# MATH 125 Study Guide - Final Guide: Row Echelon Form, Invertible Matrix, Elementary Matrix

by OC1679395

This

**preview**shows pages 1-3. to view the full**36 pages of the document.**MATH 125 (R1) Winter 2017

Additional Practice Problems for Final – Solutions

Notes: Some of the solutions here are incomplete (their main purpose is to enable you

to check your own answers). In particular, row reduction of matrices and determinant

calculations are omitted for the most part. In your exam, however, you should show all your

work, and give suitable justiﬁcation for your answers. Note also that there are often multiple

ways to solve a particular problem; in these solutions, we typically just give one approach.

1. a) Deﬁne what it means for an n×nmatrix to be invertible.

b) Find the inverse (if it exists) of the matrix

A=

1 0 1

0 1 0

2 1 1

c) For which value(s) of ais the matrix

1 0 a

0 1 a

2 1 a

invertible? Justify your answer.

Solution: (a) An n×nmatrix Ais invertible if there exists another n×nmatrix B

such that AB =In=BA, where Inis the n×nidentity matrix.

(b) We use the standard algorithm:

1 0 1 1 0 0

0 1 0 0 1 0

2 1 1 0 0 1

−→ · · · −→

1 0 0 −1−1 1

0 1 0 010

0 0 1 2 1 −1

Since the reduced row echelon form of Ais I3(the left side of the partitioned matrix),

Ais invertible and

A−1=

−1−1 1

010

2 1 −1

(the right side of the partitioned matrix).

(c) Let Bbe the given matrix. Then:

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

det(B) =

1 0 a

0 1 a

0 1 −a

(r3−2r1)

=

1a

1−a

(expand along 1st column)

=−2a.

Hence Bis invertible if and only if −2a6= 0, i.e., if and only if a6= 0.

2. Let A=

1−3 6

2−5 11

−1 1 −3

.

a) Ais invertible. Compute A−1.

b) Use A−1to solve the following system of equations:

x1−3x2+ 6x3= 1

2x1−5x2+ 11x3=−3

−x1+x2−3x3= 1

Solution: (a) We use the standard algorithm:

1−3 6 1 0 0

2−5 11 0 1 0

−1 1 −3 0 0 1

−→ · · · −→

1 0 0 4 −3−3

0 1 0 −531

0 0 1 −321

Since the reduced row echelon form of Ais I3(the left side of the partitioned matrix),

Ais invertible and

A−1=

4−3−3

−531

−321

(the right side of the partitioned matrix).

(b) The matrix equation of the system is A~x =

1

−3

1

, so by (a), the unique solution

of the system is

~x =A−1

1

−3

1

=

4−3−3

−531

−321

1

−3

1

=

10

−13

−8

.

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3. Compute the determinant of each of the following matrices:

a)

3 7 0 −6

7 9 0 3

−9−3 0 5

1 5 0 3

b)

123

−1−2−3

414

c)

2 0 0 0

−1 6 0 0

5 9 1 0

7 8 3 −3

d)

1 0 1 0

0 1 0 1

1 1 1 2

1 1 0 1

e)

12304

01001

1 1 1 1 −2

1−2 1 0 1

00102

Solution: (a) Since the matrix contains a column of zeros, its determinant is 0.

(b) Since the matrix contains a row which is a multiple of the other (r2=−r1), its

determinant is 0.

(c) Since the matrix is triangular, its determinant is the product of its diagonal entries,

i.e., (2)(6)(1)(−3) = −36.

(d) The determinant is 1 (standard calculation omitted: use cofactor expansion/row

operations/column operations or a combination of all these methods).

(e) The determinant is 5 (standard calculation omitted: use cofactor expansion/row

operations/column operations or a combination of all these methods).

4. Find the rank and nullity of

1 6 11 16 21

3 8 13 18 23

5 10 15 20 25

7 12 17 22 27

9 14 19 24 29

(Hint: The rank is 2, and it shouldn’t take many computations to see this). What is

the determinant of the matrix?

Solution: Denote the given matrix by A. Observe that each row of Ais obtained from

the previous one by adding 2 to each entry. Thus:

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