transformation x Ax is one-to-one.→g. The equation Ax = b has at least one solution for each b in R n .h. The columns of A span R n .i. The linear transformation x Ax maps R n onto R n .→j. There is an matrix C such that CA = I.n n ⨯k. There is an matrix D such that AD = I.n n ⨯l. A T is an invertible matrix.m. If , then 0A ≠()()T11T A A --=n. If A, B are all invertible, then (AB)* = B *A *o. T**T )(A )(A =p. If , then 0A ≠()()*11*A A--=q. ()*1n *A 1)(A --=-r. If , then ( L is a natural number )0A ≠()()L 11L A A--=s. ()*1n *A K)(KA --=-t. If , then 0A ≠*1A A1A =-8. Show and prove the Gram-Schmidt process.Solution:The Gram-Schmidt process:Given a basis {x 1, . . . , x p } for a subspace W of R n , define11x v = 1112222v v v v x x v ⋅⋅-= 222231111333v v v v x v v v v x x v ⋅⋅-⋅⋅-=. ..1p 1p 1p 1p p 2222p 1111p p p v v v v x v v v v x v v v v x x v ----⋅-⋅⋅⋅-⋅⋅-⋅⋅-=Then {v 1, . . . , v p } is an orthogonal basis for W. In additionSpan {v 1, . . . , v p } = {x 1, . . . , x p } for pk ≤≤1PROOFFor , let W k = Span {v 1, . . . , v p }. Set , so that Span {v 1} = Span p k ≤≤111x v ={x 1}.Suppose, for some k < p, we have constructed v 1, . . . , v k so that {v 1, . . . , v k } is an orthogonal basis for W k . Define1k w 1k 1k x proj x v k +++-=By the Orthogonal Decomposition Theorem, v k+1 is orthogonal to W k . Note that proj Wk x k+1 is in W k and hence also in W k+1. Since x k+1 is in W k+1, so is v k+1 (because W k+1 is a subspace and is closed under subtraction). Furthermore, because x k+1 is not in W k = Span {x 1, . . . , 0v 1k ≠+x p }. Hence {v 1, . . . , v k } is an orthogonal set of nonzero vectors in the (k+1)-dismensional space W k+1. By the Basis Theorem, this set is an orthogonal basis for W k+1. Hence W k+1 = Span {v 1, . . . , v k+1}. When k + 1 = p, the process stops.9. Show and prove the diagonalization theorem.Solution:diagonalization theorem:If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal.PROOFLet v 1 and v 2 be eigenvectors that correspond to distinct eigenvalues, say, and . To 1λ2λshow that , compute0v v 21=⋅ Since v 1 is an eigenvector2T 12T 11211v )(Av v )v (λv v λ==⋅()()2T 12T T 1Av v v A v ==)(221v v Tλ= 2122T 12v v λv v λ⋅==Hence , but , so ()0v v λλ2121=⋅-()0λλ21≠-0v v 21=⋅10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent.Solution:If v 1, . . . , v r are eigenvectors that correspond to distinct eignvalues λ1, . . . , λr of an n n ⨯matrix A.Suppose {v 1, . . . , v r } is linearly dependent. Since v 1 is nonzero, Theorem, Characterization of Linearly Dependent Sets, says that one of the vectors in the set is linear combination of the preceding vectors. Let p be the least index such that v p +1 is a linear combination of he preceding (linearly independent) vectors. Then there exist scalars c 1, . . . ,c p such that (1)1p p p 11v v c v c +=+⋅⋅⋅+Multiplying both sides of (1) by A and using the fact that Av k = λk v k for each k, we obtain111+=+⋅⋅⋅+p p p Av Av c Av c (2)11111++=+⋅⋅⋅+p p p p p v v c v c λλλMultiplying both sides of (1) by and subtracting the result from (2), we have1+p λ (3)0)()(11111=-+⋅⋅⋅+-++p p p p c v c λλλλSince {v 1, . . . , v p } is linearly independent, the weights in (3) are all zero. But none of thefactors are zero, because the eigenvalues are distinct. Hence for i = 1, . . . , p. 1+-p i λλ0=icBut when (1) says that , which is impossible. Hence {v 1, . . . , v r } cannot be linearly 01=+p v dependent and therefore must be linearly independent.。
