(b) Here we have to find a linear system whose solution is⎛⎝x1x2x3⎞⎠=t⎛⎝523⎞⎠=⎛⎝5t2t3t⎞⎠.From the last row we read offt=13x3so thatx1= 5t=53x3, andx2= 2t=23x3.This system can be written more nicely as3x1−5x3= 03x2−2x3= 0and corresponds to the matrix (i.e. linear transformation)T=30−503−2.8. We first write the augmented coeﬃcient matrix and then perform Gauss-Jordan elimi-nations (row operations):⎛⎝111|112k|214k2|3⎞⎠subtract row I from row IIsubtract row I from row III⎛⎝111|101k−1|103k2−1|2⎞⎠subtract row II from row Isubtract 3 times row II from row III⎛⎝10−k+ 2|001k−1|100k2−3k+ 2|−1⎞⎠Let us observe thatk2
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