Theorem 4.1: u1, u2, v are row vectors of 2×2 matrices

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Review

 Determinant is a number that represents a square matrix.

⇒ Determinant is not linear.

 Theorem 4.1: u1, u2, v are row vectors of 2×2 matrices; k1, k2 are scalars. Then

det  k1u1 + k2u2 v



= k1 det  u1 v



+ k2 det  u2 v



det

 v

k1u1 + k2u2



= k1 det  v u1



+ k2 det  v u2





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 Theorem 4.2: For a 2×2 matrix A =  A11 A12 A21 A22 , 1. det(A) 6= 0 ⇔ A is invertible.

2. A is invertible ⇔ A−1 = 1

det(A)

 A22 −A12

−A21 A11





 The area of the parallelogram determined by 2-dim vectors

 a b



and  c d



is

det  a b c d



=

det  a c b d

 .

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 determinant of A ∈ Mn×n, direct definition:

det(A)= P

πsgn(π)A1π(1)A2π(2) · · · Anπ(n), where

 the sum is taken over all permutations π

 k(π): the total number of inversions in π

 sgn(π) = (−1)k(π). 

 example:

A11 A12 A13 A21 A22 A23 A31 A32 A33

 ⇒

det(A)= A11A22A33−A11A23A32−A12A21A33+A12A23A31+ A13A21A32 − A13A22A31

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 determinant of A ∈ Mn×n, recursive definition:

1. n = 1 ⇒ A = (A11) ⇒ det(A)= A11 2. n ≥ 2 ⇒ det(A) = Pn

j=1 A1jc1j, where cij is the cofactor.

 example:

det

1 2 3 4 5 6 7 8 9

 = 1·(−1)1+1·det 5 6 8 9



+2·(−1)1+2·det 4 6 7 9



+3 · (−1)1+3·det 4 5 7 8



= (45 − 48) − 2(36 − 42) + 3(32 − 35) = 0

 det(I)=1

proof: det(In) = (−1)1+1det(In−1) = · · · = (−1)1+1det(I2) = 1

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 The volume of the parallelepiped determined by 3-dim vectors

 a

b c

,

 d e f

 and

 g h

i

 is

det

a b c d e f g h i

=

det

a d g b e h c f i

 .

 Theorem 4.3: The determinant is a linear function of one row, when the other rows are held fixed. That is,

det

a1 ...

ku + lv ...

an

= k det

 a1

...

u ...

an

+ l det

 a1

...

v ...

an

 . 

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 Corollary 4.3: If A has an all-zero row, then det(A)= 0. 

 Lemma 4.4:

det(A) = det

a1 ...

0 · · · 010 · · · 0 ...

an

= cik = (−1)i+kdet( ˜Aik), where

Aik = 1. 

 Theorem 4.4: det(A) = Pn

j=1(−1)i+jAijdet( ˜Aij) [End of Review]

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 Corollary 4.4: A has two identical rows ⇒ det(A) = 0  proof by induction in n: (i) It is true for 2×2 matrices.

(ii) Assume it is true for (n − 1) × (n − 1) matrices.

(iii) Let i be an index of a row other than identical rows.

⇒ det(A)= Pn

j=1(−1)i+jAijdet( ˜Aij)= 0 [ ˜Aij is (n − 1) × (n − 1), (ii)] 

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 Theorem 4.5: A ero1→ B ⇒ det(B) = -det(A) 

proof: det

...

ar + as ...

ar + as ...

= det

...

ar ...

ar + as ...

+ det

...

as ...

ar + as ...

= det

 ...

ar ...

ar ...

+ det

 ...

ar ...

as ...

+ det

 ...

as ...

ar ...

+ det

 ...

as ...

as ...

⇒ 0 = 0 + det(A) + det(B) + 0 

 E is an em1 ⇒ I ero1→ E ⇒ det(E)= −1

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 Theorem 4.5a: A ero2→ B ⇒ det(B) = kdet(A), where k is the scalar multiplied to a row. 

