Table of Contents
EGR 6013 Homework Set 1

— David Wagner 2009/09/13 20:20

Problem 1

1. Illustrate the use of the Gauss-Jordan reduction in obtaining the general solution of each of the following sets of equations:

√(a)

  • x_1 + 2 x_2 + 2x_3 =1,
  • 2x_1 + 2x_2 + 3x_3 = 3,
  • x_1 - x_2 + 3x_3 = 5.

delim{[}{matrix{3}{3}{ {1} {2} {2} {2} {2} {3} {1} {-1} {3} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3} }{rbrace} = delim{lbrace}{matrix{3}{1}{1 3 5} }{rbrace} = delim{[}{matrix{3}{3}{ {1} {2} {2} {0} {-2} {-1} {0} {-3} {1} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3} }{rbrace} = delim{lbrace}{matrix{3}{1}{1 1 4} }{rbrace}

Second Step.

=delim{[}{matrix{3}{3}{ {1} {2} {2} {0} {1} {1/2} {0} {-3} {1} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3} }{rbrace} = delim{lbrace}{matrix{3}{1}{1 {-1/2} {4}} }{rbrace} = delim{[}{matrix{3}{3}{ {1} {2} {2} {0} {1} {1/2} {0} {0} {5/2} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3} }{rbrace} = delim{lbrace}{matrix{3}{1}{1 {-1/2} 5/2} }{rbrace}

  • x_3=1
  • x_2 + {1/2} = {-1/2} right x_2=-1
  • x_1 -2 +2 = 1 right x_1=1

√(b)

  • 2x_1 + 0 x_2 +x_3 = 4,
  • x_1 - 2x_2 +2x_3 = 7,
  • 3x_1 + 2_x_2+0x_3 = 1.

delim{[}{matrix{3}{3}{ {2} {0} {1} {1} {-2} {2} {3} {2} {0} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3}}{rbrace} = delim{lbrace}{matrix{3}{1}{4 7 1}}{rbrace} = delim{[}{matrix{3}{3}{ {1} {0} {1/2} {1} {-2} {2} {3} {2} {0} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3}}{rbrace} = delim{lbrace}{matrix{3}{1}{2 7 1}}{rbrace} = delim{[}{matrix{3}{3}{ {1} {0} {1/2} {0} {-2} {3/2} {0} {2} {-3/2} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3}}{rbrace} = delim{lbrace}{matrix{3}{1}{2 5 -5}}{rbrace}

Second Step.

= delim{[}{matrix{3}{3}{ {1} {0} {1/2} {0} {1} {-3/4} {0} {2} {-3/2} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3}}{rbrace} = delim{lbrace}{matrix{3}{1}{2 {5/2} {-5}}}{rbrace} = delim{[}{matrix{3}{3}{ {1} {0} {1/2} {0} {1} {-3/4} {0} {0} {0} } }{]} delim{lbrace}{matrix{3}{1}{x_1 x_2 x_3}}{rbrace} = delim{lbrace}{matrix{3}{1}{2 {-5/2} {0}}}{rbrace}

  • x_3 is unconstrained. Let x_3=C_3.
  • x_2 ={-5/2}+{3/4}C_3
  • x_1 = 2-{1/2}C_3

Problem 2

2. Evaluate the following matrix products:

√(a) delim{[}{matrix{2}{2}{1 2 1 {-1}} }{]} delim{[}{ matrix{2}{3}{1 0 {1~} 1 {-1} {1~} } }{]}

matrix{2}{2}{ ~ {matrix{2}{3}{1 0 {1~} 1 {-1} {1~} } } {matrix{2}{2}{1 2 1 {-1}}} { {[}{ matrix{3}{3}{{1+2} {0-2} {1+2} {1-1} {0+1} {1-1} }{]} } } }