大学数学实验智慧树知到课后章节答案2023年下黑龙江工程学院
大学数学实验智慧树知到课后章节答案2023年下黑龙江工程学院黑龙江工程学院绪论单元测试1.MATLAB是matrix&laboratory两个词的组合，意为矩阵工厂(矩阵实验室)。
（）答案:对2.MATLAB和、c语音并称为三大数学软件。
（）答案:错3.MATLAB的基本数据单位是矩阵.( )答案:对4.matlab的优点有以下哪些？( )答案:功能丰富的应用工具箱(如信号处理工具箱、通信工具箱等) ，为用户提供了大量方便实用的处理工具。
;具有完备的图形处理功能，实现计算结果和编程的可视化;;高效的数值计算及符号计算功能，能使用户从繁杂的数学运算分析中解脱出来;;友好的用户界面及接近数学的自然化语言，使学者易于学习和掌握;5.MATLAB只能做高数、线代、概率的计算题，不能做其他的。
（）答案:错6.matlab主要应用于以下哪些领域（）答案:工程计算、控制设计;金融建模设计与分析等领域;图像处理、信号检测;信号处理与通讯7.本课程仅仅学习了matlab的冰山一角。
（）答案:对8.matlab也加入了对C，FORTRAN，C++，JAVA的支持。
（）答案:对9.本课程只研究在matlab在高等数学、概率论、线性代数三大学科中的简单应用，姑且把matlab当做一个大型的计算器。
（）答案:对10.“mathematica 可能是数学界最好的狙击枪，但MATLAB能给你一座军火库。
“这句话形容了矩阵运算、数据可视化、GUI（用户界面）设计、甚至是连接其他编程语言，MATLAB都能轻松实现。
（）答案:对第一章测试1.定义变量的命令为syms （）答案:对2.正弦函数的函数命令为sinx （）答案:错3.余弦函数的函数命令为cos(x) （）答案:对4.matlab不区分输入法的半角和全角（）答案:错5.matlab不区分函数命令的大小写（）答案:错6.指数函数的函数命令为e^x （）答案:错7.floor(x)是指对x朝-∞方向取整（）答案:对8.plot命令可以画离散数据的函数曲线图（）答案:对9.ezplot命令可以画连续函数的曲线图（）答案:对10.正切函数tanx 可以直接用plot命令画0到pi之间的图形（）答案:错第二章测试1.计算极限：（）答案:1/exp(1/3)2.计算极限：（）答案:23.计算极限：（）答案:4.计算极限：（）答案:1/35.计算极限：（）答案:6.计算极限：（）答案:-17.计算极限：（）答案:8.计算极限：（）答案:1/29.计算极限：( )答案:10.计算极限：( )答案:1第三章测试1.求由参数方程确定的函数的导数。
(完整版)哈工大选修课LINEARALGEBRA试卷及答案,推荐文档
LINEAR ALGEBRAANDITS APPLICATIONS 姓名：易学号：成绩：1. Definitions(1) Pivot position in a matrix;(2) Echelon Form;(3) Elementary operations;(4) Onto mapping and one-to-one mapping;(5) Linearly independence.2. Describe the row reduction algorithm which produces a matrix in reduced echelon form.3. Find the matrix that corresponds to the composite transformation of a scaling by 0.3, 33⨯a rotation of , and finally a translation that adds (-0.5, 2) to each point of a figure.90︒4. Find a basis for the null space of the matrix361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦5. Find a basis for Col of the matrixA 1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦6. Let and be positive numbers. Find the area of the region bounded by the ellipse a b whose equation is22221x y a b +=7. Provide twenty statements for the invertible matrix theorem.8. Show and prove the Gram-Schmidt process.9. Show and prove the diagonalization theorem.10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent.Answers:1. Definitions(1) Pivot position in a matrix:A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position.(2) Echelon Form:A rectangular matrix is in echelon form (or row echelon form) if it has the following three properties:1. All nonzero rows are above any rows of all zeros.2. Each leading entry of a row is in a column to the right of the leading entry of the row above it.3. All entries in a column below a leading entry are zeros.If a matrix in a echelon form satisfies the following additional conditions, then it is in reduced echelon form (or reduced row echelon form):4. The leading entry in each nonzero row is 1.5. Each leading 1 is the only nonzero entry in its column.