determinants · properties of determinants cofactor expansion cramer’s rule determinants...

Basic ConceptsProperties of Determinants

Cofactor ExpansionCramer’s Rule

Determinants

Linear Algebra

Francis Joseph CampenaMathematics Department

De La Salle University-Manila

Lectures in Linear Algebra



Permutations

DefinitionLet S = { 1, 2, . . . , n } be the set of the first n positiveintegers. A rearrangement σ = j1j2 . . . jn of the elements of S iscalled a permutation of the set S. Since S has n elements,there are n! permutations of the set S.

Example

If n = 3, then S = { 1, 2, 3 } . There are six possiblerearrangements of the elements of the set S. These are:123, 132, 213, 231, 312 and 321. Each of theserearrangements is a permutation of S. The first permutationhas j1 = 1, j2 = 2 and j3 = 3.




Odd and Even Permutations

DefinitionA permutation σ = j1j2 . . . ji · · · jk · · · jn is said to have aninversion if a larger number ji precedes a smaller number jk . Ifthe total number of inversions in σ is odd (even), then σ iscalled an odd (even) permutation.

ExampleLet n = 3 and let σ = 312. Then σ contains the inversions 31and 32. Since the number of inversions is even, σ is an evenpermutation. The following table shows the permutations forn = 3, the corresponding inversions and their classification asodd or even permutations.




Odd and Even Permutations

Permutation Inversions Classification123 none even132 32 odd213 21 odd231 21 and 31 even312 31 and 32 even321 32, 31 and 21 odd




Determinants

DefinitionLet A = [aij ] be an n × n matrix. The determinant of A, which isdenoted by |A| or by det A, is defined to be the real number

|A| =∑σ

(±) a1j1a2j2 · · · anjn

where the summation ranges over all the permutationsσ = j1j2 · · · jn of the column subscripts.

RemarkThe (+) sign is taken when σ is even and the (−) sign is usedwhen σ is odd.




Remarks

If n = 2, then 12 is even and 21 is odd, so that if

A =

(a11 a12a21 a22

)then |A| = a11a22 − a12a21




Remarks

If n = 3 and

A =

a11 a12 a13a21 a22 a23a31 a32 a33

then the even permutations give rise to the productsa11a22a33, a12a23a31 and a13a21a32, while the oddpermutations produce the products a11a23a32, a12a21a33 anda13a22a31. Hence, we have

|A| = a11a22a33 + a12a23a31 + a13a21a32 − a11a23a32

− a12a21a33 − a13a22a31




Examples

For the 2× 2 matrix A =

(2 −13 2

), we have

|A| = a11a22 − a12a21 = (2)(2)− (−1)(3) = 4 + 3 = 7




Examples

For the 3× 3 matrix B =

2 −1 0−1 4 2

3 0 −5

, we can compute

the determinant by first rewriting the first two columns, giving us

2 −1 0−1 4 2

3 0 −5

2 −1−1 4

3 0

We obtain

|B| = (2)(4)(−5) + (−1)(2)(3) + (0)(−1)(0)− (3)(4)(0)− (0)(2)(2)− (−5)(−1)(−1)= (−40) + (−6) + 0− 0− 0− (−5)= −41




Exercises

1. If S = {1,2,3,4,5}, find the number of inversions of thefollowing permutations:

(a) 32541 (b) 25134 (c) 42135

2. If S = {1,2,3,4}, determine if the given permutation is oddor even.

(a) 1234 (b) 2431 (c) 4312

3. Evaluate the following determinants:

(a)∣∣∣∣ −1 0

3 −5

∣∣∣∣ (b)

∣∣∣∣∣∣2 −2 10 1 0−2 −1 2

∣∣∣∣∣∣(c) ∣∣∣∣ x − 1 2

3 x − 2

∣∣∣∣Lectures in Linear Algebra



Exercises

4. For the matrix in 3(c), find the values of x for which thedeterminant is 0.

5. Evaluate |λI2 − A| where A =

[4 2−1 1

]6. If A =

[2 30 −1

], determine

(a) |A| (b) |A2| (c) |A3| (d) |A−1|




The Determinant of the Transpose

Theorem

If A = [aij ] is an n × n matrix, then |A| = |AT |.

