ADVERTISEMENT

MATH 205A,B - LINEAR ALGEBRA

FALL 2015

QUIZ 3

NAME:

Section:(Circle one)

A(8 : 00)

B(9 : 30)

Show ALL your work CAREFULLY.

Let

1

3

3

5

A =

.

1

7

(a) Solve the homogeneous equation Ax = 0.

The corresponding augmented matrix

1

3 0

1

3 0

1

3 0

1

3 0

1

3 0

1 0 0

3

5

0

3

5

0

0

14

0

0

1

0

0

1

0

0 1 0

.

1

7

0

0

4

0

0

4

0

0

4

0

0

0

0

0 0 0

The solution is x

= 0 = x

.

1

2

(b) Based on your answer to (a), determine whether the columns of A are linearly independent?

Justify your answer.

Since Ax = 0 has only the trivial solution, the zero vector can only be the trivial linear

combination of the columns of A. Hence the columns of A are linearly independent.

3

(c) Do the columns of A span

? Explain.

No. Since there are only two vectors (columns), the span of two vectors can be at

2

3

most

so the columns of A do not span

.

0

0

Alternatively, the equation Ax =

0

has no solution (CHECK!) so the vector

0

1

1

does not lie in the span of the columns of A.

Date: September 28, 2015.

1

ADVERTISEMENT

0 votes

Parent category: Education