=delim{[}{matrix{2}{3}{3 {-2} {3} 0 1 0}}{]}

√(b) delim{[}{matrix{2}{2}{1 2 3 {6}} }{]} delim{[}{ matrix{2}{2}{6 {-2} {-3} 1} }{]}=

matrix{2}{2}{ {~} { matrix{2}{2}{6 {-2} {-3} 1} } { matrix{2}{2}{1 2 3 {6} } } { delim{[}{matrix{2}{2}{ {6-6} {-2+2} {18-18} {-6+6} } } {]} } }

=delim{[}{matrix{2}{2}{0 0 0 0}}{]}

é delim{[}{matrix{1}{4}{a_1 a_2 cdots a_n}} {]} delim{lbrace}{matrix{4}{1}{b_1 b_2 vdots b_n} }{rbrace}=a_1 b_1 + a_2 b_2 + cdots +a_n b_n

√(d) delim{lbrace}{matrix{4}{1}{b_1 b_2 vdots b_n} }{rbrace} delim{[}{matrix{1}{4}{a_1 a_2 cdots a_n} }{]}=

matrix{2}{2}{ {~} { matrix{1}{4}{a_1 a_2 cdots a_n} } { matrix{4}{1}{b_1 b_2 vdots b_n} } { delim{[}{matrix{4}{4}{ {a_1 b_1} {a_2 b_1} {cdots} {a_n b_1} {a_1 b_2} {a_2 b_2} {cdots} {a_n b_2} {vdots} {~} {ddots} {vdots} {a_1 b_n} {a_2 b_n} {cdots} {a_n b_n} } } {]} } }

√(e) delim{[}{matrix{2}{2}{c_1 0 0 c_2} }{]} delim{[}{ matrix{2}{2}{a_11 a_12 a_21 a_22} }{]}= matrix{2}{2}{ {~} { matrix{2}{2}{a_11 a_12 a_21 a_22} } { matrix{2}{2}{c_1 0 0 c_2} } { delim{[} { matrix{2}{2}{ {a_11 c_1 +0} {a_12 c1+0 } {0+a_21 c_2} {0+a_22 c_2} } } {]} } }

=delim{[}{ matrix{2}{2}{ {a_11 c_1} {a_12 c1} {a_21 c_2} {a_22 c_2} } } {]}

√(f) delim{[}{ matrix{2}{2}{a_11 a_12 a_21 a_22} }{]} delim{[}{matrix{2}{2}{c_1 0 0 c_2} }{]}= matrix{2}{2}{ {~} { matrix{2}{2}{c_1 0 0 c_2} } { matrix{2}{2}{a_11 a_12 a_21 a_22} } { delim{[}{matrix{2}{2}{ {a_11 c_1 +0} {0+a_12 c_2 } {c_1 a_21+0} {0+a_22 c_2} } } {]} } }

=delim{[}{matrix{2}{2}{ {a_11 c_1 +0} {0+a_12 c_2 } {c_1 a_21+0} {0+a_22 c_2} } } {]}

Problem 3.

3. If the product A(B C) is defined, show that it is of the form
[a_ir] ( [b_rs] [c_sj] ) = delim{[}{sum{r}{}{ sum{s}{}{a_ir b_rs c_sj} } }{]}
and deduce that then A(B C) = (A B) C.

First, [a_ir] ( [b_rs] [c_sj] ) = [a_ir]delim{[}{sum{s}{}{b_rs c_sj} }{]}.

Let [d_rj] = [b_rs] [c_sj] = sum{s}{}{b_rs c_sj} and [e_is] = [a_ir] [b_rs] = sum{r}{}{a_ir b_rs}.

So, [a_ir] [b_rs] + [b_rs] [c_sj] = sum{r}{}{a_ir b_rs} +sum{s}{}{b_rs c_sj}.

