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MATRIX ALGEBRA

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MATRIX ALGEBRA

MGT 4850

Spring 2008

University of Lethbridge

- Let A,B,C be matrices of the same size m × n, 0 the m × n zero
- matrix, and c and d scalars.
- (1) (Closure Law) A + B is an m × n matrix.
- (2) (Associative Law) (A + B) + C = A + (B + C)
- (3) (Commutative Law) A + B = B + A
- (4) (Identity Law) A + 0 = A
- (5) (Inverse Law) A + (−A) = 0
- (6) (Closure Law) cA is an m × n matrix.

- (7) (Associative Law) c(dA) = (cd)A
- (8) (Distributive Law) (c + d)A = cA + dA
- (9) (Distributive Law) c(A + B) = cA + cB
- (10) (Monoidal Law) 1A = A

- Definition of Multiplication
2x − 3y + 4z = 5

as a “product” of the coefficient matrix

[2,−3, 4]

and the column matrix of unknowns

⎡ x ⎤

y │

⎣ z ⎦

[x

- Let A,B,C be matrices of the appropriate sizes so that the following multiplications make sense, I a suitably sized identity matrix, and c and d scalars.
(1) (Closure Law) The product AB is a matrix.

(2) (Associative Law) (AB)C = A(BC)

(3) (Identity Law) AI = A and IB = B

(4) (Associative Law for Scalars) c(AB) = (cA)B = A(cB)

(5) (Distributive Law) (A + B)C = AC + BC

(6) (Distributive Law) A(B + C) = AB + AC

- (skip from p.67 to p.101)

- Let A be a square matrix. Then a (two-sided) inverse for Invertible A is a square matrix B of the same size as A such that AB = I = BA. If such Matrix a B exists, then the matrix A is said to be invertible.
- Application-if we could make sense of “1/A,” then we could write the solution to the linear system Ax = b as simply x = (1/A)b.

Any nonsquare matrix is noninvertible. Square matrices are classified as either “singular,” i.e., noninvertible, or nonsingular,” i.e., invertible. Since we will mostly be concerned with two-sided inverses, the unqualified term “inverse” will be understood to mean a “two-sided inverse.” Notice that

this definition is actually symmetric in A and B. In other words, if B is an inverse for A, then A is an inverse for B.

(1) (Uniqueness) If A is invertible, then it has only one inverse, by A−1.

(2) (Double Inverse) If A is invertible, then (A−1)−1 = A.

(3) (2/3 Rule) If any two of the three matrices A, B, and AB are invertible, then so is the third, and moreover, (AB)−1 = B−1A−1.

(4) If A is invertible, then (cA)−1 = (1/c)A−1.

(5) (Inverse/Transpose) If A is invertible, then (AT )−1 = (A−1)T .

(6) (Cancellation) Suppose A is invertible. If AB = AC or BA = CA, then B = C.

skip from p.103 to p.113

- Let A be an invertible n×n matrix and b an n×1 column vector.
- Denote by Bi the matrix obtained from A by replacing the ith column of A
- by b. Then the linear system Ax = b has unique solution x = (x1, x2, . . . , xn),

- Use the Cramer’s rule to solve the system

- The coefficient matrix and right-hand-side vectors are