Section - Linear Algebra - Quiz Solution, Exercises of Linear Algebra

This is the Quiz Solution of Linear Algebra. Mainly this Course includes Zero Vector, Linearly Dependent, Statement, Vector, Linear Combination, Expressed, Trivial Solution, Inspection, Dependent, Theorem etc. Key important points of his quiz are: Section, Row Operations, Invertible, Column Vectors, Information, Condition, Entries, Problem, Augmented Matrix, Corresponding

Typology: Exercises

2012/2013

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MATH 205A,B - LINEAR ALGEBRA
WINTER 2013
QUIZ 5
NAME: Section:(Circle one) A(1 : 10) B(2 : 40)
Show ALL your work CAREFULLY.
(a) Suppose det
a b c
d e f
g h i
= 3. Find det
a d g
2b2e2h
c f i
.
det
a d g
2b2e2h
c f i
= 2 ·det
a d g
b e h
c f i
= 2 ·det
a d g
b e h
c f i
T
= 2 ·det
a b c
d e f
g h i
= 6.
(b) Use row operations to find det Awhere
A=
1a b +c
1b a +c
1c a +b
.
Is Ainvertible?
First note that if a=bor a=cthen Awill have two identical rows and thus is row
equivalent to a matrix that has a row of zeros. In such cases, det A= 0 and Ais not
invertible. Now, assume that a6=band a6=c. Then we have
A
1a b +c
0ba a b
1c a +b
1a b +c
0ba a b
0ca a c
1a b +c
0 1 1
0ca a c
1a b +c
0 1 1
0 1 1
1a b +c
0 1 1
0 0 0
.
It follows that det A= 0 and hence Ais not invertible.
Date: February 13, 2013.
1

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MATH 205A,B - LINEAR ALGEBRA

WINTER 2013

QUIZ 5

NAME: Section:(Circle one) A(1 : 10) B(2 : 40)

Show ALL your work CAREFULLY.

(a) Suppose det

a b c d e f g h i

 (^) = 3. Find det

a d g 2 b 2 e 2 h c f i

det

a d g 2 b 2 e 2 h c f i

 (^) = 2 · det

a d g b e h c f i

 (^) = 2 · det

a d g b e h c f i

T = 2 · det

a b c d e f g h i

(b) Use row operations to find det A where

A =

1 a b + c 1 b a + c 1 c a + b

Is A invertible? First note that if a = b or a = c then A will have two identical rows and thus is row equivalent to a matrix that has a row of zeros. In such cases, det A = 0 and A is not invertible. Now, assume that a 6 = b and a 6 = c. Then we have

A ∼

1 a b + c 0 b − a a − b 1 c a + b

1 a b + c 0 b − a a − b 0 c − a a − c

1 a b + c 0 1 − 1 0 c − a a − c

1 a b + c 0 1 − 1 0 1 − 1

1 a b + c 0 1 − 1 0 0 0

It follows that det A = 0 and hence A is not invertible.

Date: February 13, 2013. 1