(3) Elementary operations:Elementary operations can refer to elementary row operations or elementary column operations.There are three types of elementary matrices, which correspond to three types of row operations (respectively, column operations):1. (Replacement) Replace one row by the sum of itself anda multiple of another row.2. (Interchange) Interchange two rows.3. (scaling) Multiply all entries in a row by a nonzero constant.(4) Onto mapping and one-to-one mapping:A mapping T : R n → R m is said to be onto R m if each b in R m is the image of at least one x in R n.A mapping T : R n → R m is said to be one-to-one if each b in R m is the image of at most one x in R n.(5) Linearly independence:An indexed set of vectors {V1, . . . ,V p} in R n is said to be linearly independent if the vector equationx 1v 1+x 2v 2+ . . . +x p v p = 0Has only the trivial solution. The set {V 1, . . . ,V p } is said to be linearly dependent if there exist weights c 1, . . . ,c p , not all zero, such thatc 1v 1+c 2v 2+ . . . +c p v p = 02. Describe the row reduction algorithm which produces a matrix in reduced echelon form.Solution:Step 1:Begin with the leftmost nonzero column. This is a pivot column. The pivot position is at the top.Step 2:Select a nonzero entry in the pivot column as a pivot. If necessary, interchange rows to move this entry into the pivot position.Step 3:Use row replacement operations to create zeros in all positions below the pivot.Step 4:Cover (or ignore) the row containing the pivot position and cover all rows, if any, above it. Apply steps 1-3 to the submatrix that remains. Repeat the process until there all no more nonzero rows to modify.Step 5:Beginning with the rightmost pivot and working upward and to the left, create zeros above each pivot. If a pivot is not 1, make it 1 by scaling operation.3. Find the matrix that corresponds to the composite transformation of a scaling by 0.3, 33⨯a rotation of , and finally a translation that adds (-0.5, 2) to each point of a figure.90︒Solution:If ψ=π/2, then sin ψ=1 and cos ψ=0. Then we have⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−→−⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡110003.00003.01y x y x scale ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−→−110003.00003.010*******y x Rotate⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-−−−→−110003.00003.0100001010125.0010001y x Translate The matrix for the composite transformation is⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-10003.00003.0100001010125.0010001⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=10003.00003.0125.0001010⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡--=125.0003.003.004. Find a basis for the null space of the matrix361171223124584A ---⎡⎤⎢⎥=--⎢⎥⎢⎥--⎣⎦Solution:First, write the solution of A X=0 in parametric vector form:A ~ , ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡---000023021010002001x 1-2x 2 -x 4+3x 5=0x 3+2x 4-2x 5=00=0The general solution is x 1=2x 2+x 4-3x 5, x 3=-2x 4+2x 5, with x 2, x 4, and x 5 free.⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡-+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡+--+=⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡10203012010001222325425454254254321x x x x x x x x x x x x x x x xu v w=x 2u+x 4v+x 5w (1)Equation (1) shows that Nul A coincides with the set of all linear conbinations of u, v and w. That is, {u, v, w}generates Nul A. In fact, this construction of u, v and w automatically makes them linearly independent, because (1) shows that 0=x 2u+x 4v+x 5w only if the weights x 2, x 4, and x 5 are all zero.So {u, v, w} is a basis for Nul A.5. Find a basis for Col of the matrixA 1332-9-2-22-822307134-111-8A ⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎣⎦Solution:A ~ , so the rank of A is 3.⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡---07490012002300130001Then we have a basis for Col of the matrix:A U = , v = and w = ⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0001⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡0013⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--07496. Let and be positive numbers. Find the area of the region bounded by the ellipse a b whose equation is22221x y a b+=Solution:We claim that E is the image of the unit disk D under the linear transformation Tdetermined by the matrix A=, because if u= , x=, and x = Au, then ⎥⎦⎤⎢⎣⎡b a 00⎥⎦⎤⎢⎣⎡21u u ⎥⎦⎤⎢⎣⎡21x x u 1 = and u 2 = a x 1bx 2It follows that u is in the unit disk, with , if and only if x is in E , with12221≤+u u1)()(2221≤+x a x . Then we have{area of ellipse} = {area of T (D )}= |det A| {area of D} = ab π(1)2 = πab7. Provide twenty statements for the invertible matrix theorem.Let A be a square matrix. Then the following statements are equivalent. That is, for a n n ⨯given A, the statements are either all true or false.a. A is an invertible matrix.b. A is row equivalent to the identity matrix.n n ⨯c. A has n pivot positions.d. The equation Ax = 0 has only the trivial solution.e. The columns of A form a linearly independent set.f. The linear transformation x Ax is one-to-one.→g. The equation Ax = b has at least one solution for each b in R n .h. The columns of A span R n .i. The linear transformation x Ax maps R n onto R n .→j. There is an matrix C such that CA = I.n n ⨯k. There is an matrix D such that AD = I.n n ⨯l. A T is an invertible matrix.m. If , then 0A ≠()()T11T A A --=n. If A, B are all invertible, then (AB)* = B *A *o. T**T )(A )(A =p. If , then 0A ≠()()*11*A A--=q. ()*1n *A 1)(A --=-r. If , then ( L is a natural number )0A ≠()()L 11L A A--=s. ()*1n *A K)(KA --=-t. If , then 0A ≠*1A A1A =-8. Show and prove the Gram-Schmidt process.Solution:The Gram-Schmidt process:Given a basis {x 1, . . . , x p } for a subspace W of R n , define11x v = 1112222v v v v x x v ⋅⋅-= 222231111333v v v v x v v v v x x v ⋅⋅-⋅⋅-=. ..1p 1p 1p 1p p 2222p 1111p p p v v v v x v v v v x v v v v x x v ----⋅-⋅⋅⋅-⋅⋅-⋅⋅-=Then {v 