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Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Determinants
Linear Algebra
Francis Joseph CampenaMathematics Department
De La Salle University-Manila
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Remarks
If n = 2, then 12 is even and 21 is odd, so that if
A =
(a11 a12a21 a22
)then |A| = a11a22 − a12a21
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Examples
For the 2× 2 matrix A =
(2 −13 2
), we have
|A| = a11a22 − a12a21 = (2)(2)− (−1)(3) = 4 + 3 = 7
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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|
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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 |
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Matrices With Identical Rows/Columns
CorollaryIf A is a matrix with two identical rows (or columns), then|A| = 0.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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|.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Example
Consider the matrix A =
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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|.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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|.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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| .
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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)
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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)
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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 |.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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:
+ − + −− + − ++ − + −− + − +
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Theorem on Cofactors
Example
Consider the matrix A =
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
Example
Find the adjoint of the matrix A =
1 −1 32 1 −40 −3 −2
.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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.
Lectures in Linear Algebra
Basic ConceptsProperties of Determinants
Cofactor ExpansionCramer’s Rule
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
Lectures in Linear Algebra