Matrix 1. Algebra and Linear Economic Models
References Ch. 1 3 (Turkington); Ch. 4 5.2 (Klein). [1] Motivation One market equilibrium Model Assume perfectly competitive market: Both buyers and sellers are price-takers. Demand: Qd = a + bP , a > 0, and b < 0. Supply:
Q a S
Qs = c + dP , c 0 and d > 0.
D
P -c/d c -a/b
Linear Models-1
Production occurs only if -c/d < -a/b: -c/d(db) > -a/b(db) -cb > -ad 0 > cb ad bc ad < 0. Equilibrium Condition: Qd = Qs . Equilibrium quantity and price: Q and P . Q and P satisfies:
Q = a + bP ; Q = c + dP
(1) (2):
(1) (2)
a + bP = c + dP P =
(3) (1):
Q =a+b
c-a >0 b-d
(3)
c - a a (b - d ) b( c - a ) bc - ad = + = b-d b-d b-d b-d
(4)
Two lessons here: An economic model should assign proper signs on coefficients. When demand and supply are linear, the equilibrium price and quantity are nothing but the solutions of two linear equations.
Linear Models-2
Equilibrium model of two markets Assumptions: Two goods (coffee and tea). Both markets are perfectly competitive. Two goods are substitutable (not complementary). Each producer can produce only one good (short-run). Market 1: Qd 1 = 10 - 2 P + P2 ; 1 Qs1 = -2 + 3P ; 1 Qd 1 = Qs1 Question: Why is the coefficient of P2 in the demand for Good 1 positive? Why is the coefficient of P1 in the demand for Good 2 positive? Why is no P2 in the supply of Good 2? Market 2 Qd 2 = 15 + P - P2 ; 1 Qs 2 = -1 + 2 P2 . Qd 2 = Qs 2
Linear Models-3
At equilibrium: 1) Q1 = 10 - 2 P1 + P2 ; 2) Q1 = -2 + 3P 1 2) 1): 5) 5P - P2 = 12 . 1 6) 5):
5 (3P2 - 16) - P2 = 12
3) Q2 = 15 + P1 - P2 ; 4) Q2 = -1 + 2 P2 . 3) 4): 6) P = 3P2 - 16 . 1
7) P2 = 92
14
7) 6):
P = 3 92 1 14 - 16 = 52 14
You can solve for Q1 and Q2 . Lesson: As the number of equations increases, it becomes harder to solve a system of linear equations. Question: How can we find the equilibrium prices and quantities for multiple market models?
Linear Models-4
Use Matrices: 1) 1 Q1 + 0 Q2 + 2 P1 + ( -1) P2 = 10 ; 2) 1 Q1 + 0 Q2 - 3 P1 + 0 P2 = -2 ; 3) 0 Q1 + 1 Q2 + ( -1) P1 + 1 P2 = 15 ; 4) 0 Q1 + 1 Q2 + 0 P1 + ( -2) P2 = -1 .
1 1 0 0 0 2 -1 Q1 10 0 -3 0 Q2 -2 = P 15 1 -1 1 1 1 0 -2 P2 -1
0 2 -1 10 0 -3 0 -2 . 1 -1 1 15 1 0 -2 - 1
-1
Q1 1 1 Q 2= P 0 1 1 P 0
Linear Models-5
[2]
Matrix and Operations
Definition: Matrix A matrix, A, is a rectangular array of real numbers:
a11 a12 ... a1n a a22 ... a2 n 21 = [aij ]mn , A= : : : a m1 am 2 ... amn where i indexes row and j indexes column. A is called an mn matrix (m = # of rows; n = # of columns). Am n or [aij ]m n denote an mn matrix.
1 1 3 EX: ; [ 2 1 3]13 , [ 4]11 5 - 1 0 2 3
A 11 matrix is called "scalar". Definition: Square Matrix If m = n for an mn matrix A, A is called a square matrix.
1 2 1 EX: 3 0 0 -1 4 1 33
Linear Models-6
Definition: Transpose Let A be an mn matrix. The transpose of A is denoted by At (or A ), which is a nm matrix; and it is obtained by the following procedure. 1st column of A 1st row of At, 2nd column of A 2nd row of At... etc.
2 6 2 1 3 t EX: A = A = 1 2 . 6 2 2 3 2
Note: ( At )t = A . Definition: Symmetric Matrix Let A be a square matrix. A is called symmetric if and only if (iff) A = At.