1, . . . , v p } is an orthogonal basis for W. In additionSpan {v 1, . . . , v p } = {x 1, . . . , x p } for pk ≤≤1PROOFFor , let W k = Span {v 1, . . . , v p }. Set , so that Span {v 1} = Span p k ≤≤111x v ={x 1}.Suppose, for some k < p, we have constructed v 1, . . . , v k so that {v 1, . . . , v k } is an orthogonal basis for W k . Define1k w 1k 1k x proj x v k +++-=By the Orthogonal Decomposition Theorem, v k+1 is orthogonal to W k . Note that proj Wk x k+1 is in W k and hence also in W k+1. Since x k+1 is in W k+1, so is v k+1 (because W k+1 is a subspace and is closed under subtraction). Furthermore, because x k+1 is not in W k = Span {x 1, . . . , 0v 1k ≠+x p }. Hence {v 1, . . . , v k } is an orthogonal set of nonzero vectors in the (k+1)-dismensional space W k+1. By the Basis Theorem, this set is an orthogonal basis for W k+1. Hence W k+1 = Span {v 1, . . . , v k+1}. When k + 1 = p, the process stops.9. Show and prove the diagonalization theorem.Solution:diagonalization theorem:If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal.PROOFLet v 1 and v 2 be eigenvectors that correspond to distinct eigenvalues, say,and . To 1λ2λshow that , compute0v v 21=⋅ Since v 1 is an eigenvector2T 12T 11211v )(Av v )v (λv v λ==⋅()()2T 12T T 1Av v v A v ==)(221v v Tλ= 2122T 12v v λv v λ⋅==Hence , but , so ()0v v λλ2121=⋅-()0λλ21≠-0v v 21=⋅10. Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent.Solution:If v 1, . . . , v r are eigenvectors that correspond to distinct eignvalues λ1, . . . , λr of an n n ⨯matrix A.Suppose {v 1, . . . , v r } is linearly dependent. Since v 1 is nonzero, Theorem, Characterization of Linearly Dependent Sets, says that one of the vectors in the set is linear combination of the preceding vectors. Let p be the least index such that v p +1 is a linear combination of he preceding (linearly independent) vectors. Then there exist scalars c 1, . . . ,c p such that (1)1p p p 11v v c v c +=+⋅⋅⋅+Multiplying both sides of (1) by A and using the fact that Av k = λk v k for each k, we obtain111+=+⋅⋅⋅+p p p Av Av c Av c (2)11111++=+⋅⋅⋅+p p p p p v v c v c λλλMultiplying both sides of (1) by and subtracting the result from (2), we have1+p λ (3)0)()(11111=-+⋅⋅⋅+-++p p p p c v c λλλλSince {v 1, . . . , v p } is linearly independent, the weights in (3) are all zero. But none of the factors are zero, because the eigenvalues are distinct. Hence for i = 1, . . . , p.1+-p i λλ0=i cBut when (1) says that , which is impossible. Hence {v 1, . . . , v r } cannot be linearly 01=+p v dependent and therefore must be linearly independent.。
最新数学理科选修4-4《极坐标》完整版-经典习题及详细答案
数学理科选修4-4第一讲《极坐标》习题一．选择题1．已知⎪⎭⎫ ⎝⎛-3,5πM ，下列所给出的不能表示点的坐标的是（ ） A ．⎪⎭⎫ ⎝⎛-3,5π B ．⎪⎭⎫ ⎝⎛34,5π C ．⎪⎭⎫ ⎝⎛-32,5π D ．⎪⎭⎫ ⎝⎛--35,5π2．点()3,1-P ，则它的极坐标是（ ） A ．⎪⎭⎫ ⎝⎛3,2π B ．⎪⎭⎫ ⎝⎛34,2π C ．⎪⎭⎫ ⎝⎛-3,2π D ．⎪⎭⎫ ⎝⎛-34,2π3．极坐标方程⎪⎭⎫⎝⎛-=θπρ4cos 表示的曲线是（ ） A ．双曲线 B ．椭圆 C ．抛物线 D ．圆4．圆)sin (cos 2θθρ+=的圆心坐标是（ ）A ．⎪⎭⎫ ⎝⎛4,1π B ．⎪⎭⎫ ⎝⎛4,21π C ．⎪⎭⎫ ⎝⎛4,2π D ．⎪⎭⎫⎝⎛4,2π5．在极坐标系中，与圆θρsin 4=相切的一条直线方程为（ ） A ．2sin =θρ B ．2cos =θρ C ．4cos =θρ D ．