== EGR 6013 Homework Set 1 ==

 --- //[[honeypot@handbasket.org|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)
  * <m>x_1 + 2 x_2 + 2x_3 =1</m>,
  * <m>2x_1 + 2x_2 + 3x_3 = 3</m>,
  * <m>x_1 - x_2 + 3x_3 = 5</m>.

<m>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}</m>
=
<m>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}</m>


//Second Step.//

=<m>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}</m>
=
<m>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}</m>

  * <m>x_3=1</m>
  * <m>x_2 + {1/2} = {-1/2} right x_2=-1</m>
  * <m>x_1 -2 +2 = 1 right x_1=1</m>



√(b)
  * <m>2x_1 + 0 x_2 +x_3 = 4</m>,
  * <m>x_1 - 2x_2 +2x_3 = 7</m>,
  * <m>3x_1 + 2_x_2+0x_3 = 1</m>.

<m>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}</m>
=
<m>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}</m>
=
<m>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}</m>


//Second Step.//

=
<m>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}</m>
=
<m>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}</m>

  * <m>x_3 is unconstrained.</m> Let <m>x_3=C_3</m>.
  * <m>x_2 ={-5/2}+{3/4}C_3</m>
  * <m>x_1 = 2-{1/2}C_3</m>

==== Problem 2 ====

2. Evaluate the following matrix products:

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

<m>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} }{]}

} }

}</m>

=<m>delim{[}{matrix{2}{3}{3 {-2} {3} 0 1 0}}{]}</m>

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

<m>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} } } {]} }
}</m>

=<m>delim{[}{matrix{2}{2}{0 0 0 0}}{]}</m>

√(c) <m>delim{[}{matrix{1}{4}{a_1 a_2 cdots a_n}} {]} delim{lbrace}{matrix{4}{1}{b_1 b_2 vdots b_n} }{rbrace}</m>=<m>a_1 b_1 + a_2 b_2 + cdots +a_n b_n</m>

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


<m>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}
 } } {]} 
}
}</m>



√(e) <m>delim{[}{matrix{2}{2}{c_1 0 0 c_2} }{]} delim{[}{ matrix{2}{2}{a_11 a_12 a_21 a_22} }{]}</m>=
<m>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} 
 } 
} {]} }
}</m>

=<m>delim{[}{
matrix{2}{2}{
 {a_11 c_1} {a_12 c1} 
 {a_21 c_2} {a_22 c_2} 
 }
 } {]}</m>

√(f) <m>delim{[}{ matrix{2}{2}{a_11 a_12 a_21 a_22} }{]} delim{[}{matrix{2}{2}{c_1 0 0 c_2} }{]}</m>=
<m>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} 
 } } {]} }
}</m>

=<m>delim{[}{matrix{2}{2}{
 {a_11 c_1 +0} {0+a_12 c_2 } 
 {c_1 a_21+0} {0+a_22 c_2} 
 } } {]}</m>
==== Problem 3. ====

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

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

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

So, <m>[a_ir] [b_rs] + [b_rs] [c_sj] = sum{r}{}{a_ir b_rs} +sum{s}{}{b_rs c_sj}</m>.


----


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


----

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

Repeating this shows \\
=<m>delim{[}{sum{r}{}{a_ir sum{s}{}{ b_rs c_sj} } }{]}</m> 
=<m>delim{[}{sum{s}{}{a_ir c_sj sum{r}{}{ b_rs} } }{]}</m> 
=<m>delim{[}{a_ir  sum{s}{}{c_sj sum{r}{}{ b_rs} } }{]}</m>\\
which is \\
=<m>[a_ir]delim{[}{  sum{s}{}{ sum{r}{}{ b_rs c_sj} } }{]}</m>=[a_ir]( [b_rs] [c_sj] )

Going back to the original equation and messing with it in a slightly different manner, \\
<m>delim{[}{sum{r}{}{ sum{s}{}{a_ir b_rs c_sj} } }{]}</m> 
= <m>delim{[}{sum{s}{}{c_sj sum{r}{}{a_ir  b_rs } } }{]}</m>
= <m>delim{[}{sum{r}{}{c_sj a_ir sum{s}{}{ b_rs } } }{]}</m>
= <m>[c_sj]delim{[}{  sum{r}{}{ a_ir sum{s}{}{ b_rs } } }{]}</m>
= <m>[c_sj]delim{[}{  sum{s}{}{ sum{r}{}{a_ir b_rs } } }{]}</m>
==== 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 <m>{c_jk} = sum{k}{}{a_ik b_kj}  = delim{lbrace}{a_jk}{rbrace} ] sum{k}{}{] b_kj [}</m>.


  * Every k=1..n row vector of [c], <m>delim{]}{c_jk}{[} = sum{k}{}{a_ik b_kj} = delim{]}{a_ik}{[} delim{lbrace}{sum{k}{}{ b_kj }}{rbrace}</m>.


//Note ]r[ represents the row vector r.//

==== Problem 5. ====


<m>([A]+[B])([A-B]) = [A]^2 + [B][A] -[A][B] -[B]^2 = [A]^2 - [B]^2</m>

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
<m>[B]^-1[B][A][A]^-1 = [B]^-1[A][B][A]^-1</m>, the same as <m>[I] = [A][B]^-1[B][A]^-1</m> = <m>[I] = [A][I][A]^-1</m> = [I], a true statement.

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

<m>[A] = delim{[}{ matrix{2}{2}{ 1 1 1 {1~} } }{~]}</m>, <m>[B] = delim{[}{ matrix{2}{2}{1 2 1 2~} }{]}</m>

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

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