Table of Contents
EGR 6013 Homework Set 1 (19-30a)
— David Wagner 2009/09/16 07:53
Problem 19.
(a) A is nonsingular, so its inverse [A]-1 exits.
[A][B] = [0] ⇒ [A]-1[A][B] = [A]-1[0] = [I][B] = [0] ⇒ [B] = [0].
(b) B is nonsingular, so its inverse [B]-1 exits. [A][B] = [0] ⇒ [A][B][B]-1 = [0][B]-1 = [A][I] = [0] ⇒ [A] = [0].
Problem 22.
[A][M][B] = [C] = ![delim{[}{
matrix{3}{3}{ 2 1 1 1 1 0 0 0 1 }
}{]}
delim{[}{
M
}{]}
delim{[}{
matrix{2}{2}{ 3 1 1 1 }
}{]}
=delim{[}{
matrix{3}{2}{ 1 1 2 2 1 1 }
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_f1390705bc35613d4c57c4b703bb405e.png "delim{[}{
matrix{3}{3}{ 2 1 1 1 1 0 0 0 1 }
}{]}
delim{[}{
M
}{]}
delim{[}{
matrix{2}{2}{ 3 1 1 1 }
}{]}
=delim{[}{
matrix{3}{2}{ 1 1 2 2 1 1 }
}{]}")
,
[A]-1![delim{[}{
matrix{3}{3}{ {1} {-1} {-1} {-1} {2} {1} {0} {0} {1} }
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_b0586550646c09eebfd13758259eeb65.png "delim{[}{
matrix{3}{3}{ {1} {-1} {-1} {-1} {2} {1} {0} {0} {1} }
}{]}")
,
[B]-1 = ![delim{[}{
matrix{2}{2}{ {1/2} {-1/2} {-1/2} {3/2} }
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_1d1051ea90f89dd3e3a2f936be981a9e.png "delim{[}{
matrix{2}{2}{ {1/2} {-1/2} {-1/2} {3/2} }
}{]}")
.
[A]-1[A][M][B][B]-1
=[A]-1[C][B]-1
=[M]
=![delim{[}{
matrix{3}{3}{ {1} {-1} {-1} {-1} {2} {1} {0} {0} {1} }
}{]} delim{[}{
matrix{3}{2}{ 1 1 2 2 1 1 }
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_d6f4118bf09b509c3e84df43a43551ce.png "delim{[}{
matrix{3}{3}{ {1} {-1} {-1} {-1} {2} {1} {0} {0} {1} }
}{]} delim{[}{
matrix{3}{2}{ 1 1 2 2 1 1 }
}{]}")
= =
M=
![delim{[}{
matrix{3}{2}{0 {-2} 0 4 0 1 }
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_c76f3d6f7f25c4a2970d79c74b5463f5.png "delim{[}{
matrix{3}{2}{0 {-2} 0 4 0 1 }
}{]}")
.
Problem 25.
If |A| ≠ 0, prove that |[A]-1| = |A|-1.
[A]-1 = |A|-1Adj[A]
Since (Adj A)[A] = |A|I, post-multiplying by [A] ⇒ [A]-1[A] = |A|-1(Adj A)[A] = |A|-1|A|[I] = [I]