4cos -=θρ6、 已知点()0,0,43,2,2,2O B A ⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛--ππ则ABO ∆为（ ） A 、正三角形 B 、直角三角形 C 、锐角等腰三角形 D 、直角等腰三角形 7、)0(4≤=ρπθ表示的图形是（ ）A ．一条射线B ．一条直线C ．一条线段D ．圆8、直线αθ=与1)cos(=-αθρ的位置关系是（ ）A 、平行B 、垂直C 、相交不垂直D 、与有关，不确定9.两圆θρcos 2=,θρsin 2=的公共部分面积是（ ） A.214-πB.2-πC.12-πD.2π10.已知点1P 的球坐标是)4,,32(1πϕP ,2P 的柱坐标是)1,,5(2θP ,求21P P .A ．2B ．3C ．22D ．22二．填空题11．极坐标方程52sin 42=θρ化为直角坐标方程是12．圆心为⎪⎭⎫⎝⎛6,3πC ，半径为3的圆的极坐标方程为 13．已知直线的极坐标方程为22)4sin(=+πθρ，则极点到直线的距离是 14、在极坐标系中，点P ⎪⎭⎫⎝⎛611,2π到直线1)6sin(=-πθρ的距离等于____________。
哈工大matlab期末考试题试题及答案(95分)分解
春季学期MATLAB期末作业学院：机电工程学院专业：机械制造设计及其自动化学号：班号：姓名：2013年春季学期MATLAB 课程考查题姓名:学号：学院：机电学院专业：机械制造一、必答题：1.matlab常见的数据类型有哪些？各有什么特点?常量：具体不变的数字变量：会根据已知条件变化的数字字符串：由单引号括起来的简单文本复数：含有复数的数据2.MATLAB中有几种帮助的途径？（1）帮助浏览器：选择view菜单中的Help菜单项或选择Help菜单中的MATLAB Help菜单项可以打开帮助浏览器；（2）help命令：在命令窗口键入“help”命令可以列出帮助主题，键入“help 函数名”可以得到指定函数的在线帮助信息；（3）lookfor命令：在命令窗口键入“lookfor 关键词”可以搜索出一系列与给定关键词相关的命令和函数（4）模糊查询：输入命令的前几个字母，然后按Tab键，就可以列出所有以这几个字母开始的命令和函数。
注意：lookfor和模糊查询查到的不是详细信息，通常还需要在确定了具体函数名称后用help命令显示详细信息。
3.Matlab常见的哪三种程序控制结构及包括的相应的语句？1.顺序结构：数据输入A=input(提示信息,选项)数据输出disp(X)数据输出fprintf(fid,format,variables)暂停pause 或pause(n)2.选择结构：If语句：if expression （条件）statements1（语句组1）elsestatements2（语句组2）EndSwitch 语句：switch expression （表达式）case value1 （表达式1）statement1（语句组1）case value2 （表达式2）statement2（语句组2）... ...case valuem （表达式m）statementm（语句组m）otherwisestatement （语句组）end3.循环结构：For循环：for variable=expression(循环变量)statement（循环体）endWhile循环：while expression （条件<循环判断语句>）statement（循环体）end4.命令文件与函数文件的主要区别是什么？命令文件：不接受输入参数，没有返回值，基于工作空间中的数据进行操作，自动完成需要花费很多时间的多步操作时使用。
(完整版)哈工大matlab期末考试题试题及答案(95分)分解,推荐文档
春季学期MATLAB 期末作业学院：机电工程学院专业：机械制造设计及其自动化学号：班号：姓名：姓名:学号：学院：机电学院专业：机械制造2013 年春季学期MATLAB 课程考查题一、必答题：1.matlab 常见的数据类型有哪些？各有什么特点?常量：具体不变的数字变量：会根据已知条件变化的数字字符串：由单引号括起来的简单文本复数：含有复数的数据2.MATLAB 中有几种帮助的途径？（1）帮助浏览器：选择 view 菜单中的 Help 菜单项或选择 Help 菜单中的MATLAB Help 菜单项可以打开帮助浏览器；（2）help 命令：在命令窗口键入“help”命令可以列出帮助主题，键入“help 函数名”可以得到指定函数的在线帮助信息；（3）lookfor 命令：在命令窗口键入“lookfor关键词”可以搜索出一系列与给定关键词相关的命令和函数（4）模糊查询：输入命令的前几个字母，然后按 Tab 键，就可以列出所有以这几个字母开始的命令和函数。
注意：lookfor 和模糊查询查到的不是详细信息，通常还需要在确定了具体函数名称后用 help 命令显示详细信息。
3.Matlab 常见的哪三种程序控制结构及包括的相应的语句？1.顺序结构：数据输入A=input(提示信息,选项)数据输出disp(X)数据输出fprintf(fid,format,variables)暂停pause 或pause(n)2.选择结构：If 语句：if expression （条件）statements1（语句组1）elsestatements2（语句组2）EndSwitch 语句：switch expression （表达式）case value1 （表达式1）statement1（语句组1）case value2 （表达式2）statement2（语句组2）... ...case valuem （表达式m）statementm（语句组m）otherwisestatement （语句组）end3.循环结构：For 循环：for variable=expression(循环变量)statement（循环体）endWhile 循环：while expression （条件<循环判断语句>）statement（循环体）end4.命令文件与函数文件的主要区别是什么？命令文件：不接受输入参数，没有返回值，基于工作空间中的数据进行操作，自动完成需要花费很多时间的多步操作时使用。
哈工大物理试题及答案
哈工大物理试题及答案一、选择题（每题5分，共20分）1. 光在真空中的传播速度是：A. 300,000 km/sB. 1,000 km/sC. 3,000 km/sD. 3×10^8 m/s答案：D2. 根据牛顿第二定律，以下说法正确的是：A. 物体的质量越大，加速度越大B. 物体的质量越大，加速度越小C. 物体的质量越大，加速度不变D. 物体的质量越大，加速度越难改变答案：D3. 电磁波的波长与频率的关系是：A. 波长与频率成正比B. 波长与频率成反比C. 波长与频率无关D. 波长与频率成正比，但需要考虑介质的影响答案：B4. 根据热力学第一定律，以下说法正确的是：A. 能量可以被创造B. 能量可以被消灭C. 能量既不能被创造也不能被消灭D. 能量可以在不同形式间转换答案：D二、填空题（每题5分，共20分）1. 根据欧姆定律，电阻R、电压V和电流I之间的关系是：V =________。
答案：IR2. 光的折射定律，也称为斯涅尔定律，其公式为：n1 * sin(θ1) = ________。