3 2 1 2 2 2 EX: A = 2 6 5 = At ; B = 2 2 2 = B t . 1 5 7 2 2 2
Linear Models-7
Definition: Adding Matrices Let A and B be mn matrices. (A + B) is obtained by adding corresponding entries of A and B.
1 4 3 1 4 5 1 4 3 1 -2 3 EX: . + = ; - = 2 2 4 5 6 7 2 2 4 5 -2 -3
Definition: Let A = [aij ] be an mn matrix and c be a scalar (real number). Then, cA is obtained by multiplying all the entries of A by c: cA = [caij ] .
2 4 12 24 EX: 6 . = 3 5 18 30
Linear Models-8
Definition: Vector Any m1 matrix is called a column vector. Any 1n matrix is called a row vector. Vectors are normally denoted by lower cases (e.g., x, y, a, b).
2 EX: x = 1 ; z = ( 6 2 3) . 2
Note: An mn matrix can be viewed as a collection of m row vectors or n column vectors. a11 a12 ... a1n a1 a a22 ... a2 n a2 21 = ( a1 a2 ... a n ) , = EX: A = : : : : a m1 am 2 ... amn am where,
a1 j a 2j ... ain ) ; a j = . : amj
ai i = ( ai1 ai 2
Linear Models-9
Definition: Multiplication of Vector Suppose a and b are 1p and p1 vectors, respectively:
b1 b 2 a = ( a1 ,..., a p ); b = . : bp
Then, ab = a1b1 + a2b2 + ... + a p bp = ip=1ai bi .
4 EX: a = (1 2 3) ; b = 1 ab = 1 4 + 2 1 + 3 2 = 12 . 2
Definition: Multiplication of Matrices Let A and B are mp and pn matrices, respectively. Let
a11 a12 ... a1 p a1 b11 b12 ... b1n a a b a22 ... a2 p b22 ... b2 n 21 21 2 = = ( b b ... b ) A= ;B= 1 2 n : : : : : : : a b m1 am 2 ... amp am p1 bp 2 ... bpn
Then,
a1b1 a1b2 ... a1b n a b a2b2 ... a2b n 2 1 AB = : : : a b a b ... a b m n m n m 1 m 2
Linear Models-10
EX 1:
4 3 1 3 5 A= ; B = 2 1 . 2 4 6 1 0
1 4 + 3 2 + 5 1 1 3 + 3 1 + 5 0 15 6 AB = . = 2 4 + 4 2 + 6 1 2 3 + 4 1 + 6 0 22 10
EX 2: System of m linear equations.
a11 x1 + a12 x2 + ... + a1n xn = b1 a21 x1 + a22 x2 + ... + a2 n xn = b2 : am1 x1 + am 2 x2 + ... + amn xn = bm x1 a11 a12 ... a1n b1 x a b a22 ... a2 n 21 2 , x = Ax = b , where A = and b = 2 . : : : : : xn am1 am 2 ... amn bn EX 3:
x1 + 2 x2 = 1; x2 = 0 1 x1 + 2 x2 = 1
1 2 x1 1 . = 0 1 x2 0 0 x1 + 1 x2 = 0
Linear Models-11
Some Caution in Matrix Operations: AB may not be equal to BA (Commutative Law does not hold.)
-1 0 1 2 AB BA . EX: A = , B= 2 3 3 0
But the distributive law holds: A(B+C) = AB + AC, if AB and AC are computable. (B+C)A = BA + CA, if BA and CA are computable. AB = AC does not mean B = C .
0 1 1 1 2 5 EX: A = , B= , C= AB = AC . 0 2 3 4 3 4
AD = 0 (zero matrix) does not mean that A = 0 or D = 0 .
0 1 3 7 0 0 AD = . EX: A = , D= 0 2 0 0 0 0
Theorem:
( AB )t = B t At . ( A + B )t = At + B t .
Linear Models-12
Some Special Matrices (1)
Identity Matrix
Let I n be an nn square matrix. I n is called an identity matrix if all of the diagonal entries are ones and all of the off-diagonals are zeros.
1 0 0 1 0 EX: I 2 = ; I3 = 0 1 0 . 0 1 0 0 1
Note:
Amn I n = Amn ; I m Amn = Amn .
(2)
Zero (Null) Matrix
Let A = [aij ]mn . If aij = 0 for all i and j, A is called a zero matrix.
0 0 0 0 0 EX: A = 0 0 0 ; B = 0 0 A is a zero matrix, but not B. 0 0 0 0 1
(3)
Scalar Matrix
0 ... 0 0 ... 0 is called a scalar matrix. For any scalar , I n = : : : 0 0 ...
