November 8, 2017 - Shawn Zhong

$Determinant and Area • |■8(a_1&a_2@b_1&b_2 )| = area of parallelogram with sides a=(█(a_1@a_2 )), b=(█(b_1@b_2 )) • Proof by graph • Proof ○ Area(A_1,A_2 )=signed area of parallelogram spanned by A_1, A_2 ○ If A_1→A_2 is counter-clockwise = area ○ If A_1→A_2 is clockwise =−area ○ Then Area(A_1,A_2 )=det⁡(A_1,A_2 ), because ○ Alternating § Area(A_1,A_2 )=−Area(A_2,A_1 ) § (by definition, same area, but different orientation) ○ Linearity(Homogeneous) § Area(t⋅A_1,A_2 )=t⋅Area(A_1,A_2 ) § (Easy to prove from picture) ○ Linearity(Additive) § Area(A+B,C)=Area(A,C)+Area(B,C) § If A,C is parallel, then □ Area(A,C)=0 § If A,C is independent , then □ Area(A+sC,C)=Area(A,C), ∀A,C § Let B=t⋅A+s⋅C, then □ Area(A+B,C) □ =Area(A+t⋅A+s⋅C,C) □ =Area(A+t⋅A,C) □ =(1+t)Area(A,C) □ =Area(A,C)+t⋅Area(A,C) □ =Area(A,C)+Area(t⋅A,C) □ =Area(A,C)+Area(t⋅A+s⋅C,C) □ =Area(A,C)+Area(B,C) § Therefore Area(A+B,C)=Area(A,C)+Area(B,C) ○ Uniqueness Theorem § Area(A,B) § =det⁡〖(A,B)⋅Area(I_1,I_2 )〗 § =det⁡〖(A,B)⋅Area(unit square)〗 § =det⁡(A,B) Determinant and Volume • det⁡〖(A,B,C)=signed volume〗 of parallelepiped spanned by A,B,C Inverse of a Matrix • Setup ○ T:Rn→Rn linear ○ T has a matrix m(T)=[■8(T_11&⋯&T_1n@⋮&⋱&⋮@T_n1&⋯&T_nn )] • The following statements are equivalent ○ N(T)={0} ○ T is injective ○ T is one\-to\-one ○ T is bijective § because T:Rn→Rn § dim⁡N(T)+dim⁡range(T)=dim⁡〖Rn 〗 § ⇒dim⁡range(T)=n § ⇒R(T)=Rn ○ There is a map S:Rn→Rn with ST=TS=I • Find the inverse of 2×2 matrix ○ T=[■8(1&3@2&5)] ○ Find T^(−1), i.e. solve Tx=y ○ Note: Tx=y⟺x=T^(−1) y ○ Normal version § [■8(1&3@2&5)][█(x_1@x_2 )]=[█(y_1@y_2 )] § {█(x_1+3x_2=1⋅y_1+0⋅y_2@2x_2+5x_2=0⋅y_1+1⋅y_2 )┤ § ⇒{█(x_1=−5y_1+3y_2@x_2=2y_1−y_2 )┤ § ⇒x=T^(−1) y § where T^(−1)=[■8(−5&3@2&−1)] ○ Shorthand § [T│I] § ~[■8(1&3@2&5) │ ■8(1&0@0&1)] § ~[■8(1&3@0&−1) │ ■8(1&0@−2&1)] § ~[■8(1&0@0&−1) │ ■8(−5&3@−2&1)] § ~[■8(1&0@0&1) │ ■8(−5&3@2&−1)] § ~[I│T^(−1) ] § Therefore T^(−1)=[■8(−5&3@2&−1)] Minors and Cofactors • Theorem ○ |■8(a_11&a_12&⋯&a_1n@⋮&⋮&⋮&⋮@a_k1&a_k2&…&a_kn@⋮&⋮&⋮&⋮@a_n1&a_n2&⋯&a_nn )|=a_k1 C_k1+a_k2 C_k2+…+a_kn C_kn ○ C_kl=cofactor matrix • Cofactor Matrix C_kl=(−1)^(k+l) {█((n−1)×(n−1) determinant obtained@by deleting row k and column l @from the original determinant)} • Example ○ |■8(1&7&2@4&π&−1@3&ln⁡2&2)| ○ =3×|■8(7&2@π&−1)|−ln⁡2 |■8(1&2@4&−1)|+2|■8(1&7@4&π)| ○ =3×(−7−2π)−ln⁡2×(−9)+2×(π−28) ○ =−77−4π+9ln⁡2 • Matrix Multiplication ○ Let P=[■8(a_11&a_12&⋯&a_1n@a_21&a_22&…&a_2n@⋮&⋮&⋱&⋮@a_n1&a_n2&⋯&a_nn )][■8(C_11&C_21&⋯&C_n1@C_12&C_22&…&C_n2@⋮&⋮&⋱&⋮@C_1n&C_2n&⋯&C_nn )] ○ P_11=a_11 C_11+a_12 C_12+…+a_1n C_1n=det⁡A ○ P_21=a_21 C_11+a_21 C_12+…++a_21 C_1n=0 ○ Because we have two equal row ○ Therefore P=det⁡A [■(1&&@&⋱&@&&1)] •$

Shawn Zhong

Shawn Zhong

Math 375 – 11/7

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