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Math 341 Solutions
Midterm 1
- True/False If the statement is true, give a brief explanation; if it is false, provide a counterexample. T F If A is m × n and the equation A~x = ~b is consistent for some ~b, then the columns of A span Rm.
False. If A =
[ 1
]
, ~b =
[ 1
]
, and ~c =
[ 0
]
, then A~x = ~b is consistent, but A~x = ~c is not. T F If the augmented matrix [^ A ~b ]^ can be transformed by elementary row operations into reduced echelon form, then the equation A~x = ~b is consistent. False. Any matrix can be put in reduced row echelon form. The corresponding equation is consistent if there is no pivot in the last column. For example, if A =
[ 1
]
and
~b =
[ 0
]
, then [^ A ~b ]^ =
[ 1 0
]
is already in reduced form, but has a pivot in the rightmost column, so A~x = ~b is inconsistent. T F If A is m × n and the equation A~x = ~b is consistent for every ~b in Rm, then A must have m pivot columns. True. If A~x = ~b is consistent for all ~b, then A has a pivot in every row. Hence, A has m pivots, so it has m pivot columns. T F If an m × n matrix A has m pivot columns then the equation A~x = ~b has a unique solution for every ~b in Rm. False. If A has more columns than rows, the equation will have a free variable. For example, the 2 × 3 matrix A =
[ 1 0
]
has a pivot in every row, but A~x = ~0 has the infinite solution set {~x = (0, 0 , t) | t ∈ R}. T F If the set {~v 1 , ~v 2 , ~v 3 , ~v 4 } is linearly independent, then the set {~v 1 , ~v 2 , ~v 3 } is linearly independent. True. This is the same as saying that if {~v 1 , ~v 2 , ~v 3 } is linearly dependent, then {~v 1 , ~v 2 , ~v 3 , ~v 4 } is linearly dependent. This is straightforward: if c 1 ~v 1 + c 2 ~v 2 + c 3 ~v 3 = ~ 0 is a nontrivial dependence relation on {~v 1 , ~v 2 , ~v 3 }, then c 1 ~v 1 + c 2 ~v 2 + c 3 ~v 3 + 0~v 4 = ~0 is a nontrivial dependence relation on {~v 1 , ~v 2 , ~v 3 , ~v 4 }.
Fill in the blank If A is a 4 × 6 matrix and the transformation ~x 7 → A~x is onto, then A has 4 pivots and 2 free variables. [Onto says there’s a pivot in every row, so that’s 4 pivots and 6 − 4 = 2 free variables.]
If A is a 5 × 2 matrix and the transformation ~x 7 → A~x is one-to-one, then A has 2 pivots and 0 free variables. [One-to-one says there’s a pivot in every column, so that’s 2 pivots and 2 − 2 = 0 free variables.]
Problems
- Determine h and k so that the following system has (a) no solution, (b) a unique solution, and (c) infinitely many solutions. x 1 + 2x 2 = k 4 x 1 + hx 2 = 5 Solution: Of course we row reduce the augmented matrix of the system. [ (^1 2) k 4 h 5
]
[ (^1 2) k 0 h − 8 5 − 4 k
]
(a) There is a pivot in the last column exactly when h = 8 and k 6 =^54.
(b) The equation has a unique solution if the coefficient matrix has a pivot in every column. This happens when h 6 = 8 (k can be anything).
(c) There are infinitely many solutions when the equation is consistent and has a free variable. Here that means a row of zeros. This happens when h = 8 and k =^54.
- Let T : R^2 → R^2 be a linear transformation which sends ~u =
[ 1
]
to
[ 2
]
and ~v =
[ 3
]
to
[ 1
]
. Find the image under T of ~u + 2~v.
Solution: Since T is linear,
T (~u + 2~v) = T (~u) + 2T (~v) =
[ 2
]
[ 1
]
[ 4
]
- Are the following sets of vectors linearly dependent or linearly independent? Explain.
a)
Solution: Dependent. The set contains the zero vector.
b)
Solution: Independent. The two vectors are not scalar multiples of one another since − 5 6 = 2(−3).
c)
{[ 1
]
[ 3
]
[ 5
]
[ 7
]}
Solution: Dependent. There are more vectors (four) than entries in them (two).
- Let A =
(^) and ~b =
b 1 b 2 b 3
. Show that A~x = ~b is not consistent for all ~b
and find a condition describing those ~b for which the equation is consistent. Solution: We row reduce the augmented matrix of the system
1 3 − 4 b 1 3 2 − 6 b 2 − 5 − 1 8 b 3
1 3 − 4 b 1 0 − 7 6 b 2 − 3 b 1 0 14 12 5 b 1 + b 3
1 3 − 4 b 1 0 − 7 6 b 2 − 3 b 1 0 0 0 5 b 1 + b 3 + 2(b 2 − 3 b 1 )
The matrix has no pivot in the last column when 5b 1 +b 3 +2(b 2 − 3 b 1 ) = −b 1 +2b 2 +b 3 = 0. Hence A~x = ~b is consistent only for those ~b’s satisfying −b 1 + 2b 2 + b 3 = 0.
