David Wagner 2009/09/15 19:52

EGR 6013 Homework Set 1 (6-11)

Problem 6.

[A][A] = 0 = [a_ik][a_kj] = sum{k}{}{a_ik a_kj}. =delim{[}{matrix{2}{2}{a_11 a_12 a_21 a2_2  }}{]}
delim{[}{matrix{2}{2}{a_11 a_12 a_21 a_22   }}{]} = delim{[}{matrix{2}{2}{
{a_11^2 + a_12 a_21}  {a_11 a_12 + a_12 a_22}
{a_21 a_11 + a_22 a_21}  {a_21 a_12 +a_22^2 }
}}{]}
 =
delim{[}{matrix{2}{2}{0 0 0 0 }}{]}

  • a_11 = a_22= d

a_21 + a_21 = a_12 + a_12

  • a_21 = a_12 = c

delim{[}{matrix{2}{2}{d c d c }}{]}

Problem 7.

Show [E] = [A][C] = delim{[}{matrix{2}{2}{
{B_11 D_11 + B_12 D_21}  {B_11 D_12 + B_12 D_22}
{B_21 D_11 + B_22 D_21}  {B_21 D_12 + B_22 D_22}
}}{]}

[A] = delim{[}{matrix{2}{2}{B_11 B_12 B_21 B2_2  }}{]}, [C] = delim{[}{matrix{2}{2}{D_11 D_12 D_21 D2_2  }}{]}, [E] = delim{[}{matrix{2}{2}{F_11 F_12 F_21 F2_2  }}{]},

B_11 = delim{[}{matrix{2}{2}{a_11 a_12 a_21 a2_2 }}{]},
   B_12 = delim{[}{matrix{1}{2}{a_13 a_23}}{]}

B_21 = delim{[}{matrix{2}{1}{a_31 a_32}}{]}, B_22 = delim{[}{matrix{1}{1}{a_33~}}{~]}.

D_11 = delim{[}{matrix{2}{2}{c_11 c_12 c_21 c_22  }}{]}, D_12 = delim{[}{matrix{1}{2}{c_13 c_23}}{]}

D_21 = delim{[}{matrix{2}{1}{c_31 c_32}}{]}, D_22 = delim{[}{matrix{1}{1}{c_33~}}{~]}.

F_11 = delim{[}{matrix{2}{2}{e_11 e_12 e_21 e_22  }}{]}, F_12 = delim{[}{matrix{1}{2}{e_13 e_23}}{]}

F_21 = delim{[}{matrix{2}{1}{e_31 e_32}}{]}, F_22 = delim{[}{matrix{1}{1}{e_33~}}{~]}.

[E] = delim{[}{matrix{2}{2}{
{B_11 D_11 + B_12 D_21}  {B_11 D_12 + B_12 D_22}
{B_21 D_11 + B_22 D_21}  {B_21 D_12 + B_22 D_22}
}}{]} =[E] = delim{[}{matrix{2}{2}{
{delim{[}{matrix{2}{2}{a_11 a_12 a_21 a2_2 }}{]} delim{[}{matrix{2}{2}{c_11 c_12 c_21 c_22  ~}}{~]}
 + delim{[}{matrix{1}{2}{a_13 a_23}}{]} delim{[}{matrix{2}{1}{c_31 c_32~}}{~]}}

  {delim{[}{matrix{2}{2}{a_11 a_12 a_21 a2_2 }}{]} delim{[}{matrix{1}{2}{c_13 c_23~}}{~]}
 + delim{[}{matrix{1}{2}{a_13 a_23}}{]} delim{[}{matrix{1}{1}{c_33~}}{~]}}

{delim{[}{matrix{2}{1}{a_31 a_32}}{]} delim{[}{matrix{2}{2}{c_11 c_12 c_21 c_22  ~}}{~]}
 + delim{[}{matrix{1}{1}{a_33}}{]} delim{[}{matrix{2}{1}{c_31 c_32~}}{~]}}

  {delim{[}{matrix{2}{1}{a_31 a_32}}{]} delim{[}{matrix{1}{2}{c_13 c_23~}}{~]}
 + delim{[}{matrix{1}{1}{a_33}}{]} delim{[}{matrix{1}{1}{c_33~}}{~]}}
}}{]}

Continuing to substitute the results of these multiplications leads directly to the long form of the series definition of this matrix product.

sum{k}{}{a_ik c_kj} = [a_ik][c_kj] = [A][C].