Linear Models-13
(4)
Diagonal Matrix
1 0 ... 0 0 ... 0 2 . diag (1 , 2 ,..., n ) = : : : 0 0 ... n (5)
Triangular Matrix
A square matrix Ann = [aij ] is called an upper (lower) triangular if
aij = 0 for all i < j (i > j).
1 2 3 1 0 0 EX: A = 0 4 5 (upper); B = 2 3 0 (lower) 0 0 6 4 5 6
3 6 C = 2 1 (6)
2 1 1 4 1 0 (not triangular) 3 0 0 0 0 0
Idempotent Matrix
A matrix Ann is said to be idempotent iff AA = A .
1/ 2 -1/ 2 EX: ; -1/ 2 1/ 2 1/ 2 1 1/ 4 1/ 2 .
Linear Models-14
[3]
Inverse and Determinant
Definition: Inverse For Ann and Bnn, B is called the inverse of A iff AB = In or BA = In. The inverse of A is denoted by A-1 .
2 -5 3 5 1 0 -1 EX: A = ;B= AB = BA = B=A . -1 3 1 2 0 1
Theorem:
a b 1 d -b . A22 = A-1 = c d ad - bc - c a
If ad - bc = 0 , then, A-1 does not exist. Terminology: If a square matrix A has an inverse, A is said to be "invertible" or "nonsingular". If A does not have an inverse, A is said to be "singular."
Linear Models-15
Theorem: 1) For Ann and Bnn, if AB = In, then, BA = In. 2) A-1 is unique if it exists. 3) ( AB ) -1 = B -1 A-1 if both A and B are invertible and conformable. 4) ( A-1 ) -1 = A . 5) ( At ) -1 = ( A-1 )t . Proof: 2) Suppose that AB = AC = In. B = BIn = BAC = InC = C. 3) ( AB )( B -1 A-1 ) = ABB -1 A-1 = AI n A-1 = AA-1 = I n . 4) A-1 A = I n ( A-1 ) -1 = A .
t 5) At ( A-1 )t = ( A-1 = A)t I n = I n . ( A-1 )t is the inverse of At .
EX: A system of linear equations is given Amn xn1 = bm1 . If m = n and A is invertible,
A-1 Ax = A-1b I n x = A-1b x = A-1b .
Question: How can we find an inverse if n > 2?
Linear Models-16
Definition: Determinant of 22 Matrix
a a Let A22 = 11 12 . Then, | A | det( A) = a11a22 - a12a21 . a21 a22 2 1 EX: A = det( A) = 2 4 - 1 3 = 5 . 3 4
Definition: Determinant of 33 Matrix
a11 a12 Let A33 = a21 a22 a 31 a32 a13 a23 . Write a33
a11
a12
a13 : a11
a12
a13 a23 a33
a21 a22 a31 a32
a23 : a21 a22 a33 : a31 a32
det( A) = a11a22 a33 + a12 a23a31 + a13a21a32
-a11a23a32 - a12 a21a33 - a13a22a31 1 2 3 1 2 3 det( A) = 1i5i4 + 2i1i1 + 3i 4i3 EX: A = 4 5 1 4 5 1 -1i1i3 - 2i4i4 - 3i5i1 = 8 1 3 41 3 4
Linear Models-17
Definition: Minor and Cofactor Let Ann = [aij ] . Then, the minor of aij ( M ij ) is the (n-1)(n-1) matrix excluding the ith row and the jth column of A . The cofactor of
aij ( | Cij | ) is ( -1)i + j det( M ij ) .
Definition: Determinant of nn Matrix (Laplace Expansion) For Ann , det( A) = nj =1aij | Cij | = ai1 | Ci1 | + ai 2 | Ci 2 | +... + ain | Cin | , for any choice of i. Also, det( A) = in=1aij | Cij | = a1 j | C1 j | + a2 j | C2 j | +... + anj | Cnj | , for any choice of j.
1 2 3 A = 4 5 1 . 1 3 4
EX 1:
Choose the first row:
a11 = 1: | M 11 | = a12 = 2 : | M 12 | = 5 1 = 17 | C11 | = ( -1)1+1 | M 11 | = 17 . 3 4 4 1 = 15 | C12 | = ( -1)1+ 2 | M 12 | = -15 1 4
a13 = 3; | C13 | = 7 . det( A) = a11 | C11 | + a12 | C12 | + a13 | C13 | = 1 17 + 2 ( -15) + 3 7 = 8 .
