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Math 2270 Solutions for Practice Midterm 1
- Consider the following system: x 1 + 5x 2 = a cx 1 + dx 2 = b
What condition on c and d is needed to ensure that the system is consistent?
Solution The augmented matrix for this system is is
(2)
[
1 5 a c d b
]
[
1 5 a 0 d − 5 c b − ac
]
R∗ 2 = R 2 − cR 1.
To guarantee that the system is consistent, we cannot have zeroes in the first two entries of the last row in the second matrix. Thus, the condition d − 5 c 6 = 0 will guarantee that the system is consistent, regardless of the values of a and b.
Remark 1. A second condition that will guarantee that the system is consistent is b − ac = 0. The problem with this is that it does not really answer the question since this is a condition involving a, b, and c rather than one that involves only c and d. We can make it a condition on c by solving the equation for c to get c = ba , but there is an issue with that- in particular if a = 0, this condition does not make sense. Thus, a sufficient answer is what is written above since it is the only condition that is completely independent of the specific values of b and a.
1
- Let
(3) A =
Using row operations, find a matrix given in reduced row echelon form that is row equivalent to A.
Solution We compute:
A =
R∗ 2 = R 2 − 2 R 1
R∗ 3 = R 3 + R 2
R∗ 2 = R 2 − R 3
R∗ 2 = −R 2 , R 3 ∗ = R^3
R∗ 1 = R 1 − 2 R 2.
The last matrix in the above calculation has the desired properties.
- Let A be an m × n matrix. Let u, v ∈ Fn. Show that A(u + v) = Au + Av.
Solution Write A =
[
a 1 a 2... an
]
, where the ai’s are the column vectors
of A. Let us write u =
u 1 u 2 .. . un
and^ v^ =
v 1 v 2 .. . vn
. From the definition
of the product of a matrix with a vector we obtain:
(6) A(u + v) =
∑^ n
i=
(ui + vi)ai =
∑^ n
i=
uiai +
∑^ n
i=
viai = Au + Av,
as desired.
- Consider the inhomogeneous equation Ax = b. Prove or disprove the following statement: The solution set of Ax = b is the set of all vectors of the form w = p+vh where vh is any solution of the equation Ax = 0.
Solution The statement is false. The first issue is that the inhomogeneous equation need not be consistent in which case the solution set is empty. For example, if A = [0] and b = [1] then Ax = [0] 6 = b for every x ∈ F^1. This argument is sufficient to prove our claim. Notice, however, that even if we assume that the equation is consistent, the statement is false because we have not defined p. To guarantee the validity of the statement, p must be a solution of the equation. (Having already answered the question, we won’t verify the last statement.)
- Let T : Rn^ → Rm^ be a function defined by T (x) = Ax + b where A is an m × n matrix (with real entries) and b is a non-zero vector in Rm. Show that T is not a linear transformation.
Solution If T were linear, than T (2x) = 2T (x). Equivalently, if T is linear, then T (2x) − 2 T (x) = 0. However,
T (2x) − 2 T (x) = A(2x) + b − 2(Ax + b) = 2Ax + b − 2 Ax − 2 b = −b T (2x) − 2 T (x) 6 = 0.
Thus, T is not a linear transformation.
- Prove or disprove: If T is a linear transformation whose standard matrix is 3 × 2, then T is not an injective map.
Solution The statement is false. In particular, there exists an injective lin- ear transformation, T , with the required property. For example, let T (x 1 , x 2 ) = (x 1 , x 2 , 0). We claim that T is injective^3. To see this, as- sume that T (x 1 , x 2 ) = T (y 1 , y 2 ). Then (x 1 , x 2 , 0) = (y 1 , y 2 , 0). Hence, x 1 = y 1 and x 2 = y 2 and so (x 1 , x 2 ) = (y 1 , y 2 ). Notice also that the
standard matrix for T is A :=
(^) , since
(8) Ax =
[
x 1 x 2
]
x 1 x 2 0
(^) = T (x).
(^3) I would accept the claim that this is obvious.
