# MATH 3001 Study Guide - Midterm Guide: Identity Matrix, Gaussian Elimination, Linear CombinationExam

by OC2747395

This

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Please do not think that the info on this page is something you should have memorized (except for

the ﬁrst list...that’s important). Rather, these facts may be helpful to use in problems, but are fairly

straightforward to prove (so understand them, but don’t go overboard in memorizing them).

The following are equivalent statements (meaning they are either all true or all false) for an n×nmatrix A.

•Ais invertible (i.e. A−1exists)

•det(A)6= 0

•Acan be row reduced to the identity matrix I

•Ahas npivots

•rank(A) = n

•A~x =~

bhas a unique solution

•A~x =~

0only has the trivial solution

•Col(A) is a basis

•Null(A) = n~

0o

•The columns of Aas vectors are linearly independent

•The columns of Aas vectors are a basis for Rn

The following facts regarding the inverse of a matrix apply so long as Aand Bare invertible.

•A−1−1=A

•(kA)−1=k−1A−1for k6= 0

•AT−1=A−1T

•(AB)−1=A−1B−1

•det A−1=1

det(A)

The following facts regarding the transpose of a matrix apply for any matrices A, B and scalar c.

•ATT=A

•(A+B)T=AT+BT

•(AB)T=BTAT

•(cA)T=cAT

•det AT= det(A)

•If Ais a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose.

The following facts regarding the determinant of a matrix apply for any square matrices Aand B.

•If Aand Bare the same size, then det(AB) = det(A)·det(B)

•If Ais n×n, then det(cA) = cndet(A)

•If Ais upper or lower triangular, then det(A) = (a11)(a22)· · · (ann)

•If any row or column of a matrix Aconsists of all zeros, then det(A) = 0

•If any two columns of a matrix Aare identical (or can be written as a linear combination of other rows,

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Linear Independence of Vectors: Determine if the following collection of vectors in R3are linearly inde-

pendent. If the vectors are linearly dependent, ﬁnd a basis for the set of vectors

(a)

2

1

6

,

5

2

2

,

1

2

9

(b)

1

2

2

,

0

6

3

,

1

4

3

(a) If we let Abe the matrix with columns that are the vectors of this set, then we can row reduce it and gain

some more information about A. We have

A=

251

122

629

R1

′=R1−R2

=⇒

131

122

629

R2

′=R1−R2

=⇒

1 3 1

0 1 −1

6 2 9

R1

′=R1−3R2

=⇒

1 0 4

0 1 −1

6 2 9

R3

′=R3−6R1

=⇒

1 0 4

0 1 −1

0 2 −15

R3

′=R3−2R2

=⇒

1 0 4

0 1 −1

0 0 −13

R3

′=−

1

13 R3

=⇒

1 0 4

0 1 −1

0 0 1

R2

′=R2+R3

=⇒

104

010

001

R1

′=R1−4R3

=⇒

100

010

001

Since Arow reduces to the identity matrix, we may say that Ais “row equivalent” to I. Thus, Ais

invertible. This implies that the columns of Awritten as vectors are linearly independent. So, the set of

vectors that was given are linearly independent.

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(b) If we let Bbe the matrix with columns that are the vectors of this set, then we can row reduce it and gain

some more information about B. We have

B=

101

264

233

R2

′=R2−2R1

=⇒

101

062

233

R3

′=R3−2R1

=⇒

101

062

031

R3

′=R2−2R3

=⇒

101

062

000

R2

′=1

6R2

=⇒

1 0 1

0 1 1

3

0 0 0

We see that we have a row of zeros in the third row. Clearly, there are two pivots columns (the ﬁrst

and second columns). This means that the column space of Bhas a basis

1

2

2

,

0

6

3

.

Since this is diﬀerent from Col(B), we know that Col(B) is not a basis. Therefore, Bis not invertible.

This implies that the columns of Bwritten as vectors are linearly dependent. So, the set of vectors that was

given are linearly dependent.

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