January 22 Inverse of Matrices

Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone Text: Linear Algebra With Applications, Second Edition Otto Bretscher

34 Problem 4: Find the Rank of this matrix: | | | | | | Solution: Reduce to Row Canonical form and count the nonzero rows. | | | | | 1 4 7| | | | | | | | 0 1 2| | 0 1 2| | | | | | 0 0 0| | 0 0 0| | 1 | The rank is 2. The vector | -2 | annihilates the matrix | 1 |

Page 34 Problem 18 Page 34 Problem 18 Multiply these two matrices. | 1 2 | | 1 | | 5 | | 3 4 | | 2 | = |11 | | 5 6 | |17 | The first column plus twice the second.

Page 34 Problem 56 | 7 | | 1 | | 4 | Is the vector | 8 | a linear combination of | 2 | and | 5 | ? | 9 | | 3 | | 6 | We solve for a and b in the dependence relation. | 1 | | 4 | | 7 | a | 2 |+ b | 5 | = | 8 | | 3 | | 6 | | 9 | a b RHS a b RHS | | | 1 4 7| | 1 4 7| | | ||_1_ 0 -1 | | | | | | 0 1 2| | 0 1 2| | 0 |_1__ 2| | | | | | | | 0 0 0| | | Answer: yes., there is a dependence relation. | a | | -1 | | b | = | 2 | Check: : | 1 | | 4 | | 7 | -1 | 2 |+ 2 | 5 | = | 8 | | 3 | | 6 | | 9 |

New Material: We make some useful and perhaps obvious remarks using the terminology. We are solving AX = B. If we can solve AX = B then we know that B is a linear combination of the columns of A. The set of all solutions to AX = 0 is called the null space of A.

If we look at the Row Canonical form of [A | B] x1 x2 x3 x4... x RHS | |_1__ 0 0 | | |_1_____ 0 | | |__1____________ | | | In a matrix in Row Canonical Form. 1. Number of stair-step ones = number of non-zero rows. 2. The system AX = B has a solution if and only if there is NO stair-step one in the RHS column. 3. The Row Canonical Form of A is the initial part of the Row Canonical Form of [A | B]. 4. AX = B has a solution rank of A = Rank of [A | B]

5 Unknowns are either assigned parameters or are determined by the equations in the Row Canonical Form. Thus the number of stairstep ones plus the number of parameters equals the number of columns of A. 6.The number of parameters is called the nullity. Thus we get the equation rank(A) + nullity(A) = number of columns(A)

7.Suppose that the RCF(A) = I and we wish to solve AX = Bi for several vectors Bi. We could solve each of them separately, or, more efficiently, we could do it all at once by this technique. [A | B1 B2 B3 … Bn ] ~~~[ X1 X2 X3 … Xn] By the way these things multiply, we have A.[X1 X2 X3 … Xn] = [B1 B2 B3 … Bn].

In particular, if we start with [B1 B2 …Bn] = I, | 1 | | 0 | | 0 | | 0 | | 1 | | 0 | B1 = |. | B2 = |. |... Bn = |. | |. | |. | |. | |, | |, | |, | | 0 | | 0 | | 1 | Then [ A | I | ~~~[I, X ] where A.X = I

8. Matrix Multiplication is associative. (AB)C = A(BC) 9. Matrix Multiplication is not commutative. That is except for rare cases of A and B. AB =/= BA 10. The elementary row operations can be done by using ordinary matrix multiplication!.

This matrix multiplies row i by c. i | 1 | | 1 zeros | i | c | | 1 | | 1 | | 1 | | zeros 1 | | 1 | The matrix is all zero except for the diagonal Which has the number c in the (i, i) position and The rest of the elements of the diagonal are 1.

This matrix switches rows i and j i j | 1 | | 1 | i | | |. 1. | |. 1. | j | | | 1 | | 1 | Start with the identity matrix and switch Row i and Row j.

This matrix adds c times row i to row j i | 1 | | 1 | i | 1 | | 1 | | 1 | | 1 | j | c 1 | | 1 | Start with the identity matrix and put the number c in the (j, i) position. Equivalently, add c times row I to row j.

11. If we reduce a matrix A to row canonical form then we can find these elementary row operation matrices such that Ek Ek-1 Ek-2... E1 A = RCF(A). Thus (Ek Ek-1 Ek-2...E1) A = RCF(A). Letting P = Ek Ek-1 Ek-2 … E1 We have a matrix P such that PA = RCF(A).

12. If the RCF(A) = I, then (a) there exists a matrix B such that BA = I. (b) there exists a matrix C such that AC = I. (c) B = C (d) the matrix A has an inverse. That is, there exists a matrix W such that AW = I and WA = I W is called the inverse of A. (e) AV = 0 implies V = 0.

13. If A is square and RCF(A) =/= I, then there exists a non zero vector V such that AV = 0. Therefore A is not invertible. The combination of points 12 and 13 mean that a square matrix is invertible if and only if its Row Canonical Form is the identity matrix. If the Row Canonical form of A is not the identity matrix, then A cannot be invertible because it kills a nonzero vector V. A matrix cannot be invertible and kill a nonzero vector.

The Identity Matrix The square matrix with off diagonal entries zero and all ones on the main diagonal is called the identity matrix I. Under multiplication: I A = A and A I = A. | 1 | | 1 zeros | | 1 | I = |. | |. | | zeros. | | 1 |