SOLUTIONS TO PROBLEMS ELEMENTARY LINEAR …

SOLUTIONS TO PROBLEMSELEMENTARYLINEAR ALGEBRAK. R. MATTHEWSDEPARTMENT OF MATHEMATICSUNIVERSITY OF QUEENSLANDF irst Printing,1991 CONTENTSPROBLEMS (i) 0 0 02 4 0 R1 R2 2 4 00 0 0 R1 12R1 1 2 00 0 0 ;(ii) 0 1 31 2 4 R1 R2 1 2 40 1 3 R1 R1 2R2 1 0 20 1 3 ;(iii) 1 1 11 1 01 0 0 R2 R2 R1R3 R3 R1 1 1 00 0 10 1 1 R1 R1+R3R3 R3R2 R3 1 0 00 1 10 0 1 R2 R2+R3R3 R3 1 0 00 1 00 0 1 ;(iv) 2 0 00 0 0 4 0 0 R3 R3+ 2R1R1 12R1 1 0 00 0 00 0 0 .3. (a) 1 1 1 22 3 1 81 1 1 8 R2 R2 2R1R3 R3 R1 1 1 120 1 340 2 2 10 R1 R1 R2R3 R3+ 2R2 1 0 4 20 1 3 40 0 8 2 R3 18R3 1 0 4 20 1 3 40 0 114 R1 R1 4R3R2 R2+ 3R3 1 0 0 30 1 01940 0 114 .The augmented matrix has been converted to reduced row echelon formand we read off the unique solutionx= 3, y=194, z=14.

solutions to problems elementary linear algebra k. r. matthews department of mathematics university of queensland first printing, 1991

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  Linear, Elementary, Matthew, Algebra, Elementary linear algebra k, Elementary linear

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