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(MATH111)[2008](f)quiz2a~PPSpider^_10431.pdf
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MATH111 Quiz 2A
Page 1/3
Oct 11, 2008 by Daniel Zheng
..................................................40 mins allowed..................................................

Name: Student ID: Score:
I. [4] Find the matrix A whose inverse is A.1 = . .
. 0 1 4 1 0 .3 2 3 8 . .
. .
Solutions. [A.1 I] = . .
. 0 1 4 1 0 .3 2 3 8 1 0 0 0 1 0 0 0 1 . .
.
. .
. 1 0 0 0 1 0 0 0 1 .9/2 .2 3/2 7 4 .2 .3/2 .1 1/2 . .
. = [I A].

MATH111 Quiz 2A Page 2/3
II. [5] Let A be an n n matrix. Suppose Ak = 0 for some k> 1. Find the inverse for I . A if exists. Solutions. (I . A).1 = I + A + + Ak.1 .
MATH111 Quiz 2A Page 3/3
III. Suppose A is an m n matrix and B is an n m matrix. Show that
1.
[5] if the columns of B are linearly dependent, so are the columns of AB.

2.
[5] if m>n, the m m matrix AB can not be invertible.


Proof.
1. Notice that AB =[Ab1 Abn]. Since the columns of B are linearly depen-dent, there exists nonzero vector x =(x1 xn)T such that .ni=1 xibi = 0. Multiplying both sides by A, we have
nn
A . xibi = . xiAbi = 0
i=1 i=1
which implies the columns of AB are linearly dependent.
2. Proof. Since m>n, B has at most n pivot positions, hence its m columns are linearly dependent. From the conclusion of (1), we know that the columns of AB are also linearly dependent. Therefore by IMT, AB can not be invert-ible. Alternative. The columns of B are linearly dependent, hence the homoge-neous system Bx = 0 has nontrivial solutions. Consequently ABx = 0 as well has nontrivial solutions. Therefore, by IMT, AB cannot be invertible.