Math 245 - Linear Algebra - Exam 1 - Fall 2022 – Solutions
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Math 245 - Linear Algebra - Exam 1 - Fall 2022 – Solutions
1. True or False. Circle the correct answer. No explanation needed. (10 points)
(a) True/False: If A is a 5 × 6 matrix, then Ax = 0 always has a non-zero solution.
True. 6 vectors in R5 will be linearly dependent and so Ax = 0 will have a nonzero solution.
(b) True/False: If A is a 4 × 5 matrix, then Ax = b will always have infinitely many solutions.
False. It might not have any solutions.
(c) True/False: If v1, v2, v3 are vectors in R3 that are linearly independent, then the matrix with these vectors as its rows is invertible.
True. This is one of the many ways of saying a matrix is invertible.
(d) True/False: If v1, v2, v3 are linearly independent, and c1, c2, c3 are nonzero real numbers, then the vectors c1v1, c2v2, c3v3 are also linearly independent.
True. If C1(c1v1 ) + C2(c2v2 ) + C3(c3v3 ) = 0, then due to the linear independence of the vectors, C1c1 = C2c2 = C3c3 = 0. But since c1, c2, c3 are non-zero, then C1 = C2 = C3 = 0 which means that c1v1, c2v2, c3v3 are also linearly independent.
(e) True/False: If T : Rn → Rn is a one-to-one linear transformation then T is onto.
True. If T is one-to-one, then the nullspace of the corresponding matrix is zero. Thus, this matrix is invertible and so T is also onto.
2. Consider the vectors
u = l 3」
(a) Can one write
as a linear combination of u and v?
and
l 5」
「 4
v = l 1」
(5 points)
We need to solve the system represented by the following augmented matrix
l 」
「 2 1 4 .
This row reduces to
l」
「 .
From here one can see that the system is inconsistent and thus the given vector is not in the linear span of u amd v.
(b) For what value of k is
l 」k(1)
a linear combination of u and v.
Here we need to solve the system given by the augmented matrix
l 」
2
This (partially) row reduces to
l 」
This system will be consistent as long as k = −8.
3. Consider the following system of linear equations:
kx + y + z = 1
x + ky + z = 1
x + y + kz = 1.
This system has the associated augmented matrix
l 」
which (partially) row reduces to
「0 0 (k + 2)(k − 1) 1 − k .
(a) For what value(s) of k will this system have unique solution?
k 1 or k −2.
(b) For what value(s) of k will this system have no solution?
k = −2.
(c) For what value(s) of k will this system have more than one solution?
= 1.
4. Write down the parametric form of the solution to the system
x + 2y − z + 3w = 3
2x + 4y + 4z + 3w = 9
3x + 6y − z + 8w = 10.
The augmented form of this system is
l2(1) 「3 which row reduces to l0(1) 「0 This yields the parametric solutions |
2 4 6
2 0 0 |
−1 3 4 3 −1 8 |
9(3) 」 10
」 0(2) . |
|
0 1 0 |
5 2 − 0 |
x = − t − 2s,
y = s,
z = + t,
w = t.
5. Let T be the linear transformation associated to the matrix A = ] .
(a) T is a linear transformation from where to where?
This linear transformation is from R3 to R2 .
(b) Is T one-to-one?
3
No. There are three columns in R2 and so they cannot possibly be linearly independent. Thus, the nullspace of the matrix is nonzero and to T is not one-to-one.
(c) Is T onto? (4 points)
Yes. The first two columns span R2 and thus the column space of this matrix is R2 . Thus, the range of T is the column space – which is R2 .
6. Without using Wolfram Alpha or doing any serious row reduction, explain why each of these matrices is not invertible. Details are important.
(a) 0(1) 0 |
−1 1 2 |
3(1)J 6 (2 points) |
Row three is twice row two and so a single row operation will yield a row of zeros. Thus, the matrix cannot reduce to the identity.
(b)
J
0 0 0
(2 points)
Last row is zero. Thus, the matrix cannot reduce to the identity.
(c)
J
0 2 2
(3 points)
Row three is twice row two and so a single row operation will yield a row of zeros. Thus, the matrix cannot reduce to the identity.
(d) 1 1 |
0 1 3 |
2(0)J 6 (3 points) |
Column three is a multiple of column two and so the columns are not linearly independent.
2022-12-01