Example

Let A =

(−1 3

4 2

), so that AT =

(−1 4

3 2

). We have

|A| = (−1)(2)− (3)(4) = −2− 12 = −14|AT | = (−1)(2)− (4)(3) = −2− 12 = −14|A| = |AT |




Effect of a Type I Elementary Operation

TheoremIf the matrix B is obtained from a matrix A by interchanging anytwo rows (or columns) of A, then

|B| = −|A|

Example

Let A =

(−1 3

4 2

)be the matrix given in the preceding

example and let B =

(3 −12 4

). Note that B is obtained from

A by interchanging the columns. We get

|B| = (3)(4)− (2)(−1) = 12− (−2) = 12 + 2 = 14 = −|A|Lectures in Linear Algebra



Matrices With Identical Rows/Columns

CorollaryIf A is a matrix with two identical rows (or columns), then|A| = 0.




Matrices With Identical Rows/Columns

Example

Consider the matrix A =

1 −1 23 4 01 −1 2

Note that the first and

the third rows of A have identical entries. We have

|A| =

∣∣∣∣∣∣1 −1 23 4 01 −1 2

∣∣∣∣∣∣ = (1)(4)(2) + (−1)(0)(1) + (2)(3)(−1)

− (1)(4)(2)− (−1)(0)(1)− (2)(3)(−1)= 8 + 0 + (−6)− 8− 0− (−6)= 0




Matrices With a Row/Column of Zeros

TheoremIf A is a matrix with a row (or column) made up entirely of zeros,then |A| = 0.

Example

Let A =

1 0 −2−3 0 3−1 0 5

. We have

|A| = (1)(0)(5) + (0)(3)(−1) + (−2)(−3)(0)− (−1)(0)(−2)− (0)(3)(1)− (5)(−3)(0)= 0




Effect of a Type II Operation

TheoremIf B is obtained from a matrix A by multiplying any row (orcolumn) of A by a real number c, then |B| = c|A|.

CorollaryIf A is an n × n matrix, then for a scalar c, the matrix cA isobtained by multiplying each entry of A by c. Since this isequivalent to multiplying each row (or column) of A by c, wehave

|cA| = cn|A|

since one factor c is multiplied to the determinant for each rowof the matrix.




Effect of a Type III Operation

TheoremIf B is obtained from A by adding a multiple of one row (orcolumn) of A to another row (or column), then |B| = |A|.




Example


2 −1 41 −1 5−1 3 2

Let B be the matrix

obtained by performing the Type III operation R2 ← R2 − R3 on

A, giving us B =

2 −1 42 −4 3−1 3 2

. We have

|A| = (2)(−1)(2) + (−1)(5)(−1) + (4)(1)(3)− (−1)(−1)(4)− (3)(5)(2)− ((2)(1)(−1)= −4 + 5 + 12− 4− 30− (−2) = −19




Example, cont.

|B =

∣∣∣∣∣∣2 −1 42 −4 3−1 3 2

∣∣∣∣∣∣= (2)(−4)(2) + (−1)(3)(−1) + (4)(2)(3)− (−1)(−4)(4)− (3)(3)(2)− (2)(2)(−1)= −16 + 3 + 24− 16− 18− (−4) = −19

This shows that |A| = |B|.




Triangular Matrices

TheoremIf A is an upper (or lower) triangular matrix, then

|A| = a11a22 · · · ann

i.e. the determinant is the product of the entries along the maindiagonal of the matrix.

ExampleSince a diagonal matrix is both upper and lower triangular, wehave

|A| =

∣∣∣∣∣∣2 0 00 −3 00 0 4

∣∣∣∣∣∣ = (2)(−3)(4) = −24




Determinant of a Product

TheoremIf A and B are n × n matrices, then |AB| = |A| · |B|.

Example

Let A be a nonsingular matrix. Then A−1 exists and AA−1 = In.Since In is a diagonal matrix, we have |In| = 1. Using the abovetheorem, we get

|AA−1| = |A||A−1| = |In| = 1

Corollary

If A is a nonsingular matrix, then |A| 6= 0, and |A−1| = 1|A| .




Determinants of Larger Matrices

Since the determinant of an upper or lower triangular matrix iseasy to compute, you may compute the determinant of a largematrix by first transforming it to an upper triangular matrixusing elementary row operations.With each operation performed, we monitor the effect on thedeterminant.




