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Unformatted text preview: Name: PID: Discussion Section No: Time: TA: Midterm 1, Math 20F  Winter 2008 Duration: 100 minutes This is an open book exam. Calculators are not allowed. To get full credit you should support your answers. 1. Consider the system of linear equations x 1 3 x 2 = 4 3 x 1 6 x 2 = 3 2 x 1 hx 2 = b where h and b are scalars. (a) (1 point) Explain why the system cannot have infinitely many solutions. Solution: Geometrically the first two equations can only be satisfied by a unique point, since they represent lines not parallel to each other. Third equation (depending on the choice of h and b ) can be a line that either goes through this intersection point in which case the system has a “unique solution”. If the line represented by the third equation does not pass through the intersection of the first two, the system is “inconsistent”. There are only two possibilities. This can also be seen by rowreducing the augmented matrix into an echelon form and verifying that the system has no free variables. 1 3 4 3 6 3 2 h b → r 2 := r 2 3 r 1 , r 3 := r 3 2 r 1 r 2 := r 2 / 3 1 3 4 1 3 h + 6 b 8 If h = 6, then the entry (2 , 2) is a pivot entry. Otherwise the entry (3 , 2) is nonzero, but can be made zero by applying one more rowreplacement operation. Therefore the first two columns (the columns corresponding to variables) are pivot columns. There are no free variables; the solution may be unique or may not exist at all. Answer for the other type: The only difference in the question is that the third equation is 4 x 1 + hx 2 = b . Same reasoning applies both geometrically and algebraically. Row reduction again reveals that there cannot be free variables. (b) (2 points) For what values of h and b is the system inconsistent? Solution The system is inconsistent whenever the last column of the augmented matrix is a pivot column. From the rowreduced matrix in (a) if h = 6, then the last column is a pivot column whenever b 6 = 8. If h 6 = 6, then applying one more rowreplacement operation yields 1 3 4 1 3 h + 6 b 8 → r 3 := r 3 + ( h 6) r 2 1 3 4 1 3 b 8 3( h 6) ....
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This note was uploaded on 04/23/2008 for the course MATH 20F taught by Professor Buss during the Winter '03 term at UCSD.
 Winter '03
 BUSS
 Math, Linear Algebra, Algebra

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