Linear Models-18
EX 2:
1 2 A= 3 4
2 6 0 3
1 1 0 1
2 2 1 2 1 det( A) = 3( -1)3+1 6 1 1 . 0 3 1 4 4
Theorem:
det( I n ) = 1.
Theorem: If all of the entries in the ith row (jth column) of Ann are zero, then,
det( A) = 0 .
Theorem: If Ann = [aij ] is a triangular matrix, det( A) = a11a22 iiiiiann .
1 2 3 1 0 0 EX: A = 0 4 5 ; B = 2 3 0 0 0 6 4 5 6
det( A) = 1 4 6 = 24; det( B ) = 1 3 6 = 18
.
Linear Models-19
Theorem: Elementary Row (Column) Operations (a) (b) (c) If multiplying a single row (column) of A by a constant k results in B, det( B ) = k det( A) . If adding a multiple of one row (column) of A to another row (column) results in B, det( B ) = det( A) . If B results when two rows (columns) of A are interchanged, then, det( B ) = - det( A) . EX.a:
1 2 3 4 2 4 6 8 ;B= A= det( B ) = 2 det( A) . somthing same as A
1 2 2 4 A = something ; B = same as 3 6 4 8
1 2 3 2 4 6 A = 2 1 1; B = 4 2 2 . 1 1 2 1 1 2 1 2 3
A det( B ) = 2 det( A) .
1 2 3
det( B ) = 2 4 2 2 = 2 2 2 1 1 = 4 det( A) . 1 1 2 1 1 2
Linear Models-20
EX.b:
1 2 3 4 1 2 3 4 A = 2 3 4 5 ; B = 1 1 1 1 det( B ) = det( A) . something same as A
r2 of B = r2 of A r1 of A = (-1)r1 of A + r2 of A. 1 2 A= 3 4 EX.c:
1 2 3 4 5 6 A = 4 5 6 ; B = 1 2 3 det( B ) = - det( A) . something same as A
2 1 2 3 somthing ; B = 4 3 5 4
1 1 something det( B ) = det( A) . 1 1
EX:
1 2 3 A = 0 1 4 det( A) = -2 (Check this!) 1 2 1 4 8 12 B = 0 1 4 (r1 of B = 4r1 of A) det( B ) = 4 ( -2) = -8 . 1 2 1
Linear Models-21
1 2 3 C = -2 -3 -2 (r2 of B = r2 of A 2r1 of A) 1 2 1
det(C ) = -2 .
0 1 4 D = 1 2 3 (r1 and r2 of A are interchanged) 1 2 1
det( B ) = ( -1) ( -2) = 2 .
3 6 EX: C = 2 1
2 1 1 4 1 0 (not triangular) 3 0 0 0 0 0 2 4 3 0 1 1 0 0 3 1 6 0 = ( -1) ( -1) 2 0 1 0 1 1 0 0 2 4 3 0 3 6 = 3. 2 1
1 0 det(C ) = ( -1) 0 0
Linear Models-22
Theorem: Suppose that two rows (columns) of Ann are identical. Then,
det( A) = 0 .
Proof: Suppose that r1 = r2. If you subtract r1 from r2, the second row becomes zero. Thus, the determinant of this new matrix should be zero. Since this transformation does not alter the determinant of A, it must be the case that det( A) = 0 . Theorem: Expansion Using Alien Cofactors For Ann , nj =1ahj | Cij | = ah1 | Ci1 | + ah 2 | Ci 2 | +... + ahn | Cin | = 0 , for any choice of h i . Also, for any choice of h j , in=1aih | Cij | = a1h | C1 j | + a2 h | C2 j | +... + anh | Cnj | = 0 . Proof: Consider a matrix, Bnn , which is identical to A except that the ith row of A is replaced by the hth row of A. Since the two rows of B are identical, det( B ) = 0 . The Laplace extension of B using the ith row of B will give us: nj =1ahj | Cij | = ah1 | Ci1 | + ah 2 | Ci 2 | +... + ahn | Cin | = det( B ) = 0 .
Linear Models-23
1 3 EX: A = 2 0
2 2 4 2
3 0 5 3
4 3 3 1 ; B = 7 2 4 0
2 2 4 2
0 0 5 3
1 1 det( B ) = 0 . 7 4
For A and B, the cofactors corresponding to the first row entries (say, | C11 |, | C12 |,| C13 |, | C14 | ) are the same. a21 | C11 | + a22 | C12 | + a23 | C13 | + a24 | C14 | = det( B) = 0 Theorem: For any Ann and Bnn , det( AB ) = det( A) det( B ) .
3 1 -1 3 det( A) = 1; det( B ) = -23 . EX: A...