Shawn Zhong

Shawn Zhong

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Shawn Zhong

钟万祥
  • Tutorials
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Home / 2017 / Page 13

Math 375 – 11/6

  • Nov 07, 2017
  • Shawn
  • Math 375
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Find the Inverse of Matrix • Gauss-Jordan Elimination ○ (A│I)~(I│A^(−1) ) • Example ○ (■8(1&2&4@3&5&−7@0&0&1)│■(1&&@&1&@&&1))→(■8(1&2&4@0&−1&−13@0&0&1)│■(1&0&0@−3&1&0@0&0&1)) ○ →(■8(1&2&4@0&−1&−13@0&0&1)│■(1&0&0@−3&1&0@0&0&1))→(■8(1&2&4@0&−1&0@0&0&1)│■(1&0&0@−3&1&13@0&0&1)) ○ →(■8(1&2&0@0&−1&0@0&0&1)│■(1&0&−4@−3&1&13@0&0&1))→(■8(1&0&0@0&−1&0@0&0&1)│■(−5&2&22@−3&1&13@0&0&1)) ○ →(■8(1&0&0@0&1&0@0&0&1)│■(−5&2&22@3&−1&−13@0&0&1)) ○ Therefore (■8(1&2&4@3&5&−7@0&0&1))^(−1)=(■(−5&2&22@3&−1&−13@0&0&1)) Question 1 • Recall that the determinant is a polynomial in the entries of the matrix. • Find the coefficient of t^3 in the following polynomial |■8(2&3&−7&t@5&t&a&b@t&−1&0&55@1/2&3&c&−π)| • Answer: By cofactor expansion, the coefficient is c Question 2 • Suppose A is an orthogonal matrix, meaning A is invertible and A^(−1)=A^T • What possible value could the determinant of A have? • Answer: ○ |A^(−1) |=|A^T | ○ ⇒1/|A| =|A| ○ ⇒|A|=±1 Question 3 • Let V be the vector space of all (real) polynomials of degree 2 or less. • Using the basis 1,x,x^2, find the matrix of the linear map T:V→V given by • (Tf)(x)=f(x+2) for all f∈V and x∈R • Answer: ○ T(1)=1 ○ T(x)=2+x ○ T(x^2 )=4+4x+x^2 ○ ⇒M(T)=■(&■8(1&x&x^2 )@■8(1@x@x^2 )&(■8(1&2&4@0&1&4@0&0&1)) ) Question 4 • Let x,y,z,w be real numbers. • Compute the determinant of the following matrix • Answer: ○ |■8(1&x&x^2&x^3@1&y&y^2&y^3@1&z&z^2&z^3@1&w&w^2&w^3 )|=(w−z)(w−y)(w−x)(z−y)(z−x)(y−x)
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Math 375 – Homework 9

  • Nov 07, 2017
  • Shawn
  • Math 375
  • No comments yet
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Math 375 – 11/2

