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Math 232 First Midterm Solution February 5, 2007
- We consider the following system of equations:
x 1 +x 2 +2x 3 − 4 x 4 = 1 x 1 +2x 2 +x 3 +x 4 = 2 2 x 1 +4x 2 +2x 3 −x 4 = 1
(a) (1 point) Write down the augmented matrix corresponding to this system.
Answer
(b) (3 points) Determine the reduced row echelon form of this matrix. Show your work. (use the back of the previous page if you need more room)
Answer
(c) (2 points) Write down the full solution set to the system in parametric form.
Answer
From part (b) we see that the given system is equivalent to
x 1 = 9 − 3 x 3 x 2 = −4 + x 3 x 3 free variable x 4 = 1
Hence, in parametric form, the set of solution vectors
x 1 x 2 x 3 x 4
is:
+^ x^3
:^ x^3 ∈^ R
(d) (2 points) Let b =
(^) and a 1 =
(^) , a 2 =
(^) , a 3 =
(^) , a 4 =
. Can you
write b as a linear combination of a 1 ,... , a 4? Motivate your answer and, if it is “yes”, give such a linear combination. [HINT: Relate the given vectors to the system given in (a)]
Answer
In vector notation, the system in this question is
x 1 a 1 + x 2 a 2 + x 3 a 3 + x 4 a 4 = b
hence, any solution (x 1 , x 2 , x 3 , x 4 to this system gives a way of expressing b as a linear combination of the ai. We have determined all solutions in (c), so we can for instance write: 9 a 1 − 4 a 2 + 0a 3 + 1a 4 = b
- (a) (4 points) Determine the inverse of the matrix. Show your work.
M =
Answer
Hence,
M −^1 =
(b) (3 points) Prove that, for an invertible n × n matrix A, the linear transformation T : Rn^ → Rn^ defined by T (x) = Ax, is onto.
Answer
A transformation is called onto if for any vector b ∈ Rn^ (the codomain), we can find a vector x in the domain such that T (x) = b. Therefore, in order to show that T is onto, we need to show that for any b ∈ Rn, the equation Ax = b has a solution. If A is invertible, we can take x = A−^1 b. Then Ax = A(A−^1 b) = AA−^1 b = Inb = b, so indeed T is onto if A is invertible.
- Mark true or false and give a reason. (a) (2 points) A linear transformation T : Rn^ → Rm^ can never be one-to-one if n > m.
Answer
True. Let A be the standard matrix of T , so that T (x) = Ax. The matrix A will be m × n. If n > m, this means that the row echelon form of A will have columns without pivot. Hence, the system Ax = 0 has free variables and therefore a non-trivial solution, say v 6 = 0. It follows that T (v) = Av = 0 and T ( 0 ) = A 0 = 0. It follows that two distinct vectors have the same image under T and therefore T is not one-to-one.
(b) (2 points) The following map is linear:
(^ R^2 →^ R^2
x 1 x 2
x 1 + x 2 x 1 − x 2
Answer
True. We call the transformation T. We need to show that for all v, w ∈ R^2 and c ∈ R we have T (v + bw) = T (v) + T (w) and T (cv) = cT (v):
T (v+w) = T
v 1 + w 1 v 2 + w 2
(v 1 + w 1 ) + (v 2 + w 2 ) (v 1 + w 1 ) − (v 2 + w 2 )
v 1 + v 2 v 1 − v 2
w 1 + w 2 w 1 − w 2
= T (v)+T (w)
T (cv) = T
cv 1 cv 2
cv 1 + cv 2 cv 1 − cv 2
= c
v 1 + v 2 v 1 − v 2
= cT (v)
(c) (2 points) A matrix transformation is always an linear transformation.
Answer
True. A matrix transformation is of the form T (x) = Ax. The fact that A(v + w) = Av + Aw and A(cv) = cAv are properties of matrix multiplication.
