# Math midterm 1 solution

(28 Points.) The following are True/False questions. For this problem only, you do not have to show any work. There will be no partial credit given for this problem. For this problem:

- A correct answer gives 4 points.
- An incorrect answer gives 0 points.
- If you leave the space blank, you receive 2 points.

F (a) Five vectors in R^{5 }always span R^{5}.

F (b) If *T *: R^{n }→ R^{m }and *U *: R^{m }→ R^{p }are both linear transformations, then the standard matrix for the linear transformation *U *â—¦*T *is *n *× *p*.

__T __(c) If the Row Echelon form of a matrix *A *has a pivot in every column, then whenever *A*~*x *= ~*b *has a solution, it must be unique.

__T __(d) The sum of two solutions to the homogeneous problem *A*~*x *= ~0 is also a solution.

__T __(e) If a linear transformation *T *: R^{n }→ R^{m }is both one-to-one and onto, then *n *must equal *m*.

F (f) Two vectors whose entries are all positive must be linearly independent.

__T __(g) Assume*T *: R^{n }→ R^{m }is a linear transformation and~v_{1},. . . ,~v_{p }are vectors in R^{n}. If {*T*(~v_{1}),. . . ,*T*(~v_{p})} are linearly independent, then {~v_{1},. . . ,~v_{p}} are also linearly independent.

Each statement below is either true (in all cases) or false (in at least one case).

If false, construct a counterexample. If true, give a justification that carefully notes everywhere you use the definition of “linear (in)dependence.”

- (12 Points.) If ~v
_{1 }and ~v_{2 }are linearly independent and ~v_{1},~v_{2},~v_{3}are linearly dependent, then ~v_{3 }is in Span ~v_{1},~v_{2}. - (12 Points.) If
*u*~_{1 }and*u*~_{2 }are linearly dependent vectors in R^{2}, then one of these vectors must be a constant multiple of the other.

*(a) *True. *We want to show that *~v_{1 }*can be written as a linear combination of *v_{1 }*and *v_{2}*:*

~v_{1},~v_{2},~v_{3} *being linearly dependent means that there are numbers a*_{1}*, a*_{2}*, and a*_{3 }not all zero *so that*

*a*_{1}~v_{1}+ *a*_{2}~v_{2}+ *a*_{3}~v_{3 }= 0.

*Note that a*1 , 0 *because otherwise we would have*

*a*_{2}~v_{2}+ *a*_{3}~v_{3 }= 0,

*where at least one of a*_{2 }*or a*_{3 }*is nonzero – which can’t happen because *~v_{1 }*and *~v_{2 }*are linearly independent.*

*So we’re safe to divide through by a*_{1}*, giving:*

*a*2 *a*3

~v_{1}+ ~v_{2}+ ~v_{3 }= 0, *a*1 *a*1

*and thus:*

*a*2 *a*3

~v_{1 }=− ~v_{2}− ~v_{3}; *a*1 *a*1

*and so *~v_{1 }*is in Span *~v_{1},~v_{2} *.*

*(b) *True. *Because u*_{1 }*and u*_{2 }*are linearly dependent, there are c*_{1 }*and c*_{2 }not both zero *so that*

*c*_{1}*u*~_{1}+ *c*_{2}*u*~_{2 }= 0.

*If c*_{1 }, 0*, then*

*c*2 *u*~_{1 }=− *u*~_{2}.

*c*1

*So u*_{1 }*is a constant multiple of u*_{2}*. If, on the other hand, c*_{1 }= 0*, then c*_{2 }, 0*, in which case:*

*c*1

*u*~_{2 }=− *u*~_{1},

*c*2

*and thus u*_{2 }*is a constant multiple of u*_{1}*.*

Consider the Linear Transformation *T *: R^{4 }→ R^{3 }whose standard matrix is:

- (18 Points.) Is
*T*onto? - (6 Points.) If
*T*is onto, solve*T*(~*x*) = ~y for a*nonzero*vector ~y of your choice. If it is not onto, find a vector ~y for which ~y is not in the Range of*T*.

Reminder: you must show work to receive credit.

*Let’s solve this entire problem in one computation (you don’t have to, but it’s more e*ffi*cient to do so). In particular, we solve, for any *~*b:*

(24 Points.) Consider the following system of equations (where *a *and *b *are constants):

−2*x*_{1 }+ 4*x*_{2 }= 1

4*x*_{1 }− *ax*_{2 }= *b*

- Choose
*a*and*b*so that the system has a unique solution. - Choose
*a*and*b*so that the system has no solutions. - Choose
*a*and*b*so that the system has many solutions.

*Let’s row reduce: *ï£¯ï£¯ï£¯ï£°−ï£».

*For a unique solution, choose any a and b so that both the entries that contain a and b are non-zero (so that our system has 2 pivots); a*=*b*= 0*will work.**For no solutions, choose a*= 8*and any b such that*2 +*b*, 0*; b*= 0*will work (the second equation would then be*0*x*_{1}+ 0*x*_{2 }= 2*, which has no solution).**For many solutions, choose a*= 8*and b*=−2*, making the final row zero, thus giving us one free variable for this system.*

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