Example

Consider the matrix

A =

1 1 3 −11 1 1 11 −2 1 −14 1 8 −1

We evaluate |A| by converting the matrix to an upper triangularmatrix, as follows:

|A| =

∣∣∣∣∣∣∣∣1 1 3 −11 1 1 11 −2 1 −14 1 8 −1

∣∣∣∣∣∣∣∣ =∣∣∣∣∣∣∣∣

1 1 3 −10 0 −2 20 −3 −2 00 −3 −4 3

∣∣∣∣∣∣∣∣ (Type III operations)




Example, cont.

|A| = −

∣∣∣∣∣∣∣∣1 1 3 −10 −3 −2 00 0 −2 20 −3 −4 3

∣∣∣∣∣∣∣∣ (Interchange of two rows)

= −(−3)(−2)

∣∣∣∣∣∣∣∣1 1 3 −10 1 2/3 00 0 1 −10 −3 −4 3

∣∣∣∣∣∣∣∣ (Type II operations)

= −6

∣∣∣∣∣∣∣∣1 1 3 −10 1 2/3 00 0 1 −10 0 −2 3

∣∣∣∣∣∣∣∣ (Type III operation)




Example, cont.

|A| = −6

∣∣∣∣∣∣∣∣1 1 3 −10 1 2/3 00 0 1 −10 0 0 1

∣∣∣∣∣∣∣∣ (Type III operation)

= −6(1)(1)(1)(1) = −6




Cofactors

DefinitionLet A = [aij ] be an n × n matrix.(a) The submatrix Mij is the (n − 1)× (n − 1) matrix obtained

after deleting the entries from the i th row and from the j thcolumn of A.

(b) The determinant |Mij | of the submatrix Mij is called theminor of the entry aij of A.

(c) The cofactor of aij is the quantity Aij = (−1)i+j |Mij |.




Remarks

(a) A common error in computing the cofactors is in failing toinclude the factor (−1)i+j , which assumes values of ±1depending on whether the sum i + j is even or odd. Wehave

Aij =

{|Mij | if i + j is even−|Mij | if i + j is odd

(b) If the signs of (−1)i+j are arranged in the (i , j)-position of amatrix-like formation, then the signs will alternate in eachrow or column. For n = 4, following pattern is obtained:

+ − + −− + − ++ − + −− + − +




Example

Consider the matrix

A =

2 −1 42 −4 3−1 3 2

The submatrix M23 is the 2× 2 matrix given by

M23 =

(2 −1−1 3

)The minor |M23| is given by

|M23| =∣∣∣∣ 2 −1−1 3

∣∣∣∣ = (2)(3)− (−1)(−1) = 5

The cofactor A23 is computed as

A23 = (−1)2+3|M23| = (−1)(5) = −5Lectures in Linear Algebra



Cofactor Expansion

TheoremIf A = [aij ] is an n × n matrix, then

(i) (Cofactor expansion along the ith row)

|A| =n∑

k=1

aikAik = ai1Ai1 + ai2Ai2 + . . .+ ainAin

(ii) (Cofactor expansion along the jth column)

|A| =n∑

k=1

akjAkj = a1jA1j + a2jA2j + . . .+ anjAnj




Example

If we have the matrix A =

1 −1 32 1 −40 −3 −2

, then we get

|M11| =

∣∣∣∣ 1 −4−3 −2

∣∣∣∣ = 1(−2)− (−3)(−4) = −14

|M12| =

∣∣∣∣ 2 −40 −2

∣∣∣∣ = 2(−2)− 0(−4) = −4

|M13| =

∣∣∣∣ 2 10 −3

∣∣∣∣ = 2(−3)− 0(1) = −6




Example

If we use cofactor expansion along the first row, then we have

|A| = a11A11 + a12A12 + a13A13

= a11|M11| − a12|M12|+ a13|M13|= (1)(−14)− (−1)(−4) + (3)(−6) = −36




Example, cont.

On the other hand, if we choose to do cofactor expansion alongthe second column, we will need the following:

|M22| =

∣∣∣∣ 1 30 −2

∣∣∣∣ = 1(−2)− 0(3) = −2

|M32| =

∣∣∣∣ 1 32 −4

∣∣∣∣ = 1(−4)− 2(3) = −10

together with the value of |M12| which was already computedearlier.