 E is an em2 ⇒ I ero2→ E ⇒ det(E) = k

 Theorem 4.6: A ero3→ B ⇒ det(B) = det(A) 

proof: det

...

ar ...

as + kar ...

= det

 ...

ar ...

as ...

+ kdet

 ...

ar ...

ar ...

⇒ det(B) = det(A) + 0 

 E is an em3 ⇒ I ero3→ E ⇒ det(E)= 1

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 Corollary 4.6: rank(A)< n ⇒ det(A)= 0 

proof: rank(A)< n ⇒ n rows are linearly dependent.

⇒ The ith row is a1 = P

j6=i kjaj

⇒ det

a1 ...

k1a1 + · · · + knan ...

an

= k1det

 a1

...

a1 ...

an

+· · ·+kndet

 a1

...

an ...

an

⇒ det(A)= 0 + · · · + 0 

 Its converse is also true.

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Properties of determinant

 Theorem 4.7: det(AB) = det(A)det(B) 

proof: (i) If rank(A)< n ⇒ rank(AB)≤rank(A)< n

⇒ det(AB) = 0 = det(A)det(B).

Now assume that rank(A) = n.

(ii) If A is an em1, then B ero1→ AB.

⇒ det(AB) = -det(B) = det(A)det(B) [Thm 4.5]

(iii) If A is an em2, then B ero2→ AB.

⇒ det(AB) = kdet(B) = det(A)det(B) [Thm 4.5a]

(iv) If A is an em3, then B ero3→ AB.

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⇒ det(AB) = det(B) = det(A)det(B) [Thm 4.6]

(v) In general A = Em · · · E1 [A is invertible, Corol 3.6.3]

⇒ det(AB) = det(Em · · · E1B).

= det(Em)det(Em−1 · · · E1B)

= det(Em)det(Em−1)det(Em−2 · · · E1B)

= det(Em)det(Em−1)· · · det(E2)det(E1) det(B)

= det(Em)det(Em−1)· · · det(E2E1) det(B)

= det(Em)det(Em−1)· · · det(E3E2E1) det(B)

= det(Em · · · E1) det(B). 

 Corollary 4.7:

1. A is invertible ⇒ det(A)6= 0

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2. det(A−1) = (det(A))−1

proof: “1”: det(A) = det(Em)· · · det(E1)6= 0[Corol 3.6.3]

“2”: det(A)det(A−1) = det(AA−1) = det(I) = 1 

 Theorem 4.8: det(At) = det(A) 

proof: (i) If rank(A)< n, rank(At) = rank(A)

⇒ det(At) = det(A) = 0.

(ii) If rank(A)= n ⇒ A = Em · · · E1

⇒ det(At) = det(E1t · · · Emt ) = det(E1t) · · · det(Emt )

= det(E1) · · · det(Em) = det(Em · · · E1) 

 Now the determinant can be expanded along i-th “column”:

det(A) = det(At) = Pn

i=1(−1)i+jAijdet( ˜Aij)

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 Theorem 4.9 (Cramer’s rule): Ax = b is a system; det(A)6= 0;

and x = (x1, · · · , xn)t. Then, ci denoting the ith column of A, xk = 1

det(A)det(c1, c2, · · · , b, · · · , cn), where the k-th column is replaced by b. 

proof: Let Xk = (e1, e2, · · · , x, · · · , en), where the k-th column is x.

For example,

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⇒ ( ˜Xk)ik has an all-zero column except that ( ˜Xk)kk = In−1.

⇒ det(( ˜Xk)kk) = 1, and det(( ˜Xk)ik) = 0, i 6= k.

⇒ det(Xk) = Pn

i=1(−1)i+kXikdet(( ˜Xk)ik) = Xkk = xk

AXk = (Ae1, Ae2, · · · , Ax, · · · , Aen) = (c1, c2, · · · , b, · · · , cn)

⇒det(A)det(Xk) = det(c1, c2, · · · , b, · · · , cn)

⇒ det(A)xk = det(c1, c2, · · · , b, · · · , cn) 

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