  • Nov 03, 2017
  • Shawn
  • Math 375
  • No comments yet
Uniqueness Theorem • Theorem ○ Suppose f(A_1,…,A_n ) is a function of A_1,…,A_n∈Rn ○ That satisfies Linearity and Alternating § f(B+C,A_2,…,A_n )=f(B,A_2,…,A_n )+f(C,A_2,…,A_n ) § f(t⋅A_1,A_2,…,A_n )=t⋅f(A_1,A_2,…,A_n ) § f(A_1,A_2,…,A_i,..,A_j,…A_n )=−f(A_1,A_2,…,A_j,..,A_i,…A_n ) ○ Then f(A_1,…,A_n )=det⁡(A_1,…,A_n )⋅f(I_1,…,I_n ) where § I_1=[1,0,0,…,0] § I_2=[0,1,0,…,0] § ⋮ § I_n=[0,0,0,…,1] • Proof ○ f(A_1,…,A_n ) ○ =f(a_11 I_1+a_12 I_2+…+a_1n I_n,…,a_n1 I_1+a_n2 I_2+…+a_nn I_n ) ○ =∑_█(1≤i_1,i_2,…,i_n≤n@all different)^n▒a_(1i_1 ) a_(2i_2 )…a_(ni_n )⋅f(I_(i_1 ),I_(i_2 ),…,I_(i_n ) ) ○ =∑_█(1≤i_1,i_2,…,i_n≤n@all different)^n▒a_(1i_1 ) a_(2i_2 )…a_(ni_n )⋅sign(i_1,…,i_n )⋅f(I_1,I_2,…,I_n ) ○ =f(I_1,I_2,…,I_n )⋅∑_█(1≤i_1,i_2,…,i_n≤n@all different)^n▒a_(1i_1 ) a_(2i_2 )…a_(ni_n )⋅sign(i_1,…,i_n ) ○ =f(I_1,I_2,…,I_n )⋅det⁡(A_1,…,A_n ) • Example ○ |■8(A_(k×k)&0@C_(l×k)&B_(l×l) )|=|■(a_11&…&a_1k&&&@⋮&⋱&⋮&&&@a_k1&…&a_kk&&&@c_11&…&c_1k&b_11&…&b_1l@⋮&⋱&⋮&⋮&⋱&⋮@c_l1&…&c_lk&b_l1&…&b_ll )|=det⁡A⋅det⁡B ○ Consider a function f that satisfies the Uniqueness Theorem § f((A_1 ) ̅+(A_1 ) ̅ ̅,A_2,…,A_n )=f((A_1 ) ̅,A_2,…,A_n )+d((A_1 ) ̅ ̅,A_2,…,A_n ) § f(tA_1,A_2,…,A_n )=f(A_1,A_2,…,A_n ) § f(A_1,A_2,…,A_i,..,A_j,…A_n )=f(A_1,A_2,…,A_j,..,A_i,…A_n ) ○ Let f_BC (A_1,…,A_k )=|■8(A_(k×k)&0@C_(l×k)&B_(l×l) )| with B,C fixed, and A as variable § f_BC (A_1,…,A_k ) § =det⁡(A_1,…,A_k ) f_BC (I_1,…,I_k ) § =det⁡A⋅|■(1&&&&&@&⋱&&&&@&&1&&&@c_11&…&c_1k&b_11&…&b_1l@⋮&⋱&⋮&⋮&⋱&⋮@c_l1&…&c_lk&b_l1&…&b_ll )| § =det⁡A⋅|■(1&&&&&@&⋱&&&&@&&1&&&@&&&b_11&…&b_1l@&&&⋮&⋱&⋮@&&&b_l1&…&b_ll )| § =det⁡A⋅|■(I&@&B)| ○ Let g(B)=|■(I&@&B)| that satisfies the Uniqueness Theorem § g(B)=det⁡B⋅g(I)=det⁡B⋅|■(1&&@&⋱&@&&1)|=det⁡B ○ Therefore |■8(A_(k×k)&0@0&B_(l×l) )|=det⁡A⋅det⁡B Properties of Determinant • det⁡〖(AB)=det⁡A⋅det⁡B 〗 (where A_(n×n), B_(n×n)) ○ det⁡A⋅det⁡B ○ =|■8(A&0@I&B)| ○ =|■8(0&−AB@I&B)| ○ =(−1)^(n^2 ) |■8(I&B@0&−AB)| ○ =(−1)^(n^2 ) det⁡I⋅det⁡(−AB) ○ =(−1)^(n^2 )⋅det⁡(−AB) ○ =(−1)^(n^2 ) (−1)^n det⁡(AB) ○ =(−1)^(n^2+n) det⁡(AB) ○ =det⁡(AB) • Power of Determinants ○ det⁡(A^n )=det⁡(A⋅A…A)=det⁡〖(A)⋅〗 det⁡(A)…det⁡(A)=(det⁡A )^n • Determinant of Inverse ○ If A has an inverse(A^(−1)), and det⁡A≠0, then ○ A^(−1) A=I ○ ⇒det⁡〖A^(−1) 〗⋅det⁡A=det⁡I=1 ○ ⇒det⁡〖A^(−1)=1/det⁡A 〗 • Matrix Product and Determinant ○ |■8(A_(n×n)&0@I&B_(n×n) )| ○ =|■(a_11&…&a_1n&&&@⋮&⋱&⋮&&&@a_n1&…&a_nn&&&@1&&&b_11&…&b_1n@&⋱&&⋮&⋱&⋮@&&1&b_n1&…&b_nn )| ○ =|■(0&…&a_1n&−a_11 b_11&…&−a_11 b_1n@⋮&⋱&⋮&⋮&⋱&⋮@0&…&a_nn&−a_n1 b_11&…&−a_n1 b_1n@1&…&0&b_11&…&b_1n@⋮&⋱&⋮&⋮&⋱&⋮@0&…&1&b_n1&…&b_n )|=… ○ =|■(0&…&0&−∑_(i=1)^n▒〖a_1i b_i1 〗&…&−∑_(i=1)^n▒〖a_1i b_in 〗@⋮&⋱&⋮&⋮&⋱&⋮@0&…&0&−∑_(i=1)^n▒〖a_ni b_i1 〗&…&−∑_(i=1)^n▒〖a_ni b_in 〗@1&…&0&b_11&…&b_1n@⋮&⋱&⋮&⋮&⋱&⋮@0&…&1&b_n1&…&b_nn )| ○ =|■8(0&−AB@I&B)|
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Math 375 – Homework 8

  • Nov 02, 2017
  • Shawn
  • Math 375
  • 1 comment
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Math 375 – 11/1

  • Nov 02, 2017
  • Shawn
  • Math 375
  • No comments yet
Understanding of Determinant in Terms of Volumes • The volume of this parallelepiped is the absolute value of the determinant of the matrix formed by the rows constructed from the vectors r1, r2, and r3. • Negative determinant = flip the original image
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