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Math 4 Winter 2008(Instructor: Professor R. C. Reilly)Answer Key for Practice Midterm #1(1) (a) LetI3denote the 3-by-3 identity matrix. We use Gauss-Jordan Elimination on the matrix [A|I3] to obtainA-1. You should be able to figure out the row operations being used, so I won’t list them out here. (I do list themout in some of the other problems.)[A|I3] =2-14100060010-123001→-1230010600102-14100→-1230010600100310102→-1230010310102060010→-123001031010200-20-21-4→-12300103101020011/10-1/201/5→-120-3/103/202/503001/200011/10-1/201/5→-120-3/103/202/501001/600011/10-1/201/5→-100-3/10-11/602/501001/600011/10-1/201/5→1003/1011/60-2/501001/600011/10-1/201/5From this one reads offA-1=3/1011/60-2/501/601/10-1/201/5(b) We learned in the lecture that whenA-1exists, then the solution toAx=bisA-1b. In this case thatmeansx=3/1011/60-2/501/601/10-1/201/53-10=43/60-1/67/20(2) Let us transform the augmented matrixC= [A|b] =-124-243-6513to reduced row-echelon form bythe usual elementary row operations:-124-243-6513(1)→-124-240017-515(2)→-124-24001-5/1715/17(3)→-120-14/178/17001-5/1715/17(4)→1-2014/17-8/17001-5/1715/17The steps in the preceding row-reduction can be summarized as follows:Step (1):R2→R2+ 3R1. Note: By the end of this step it is clear that the original system is consistent and thatthere will be two free variables in the final solution, namelyx2andx4.Step (2):R2→R2/17Step (3):R1=R1+ (-4)R2. This is the first part of the ‘Jordan’ half of ’Gauss-Jordan Elimination’.Step (4):R1→(-1)R1Note: This completes the ‘Jordan’ half; the final matrix is in reduced row-echelon form.When one translates the final matrix into equations, one getsx1=-817+ 2x2-1417x4,x3=1517+517x4.1
Thus, when put into the ‘standard form’, the general solution isx=x1x2x3x4=-817