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Linear Algebra Quiz 3 for MATH 205A,B - Solutions, Exercises of Linear Algebra

The solutions to quiz 3 of the linear algebra course math 205a,b, held in winter 2013. It includes the parametric form of the solutions to the homogeneous equation a⃗x = ⃗0, the determination of linear independence of the columns of matrix a, and the explanation of why the columns of a do not span r3.

Typology: Exercises

2012/2013

Uploaded on 02/27/2013

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MATH 205A,B - LINEAR ALGEBRA
WINTER 2013
QUIZ 3
NAME: Section:(Circle one) A(1 : 10) B(2 : 40)
Show ALL your work CAREFULLY.
Let
A=
121
1 2 2
241
.
(a) Express the solutions to the homogeneous equation A~x =~
0 in parametric form.
The coefficient matrix Acan be reduced as follows.
121
1 2 2
241
121
0 0 1
241
121
0 0 1
0 0 1
121
0 0 1
0 0 0
12 0
001
000
.
The solutions to A~x =~
0are
~x =
x1
x2
x3
=
2x2
x2
0
=x2
2
1
0
where x2is the parameter.
(b) Based on your answer to (a), determine whether the columns of Aare linearly independent?
Justify your answer.
The columns of Aare NOT linearly independent since ~
0can be written as a non-trivial
linear combinations of the columns ~a1, ~a2,~a3. For instance, ~
0 = (2)~a1+ (1)~a2+ (0)~a3(take
x2= 1).
(c) Do the columns of Aspan R3? Explain.
The columns of Ado NOT span R3. For any vector ~
b=
b1
b2
b3
where b36= 0, the system
A~x =~
bdoes not have a solution because the row reduced echelon form of Ahas a row
of zeroes and thus the system would be inconsistent.
Date: January 25, 2013.
1

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Download Linear Algebra Quiz 3 for MATH 205A,B - Solutions and more Exercises Linear Algebra in PDF only on Docsity!

MATH 205A,B - LINEAR ALGEBRA

WINTER 2013

QUIZ 3

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

Show ALL your work CAREFULLY.

Let

A =

(a) Express the solutions to the homogeneous equation A~x = ~0 in parametric form. The coefficient matrix  A can be reduced as follows.

The solutions to A~x = ~ 0 are

~x =

x 1 x 2 x 3

2 x 2 x 2 0

 (^) = x 2

where x 2 is the parameter. (b) Based on your answer to (a), determine whether the columns of A are linearly independent? Justify your answer. The columns of A are NOT linearly independent since ~ 0 can be written as a non-trivial linear combinations of the columns ~a 1 , ~a 2 , ~a 3. For instance, ~0 = (2)~a 1 + (1)~a 2 + (0)~a 3 (take x 2 = 1). (c) Do the columns of A span R^3? Explain.

The columns of A do NOT span R^3. For any vector ~b =

b 1 b 2 b 3

 (^) where b 3 6 = 0, the system

A~x = ~b does not have a solution because the row reduced echelon form of A has a row of zeroes and thus the system would be inconsistent.

Date: January 25, 2013. 1