答案：n2 * sin(θ2)3. 根据库仑定律，两个点电荷之间的力F与它们的电荷量q1和q2以及它们之间的距离r的关系是：F = ________。
答案：k * q1 * q2 / r^24. 根据牛顿第三定律，作用力和反作用力的关系是：它们大小相等、方向相反、作用在_______。
答案：不同的物体上三、简答题（每题10分，共20分）1. 简述麦克斯韦方程组的四个方程及其物理意义。
答案：麦克斯韦方程组包括高斯定律、高斯磁定律、法拉第电磁感应定律和安培环路定律。
高斯定律描述了电场线不闭合的性质，高斯磁定律表明磁场线也是闭合的，法拉第电磁感应定律说明了变化的磁场可以产生电场，安培环路定律描述了电流和磁场之间的关系。
2. 解释什么是波粒二象性，并举例说明。
答案：波粒二象性是指微观粒子如电子、光子等既表现出波动性，又表现出粒子性。
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71
2 4 5 8 4
5. Find a basis for Col A of thematrix
1 3 3 2 -9 -2 -2 2 -8 2 NhomakorabeaA
2
3
0
7
1
3
4 -1 11 -8
6. Let a and b be positive numbers. Find the area of the region bounded by the ellipse
(5) Linearly independence: An indexed set of vectors {V1, . . . ,Vp} in Rn is said to be linearly independent if the vector equation
a rotation of 90 , and finally a translation that adds (-0.5, 2) to each point of a figure.
4. Find a basis for the null space of the matrix
A
3 1
6 2
1 1 23
(2) Echelon Form: A rectangular matrix is in echelon form (or row echelon form) if it has the following three properties: 1. All nonzero rows are above any rows of all zeros. 2.Each leading entry of a row is in a column to the right of the leading entry of the row above it. 3. All entries in a column below a leading entry are zeros. If a matrix in a echelon form satisfies the following additional conditions, then it is in reduced echelon form (or reduced row echelon form): 4. The leading entry in each nonzero row is 1. 5. Each leading 1 is the only nonzero entry in its column.
(5) Linearly independence.
2. Describe the row reduction algorithm which produces a matrix in reduced echelonform.
3.Find the 3 3 matrix that corresponds to the composite transformation of a scaling by 0.3,
(4) Onto mapping and one-to-one mapping: A mapping T : Rn → Rm is said to be onto Rm if each b in Rm is the image of at least one x in Rn. A mapping T : Rn → Rm is said to be one-to-one if each b in Rm is the image of at most one x in Rn.
whose equation is
x2 y2 a2 b2 1
7. Provide twenty statements for the invertible matrix theorem. 8. Show and prove the Gram-Schmidt process. 9. Show and prove the diagonalization theorem. 10.Prove that the eigenvectors corresponding to distinct eigenvalues are linearly independent.
Answers: 1. Definitions (1) Pivot position in a matrix: A pivot position in a matrix A is a location in A that corresponds to a leading 1 in the reduced echelon form of A. A pivot column is a column of A that contains a pivot position.
LINEAR ALGEBRA AND
ITS APPLICATIONS
姓名：易 学号： 成绩：
1. Definitions
(1) Pivot position in a matrix;
(2) Echelon Form;
(3) Elementary operations;
(4) Onto mapping and one-to-one mapping;
(3) Elementary operations: Elementary operations can refer to elementary row operations or elementary column operations. There are three types of elementary matrices, which correspond to three types of row operations (respectively, column operations): 1. (Replacement) Replace one row by the sum of itself anda multiple of another row. 2. (Interchange) Interchange two rows. 3. (scaling) Multiply all entries in a row by a nonzero constant.