**Terminology:**- diagonal matrix, zero matrix, matrix-matrix product, commute, transpose, matrix inverse, invertible, singular and nonsingular, determinant, inner (scalar or dot) product, outer product.
**Objectives:**- learn how to combine matrices linearly; learn about the the matrix-matrix product and its properties; learn about the matrix transpose and its properties; learn about matrix inverses and how to find them; learn how to use matrix algebra to simplify matrix expressions.
**Reading Assignment:**- Chapter 2, Sections 2.1-2.2 (pages 97-117).
**Lesson Outline****Key Ideas and Discussion:**- Matrices can be linearly combined just like vectors can
be linearly combined; after all,
*n*-vectors are just matrices. However, multiplication of matrices is more complicated. An easy way to multiply two matrices*A*and*B*, is to use think of*B*as a group of columns and use the the matrix-vector product on each column. The columns of*C*=*AB*are just products of*A*with the respective columns of*B*. For this to work properly, the number of columns of*A*must be the same as the number of rows of*B*. Another way to compute*C*is element by element;*c*_{ij}is the inner product of row*i*of*A*with column*j*of*B*. An unfortunate fact about the matrix product is that usually (most matrices do not commute). You are probably very used to using*ab*=*ba*when working with scalars*a*and*b*, but when you are doing matrix algebra, it is very important to remember that you cannot assume*AB*=*BA*. One consequence of this is that the transpose of a product of matrices is the product of the transposes of the individual matrices, but with the*order*of the terms*reversed*.Inverses of matrices can be used to find solutions to linear systems. The scalar equation

*ax*=*b*has the solution*x*=*b*/*a*; the correct analogue for the solution to a linear system is . An matrix*A*is invertible (*A*^{-1}, the unique inverse, exists) if and only if has a unique solution for all in*R*^{n}. The inverse of a product of square matrices is the product of the inverses of the terms, with the*order*of the terms*reversed*. There is a simple formula for inverses of matrices. For larger matrices, elementary row operations can be used to find inverses. The whole process is equivalent to the simultaneous solution of a set of*n*linear systems whose right-hand sides are the columns of the identity matrix*I*. In practice, you take*A*and augment it with*I*. Then reduce.*A*completely (if possible), performing the same row operations on*I*. When you are finished, the part of the augmented*A*that was originally occupied by*I*, now contains*A*^{-1}. **Practice Problems:**- 2.1.1, 5, 9, 11, 15, 25 (pp. 107-109);
2.2.1, 5, 7, 17, 19, 31, 33 (pp. 117-119).
**Assignment**

1999-09-22