Example, cont.

We get

|A| = a12A12 + a22A22 + a32A32

= −a12|M12|+ a22|M22| − a32|M32|= −(−1)(−4) + (1)(−2) + (−3)(−10) = −36

Observe that the value of |A| is the same for the twoexpansions.




Theorem on Cofactors

TheoremIf A = [aij ] is an n × n matrix, then

ai1Ak1 + ai2Ak2 + · · ·+ ainAkn = 0, if i 6= ka1jA1k + a2jA2k + · · ·+ anjAnk = 0, if j 6= k




Theorem on Cofactors

Example


1 −1 32 1 −40 −3 −2

from the preceding

example. If we multiply the entries of the first row with thecofactors of the second row, we obtain

a11A21 + a12A22 + a13A23

= 1[(−1)(−2)− (−3)(3)]− (−1)[(1)(−2)− (0)(3)] + 3[(1)(−3)− (0)(−1)]= 1(2 + 9) + 1(−2− 0) + 3(−3− 0)= 1(11) + 1(−2) + 3(−3) = 0




The Adjoint of a Matrix

DefinitionLet A = [aij ] be an n × n matrix. The adjoint of A, denoted byadj A, is the n × n matrix whose (i , j)-entry is Aji .

RemarkThe adjoint is constructed by first forming the matrix whose(i , j)-entry is the cofactor Aij , then taking the transpose of thematrix.




Example

Find the adjoint of the matrix A =

1 −1 32 1 −40 −3 −2

.




An Alternative Way to Find A−1

TheoremLet A be an n × n matrix. Then(a) A(adj A) = |A|In

(b) If |A| 6= 0, then A is nonsingular, and A−1 =1|A|

(adj A).

Example

Find the inverse of the matrix A =

1 −1 32 1 −40 −3 −2

by using the

adjoint.




Exercises

Consider the matrices

A =

1 −2 53 −1 42 5 −3

B =

0 2 1 32 −1 3 4−2 1 5 2

0 1 0 2

1. Find the cofactors of the first row entries of A.2. Use the determinant to determine if A is singular or

nonsingular.3. Verify that A(adj A) = |A|I3.4. If A is nonsingular, use the adjoint to find its inverse.5. Use the adjoint to find the inverse of the matrix B above.6. Suppose a square matrix D is a symmetric matrix, prove that

the adjoint of D i also symmetric.7. Suppose a square matrix D is a nonsingular upper triangular

matrix, show that D−1 is also upper triangular.Lectures in Linear Algebra



Cramer’s Rule

Consider the linear system

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2...

...an1x1 + an2x2 + · · ·+ annxn = bn

of n equations in n unknowns, with A = [aij ] as the coefficientmatrix.




Cramer’s Rule Cont.

If |A| 6= 0, then the unique solution to the system is given by

x1 =|A1||A|

, x2 =|A2||A|

, · · · , xn =|An||A|

where Aj is the matrix obtained by replacing the j th column of Aby

B =

b1b2...

bn




Example

Use Cramer’s rule to solve the linear system

2x + y + z = 63x + 2y − 2z = −2

x + y + 2z = 4

The coefficient matrix is A =

2 1 13 2 −21 1 2

and its

determinant is

|A| =

∣∣∣∣∣∣2 1 13 2 −21 1 2

∣∣∣∣∣∣ = 2(4 + 2)− 1(6 + 2) + 1(3− 2) = 5 6= 0

which shows that a solution can be found using Cramer’s rule.Lectures in Linear Algebra



Properties of Determinants

TheoremConsider a square matrix A of order n, the following statementsare equivalent1. A is non-singular.2. AX = 0 has only the trivial solution.3. A is row equivalent to In.4. The linear system AX = B has a unique solution for every

n × 1 matrix B.5. |A| 6= 0.




Exercises

In each of the following linear systems, determine if thecoefficient matrix is nonsingular. If so, use Cramer’s rule toobtain the solution.

1.

x + 2y + 3z = 6x + z = 2

x + y − z = 1

2.

2x − y − z = 1x − 2y + 3z = 43x + y − 4z = −1

3.

x − y + 4z = −2−8x + 3y + z = 0

2x − y + z = 6

4. x + y + z − 2w = −4

2y + z + 3w = 42x + y − z + 2w = 5

x − y + w = 4


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