Now, all [a_ir] is constant over the index s, so [a_ir] ( [b_rs] [c_sj] ) = delim{[}{sum{s}{}{ [a_ir] b_rs c_sj} }{]} = delim{[}{sum{s}{}{ a_ir b_rs c_sj} }{]}

[a_ir]is constant over the sum indexed by s, so
delim{[}{sum{r}{}{ sum{s}{}{a_ir b_rs c_sj} } }{]} = delim{[}{sum{r}{}{a_ir sum{s}{}{ b_rs c_sj} } }{]}

Repeating this shows
=delim{[}{sum{r}{}{a_ir sum{s}{}{ b_rs c_sj} } }{]} =delim{[}{sum{s}{}{a_ir c_sj sum{r}{}{ b_rs} } }{]} =delim{[}{a_ir sum{s}{}{c_sj sum{r}{}{ b_rs} } }{]}
which is
=[a_ir]delim{[}{ sum{s}{}{ sum{r}{}{ b_rs c_sj} } }{]}=[a_ir]( [b_rs] [c_sj] )

Going back to the original equation and messing with it in a slightly different manner,
delim{[}{sum{r}{}{ sum{s}{}{a_ir b_rs c_sj} } }{]} = delim{[}{sum{s}{}{c_sj sum{r}{}{a_ir b_rs } } }{]} = delim{[}{sum{r}{}{c_sj a_ir sum{s}{}{ b_rs } } }{]} = [c_sj]delim{[}{ sum{r}{}{ a_ir sum{s}{}{ b_rs } } }{]} = [c_sj]delim{[}{ sum{s}{}{ sum{r}{}{a_ir b_rs } } }{]}

Problem 4.

Each column of c is a linear combination of the corresponding column in [a] multiplied by the sum of the corresponding row elements in [b]. Similarly, each row vector of c is a linear combination of the corresponding row in a multiplied by the sum of the corresponding column elements in b. I don't see how to “prove” this since it is just a restatement of the definition of matrix multiplication.

  • Every j=1..n column vector {c_jk} = sum{k}{}{a_ik b_kj} = delim{lbrace}{a_jk}{rbrace} ] sum{k}{}{] b_kj [}.
  • Every k=1..n row vector of [c], delim{]}{c_jk}{[} = sum{k}{}{a_ik b_kj} = delim{]}{a_ik}{[} delim{lbrace}{sum{k}{}{ b_kj }}{rbrace}.

Note ]r[ represents the row vector r.

Problem 5.

([A]+[B])([A-B]) = [A]^2 + [B][A] -[A][B] -[B]^2 = [A]^2 - [B]^2

This is true when [B][A] -[A][B] = 0, equivalent to [B][A] = [A][B].

If [A] and [B] are invertible, premultiplying by [B] and postmultiplying both sides by [A] gives [B]^-1[B][A][A]^-1 = [B]^-1[A][B][A]^-1, the same as [I] = [A][B]^-1[B][A]^-1 = [I] = [A][I][A]^-1 = [I], a true statement.

Thus, this is true when both [A] and [B] have inverses.

[A] = delim{[}{ matrix{2}{2}{ 1 1 1 {1~} } }{~]}, [B] = delim{[}{ matrix{2}{2}{1 2 1 2~} }{]}

([A]+[B])([A-B]) =( delim{[}{ matrix{2}{2}{1 1 1 1~} }{]} + delim{[}{ matrix{2}{2}{1 2 1 2~} }{]} ) ( delim{[}{ matrix{2}{2}{1 1 1 1 ~} }{]} - delim{[}{ matrix{2}{2}{1 2 1 2~} }{]} ) =delim{[}{ matrix{2}{2}{2 3 2 3} }{]} delim{[}{ matrix{2}{2}{0 {-1} 0 {-1~}} }{]} =delim{[}{ matrix{2}{2}{0 {-5} 0 {-5}} }{]}

[A]^2 - [B]^2 =delim{[}{ matrix{2}{2}{1 1 1 1~} }{~]}^2 - delim{[}{ matrix{2}{2}{1 2 1 2~} }{]} ^2 =delim{[}{ matrix{2}{2}{2 2 2 2~} }{~]} - delim{[}{ matrix{2}{2}{3 6 3 6~} }{]} =delim{[}{ matrix{2}{2}{{-1} {-4} {-1} {-4}~} }{]}


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