Problem 10.

Prove |[A][B]| = |[A]||[B]|.

[A][B] =delim{[}{matrix{2}{2}{ {a_11 b_11 + a_12 b_21}  {a_11 b_12 + a_12 b_22} {a_21 b_11 + a_22 b_21}  {a_21 b_12 +a_22 b_22 } }}{]}

|[A][B]| = delim{|}{matrix{2}{2}{ {a_11 b_11 + a_12 b_21}  {a_11 b_12 + a_12 b_22} {a_21 b_11 + a_22 b_21}  {a_21 b_12 +a_22 b_22 } }}{|} = ({a_11 b_11 + a_12 b_21}) ( {a_21 b_12 +a_22 b_22 } )
- ({a_11 b_12 + a_12 b_22}) ({a_21 b_11 + a_22 b_21} ) = a_11 b_11 a_21 b_12 + a_11 b_11 a_22 b_22 + a_12 b_21 a_21 b_12 + a_12 b_21 a_22 b_22
-a_11 b_12 a_21 b_11 - a_11 b_12 a_22 b_21 - a_12 b_22 - a_21 b_11 -a_12 b_22 a_22 b_21

= a_11 a_22 b_11 b_22
- a_11 a_22 b_12 b_21
- a_12 a_21b_11 b_22
+a_12 a_21 b_12 b_21

= (a_11 a_22 - a_12 a_21)(b_11 b_22 - b_12 b_21) = |A||B|

Problem 11.

delim{[}{ matrix{3}{3}{
 3 1 {-lambda}
 4 {-2} {-3}  
 {2 lambda} 4 lambda
} }{]}
delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace}
=
delim{lbrace}{ matrix{3}{1}{ 0 0 0 } }{rbrace}

delim{[}{ matrix{3}{3}{
 3 1 {0}
 4 {-2} {-3}  
 0 4 0
} }{]}
delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace}
=
lambda 
delim{[}{ matrix{3}{3}{
 0 0 {1}
 0 0 0  
 {-2} 0 {-1}
} }{]}

delim{lbrace}{ matrix{3}{1}{ x_1 x_2 x_3 } }{rbrace}

x_2 is arbitrary and has no effect on λ.

delim{[}{ matrix{2}{2}{
 3  {0}
 0  0
} }{]}
delim{lbrace}{ matrix{2}{1}{ x_1 x_3 } }{rbrace}
=
lambda delim{[}{ matrix{2}{2}{
 0  {1}
 {-2 }  {-1}
} }{]}
delim{lbrace}{ matrix{2}{1}{ x_1 x_3 } }{rbrace}

delim{[}{ matrix{2}{2}{
 {-1/2}  {-1/2}
 1  0
} }{]}
delim{[}
{ matrix{2}{2}{
 3  {0}
 0  0
} }{]}
delim{lbrace}{ matrix{2}{1}{ x_1 x_3 } }{rbrace}
=

lambda
delim{[}{ matrix{2}{2}{
 {-1/2}  {-1/2}
 1  0
} }{]}

delim{[}{ matrix{2}{2}{
 0  {1}
 {-2}  {-1}
} }{]}
delim{lbrace}{ matrix{2}{1}{ x_1 x_3 } }{rbrace}

delim{[}{ matrix{2}{2}{
 {-3/2}  {0}
 3  0
} }{]}
delim{lbrace}{ matrix{2}{1}{ x_1 x_3 } }{rbrace}
=
lambda
delim{lbrace}{ matrix{2}{1}{ x_1 x_3 } }{rbrace}

Now it seems x_3 is messed up, and all that is left is λ=-3/2 and the trivial solution λ=0. For λ=-3/2, x_1=-C_3/2, x_3=C_3.

Going back to equation 2, 4 x_1 - 2x_2-3x_3=0= -2 C_3 - 2x_2-3C_3=02x_2=-2 C_3 -3C_3=0x_2={-5/2}C_3


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