A First Course in Linear Algebra by Robert A Beezer By Robert A Beezer

A primary direction in Linear Algebra is an advent to the elemental strategies of linear algebra, in addition to an creation to the thoughts of formal arithmetic. It starts with structures of equations and matrix algebra earlier than stepping into the speculation of summary vector areas, eigenvalues, linear ameliorations and matrix representations. It has various labored examples and workouts, besides distinctive statements of definitions and whole proofs of each theorem, making it excellent for self sufficient research.

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Extra resources for A First Course in Linear Algebra

Example text

Example NRREF A matrix not in reduced row-echelon form The matrix E is not in reduced row-echelon form, as it fails each of the four requirements once.   1 0 −3 0 6 0 7 −5 9 0 0 0 5 0 1 0 3 −7   0 0 0 0 0 0 0 0 0 E=  0 1 0 0 0 0 0 −4 2  0 0 0 0 0 0 1 7 3 0 0 0 0 0 0 0 0 0 Our next theorem has a “constructive” proof. Learn about the meaning of this term in Proof Technique C. Theorem REMEF Row-Equivalent Matrix in Echelon Form Suppose A is a matrix. Then there is a matrix B so that §R R E F Beezer: A First Course in Linear Algebra 30 1.

Example FDV Free and dependent variables Consider the system of five equations in five variables, x1 − x2 − 2x3 + x4 + 11x5 = 13 x1 − x2 + x3 + x4 + 5x5 = 16 2x1 − 2x2 + x4 + 10x5 = 21 2x1 − 2x2 − x3 + 3x4 + 20x5 = 38 2x1 − 2x2 + x3 + x4 + 8x5 = 22 whose augmented matrix row-reduces  1 −1 0 0   0 0 0 0 0 0 to 0 1 0 0 0 0 0 1 0 0 3 −2 4 0 0  6 1   9 0 0 Columns 1, 3 and 4 ae pivot columns, so D = {1, 3, 4}. From this we know that the variables x1 , x3 and x4 will be dependent variables, and each of the r = 3 nonzero rows of the row-reduced matrix will yield an expression for one of these three variables.

The choice of the letters D and F refer to our upcoming definition of dependent and free variables (Definition IDV). An example will help us begin to get comfortable with this aspect of reduced row-echelon form. Example RREFN Reduced row-echelon For the 5 × 9 matrix  1 5 0 0 0 0 1 0  B= 0 0 0 1   0 0 0 0 0 0 0 0 45 form notation 2 4 3 0 0 8 7 9 0 0 0 0 0 1 0  5 −1 2 0  3 −6   4 2 0 0 §T S S Beezer: A First Course in Linear Algebra 46 in reduced row-echelon form we have r=4 d1 = 1 d2 = 3 d3 = 4 d4 = 7 f1 = 2 f2 = 5 f3 = 6 f4 = 8 f5 = 9 Notice that the sets D = {d1 , d2 , d3 , d4 } = {1, 3, 4, 7} F = {f1 , f2 , f3 , f4 , f5 } = {2, 5, 6, 8, 9} have nothing in common and together account for all of the columns of B (we say it is a partition of the set of column indices).