Analyttic Geometry and Matrix Midterm Exam 2 December 16, 2015
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1. (10 points) Find the general solutions of system whose augmented matrix is given as follows.
1 −3 0 −1 0 −2
0 1 0 0 −4 1
0 0 0 1 9 4
0 0 0 0 0 0
Solution: Since the system of equations is equivalent to the following x1− 3x2− x4 = −2
x2− 4x5 = 1 x4+ 9x5 = 4
⇒
x4 = 4 − 9x5 x2 = 1 + 4x5 x1 = 3x2+ x4− 2
= 3(1 + 4x5) + 4 − 9x5− 2
= 3x5+ 5
Hence, the set of solutions is of the form {(3x5+ 5, 1 + 4x5, x3, 4 − 9x5, x5) | x3, x5∈ R},
i.e. the general solution is of the form x5(3, 4, 0, −9, 1) + x3(0, 0, 1, 0, 0) + (5, 1, 0, 4, 0), where x3and x5are arbitrary real numbers.
2. (10 points) Find x1, x2, and x3such that b = x1a1+ x2a2+ x3a3, where
a1=
1 0 1
, a2=
−2 3
−2
, a3=
4 7 5
, b =
11
−5 9
.
Solution: Since
1 −2 4 11 0 3 7 −5 1 −2 5 9
R3→ R3− R1
−→
1 −2 4 11 0 3 7 −5 0 0 1 −2
⇒
x3 = −2
3x2 = −5 − 7x3= 9 ⇒ x2= 3 x1 = 11 + 2x2− 4x3= 25
3. (10 points) Let A =
2 −5 8
−2 −7 1
4 2 7
. Describe all solutions of Ax = 0 in parametric vector form.
Analytic Geometry and Matrix Midterm Exam 2 (Continued) December 16, 2015
Solution: Since
2 −5 8
−2 −7 1 4 2 7
R2→ R2+ R1
−→
R3→ R3− 2R1
2 −5 8
0 −12 9 0 12 −9
R3→ R3+ R2
−→
2 −5 8 0 −12 9
0 0 0
Thus, we have
12x2 = 9x3 ⇒ x2=3 4x3 2x1 = 5x2− 8x3= 15
4 x3− 8xx ⇒ x1= −17 8 x3 and the solutions are of the form x3(−17
8 ,3
4, 1), where x3is an arbitrary real number.
4. (10 points) Determine if the columns of the matrix
0 −3 9
2 1 −7
−1 4 −5
1 −4 −2
form a linearly independent set. Justify your answer.
Solution: Since
0 −3 9 2 1 −7
−1 4 −5 1 −4 −2
R4↔ R1
−→
1 −4 −2 2 1 −7
−1 4 −5 0 −3 9
R2→ R2− 2R1 R3→ R3+ R1
−→
1 −4 −2 0 9 −3 0 0 −7 0 −3 9
R2↔ R4
−→
1 −4 −2 0 −3 9 0 0 −7 0 9 −3
R4→ R4+ 3R2
−→
1 −4 −2 0 −3 9 0 0 −7
0 0 24
R4→ R4+24 7 R3
−→
1 −4 −2 0 −3 9 0 0 −7
0 0 0
has 3 pivot positions, the equation 0 = Ax = A
x1 x2 x3
has only trivial solutions, i.e. column vectors are linearly independent.
5. Let T (x1, x2) = (3x1+ x2, 5x1+ 7x2, x1+ 3x2).
(a) (10 points) Find the standard matrix A for the linear transformation T.
(b) (10 points) Show that T is a one-to-one linear transformation.
(c) (10 points) Determine the range of T.
Solution:
Analytic Geometry and Matrix Midterm Exam 2 (Continued) December 16, 2015
(a)
A=
3 1 5 7 1 3
(b) Since
3 1 5 7 1 3
R3↔ R1
−→
1 3 5 7 3 1
R2→ R2− 5R1
−→
R3→ R3− 3R1
1 3
0 −8 0 −8
R3→ R3− R2
−→
1 3
0 −8
0 0
has 2 pivot positions, column vectors
3 5 1
and
1 7 3
are linearly independent which implies that T is a one-to-one linear transformation.
(c) For each
b1 b2 b3
∈ R3, since
3 1 b1 5 7 b2 1 3 b3
R3↔ R1
−→
1 3 b3 5 7 b2 3 1 b1
R2→ R2− 5R1
−→
R3→ R3− 3R1
1 3 b3
0 −8 b2− 5b3 0 −8 b1− 3b3
R3→ R3− R2
−→
1 3 b3
0 −8 b2− 5b3 0 0 b1− b2+ 2b3
the range of T is
Range(T ) = {
b1 b2 b3
∈ R3| b1− b2+ 2b3= 0}.
6. Let
A=
1 −2 −1
−2 2 0
4 −1 3
. (a) (10 points) How many rows of A contain a pivot positions?
(b) (10 points) Can each vector in R3be written as a linear combination of the columns of the matrix A? Why or why not?
(c) (10 points) Find the subset of R3spanned by the columns of A.
Solution:
(a) Since
1 −2 −1
−2 2 0
4 −1 3
R2→ R2+ 2R1
−→
R3→ R3− 4R1
1 −2 −1 0 −2 −2
0 7 7
R2→ −1 2R2
−→
R3→ R3− 3R1
1 −2 −1
0 1 1
0 7 7
Analytic Geometry and Matrix Midterm Exam 2 (Continued) December 16, 2015
R3→ R3− 7R2
−→
1 −2 −1
0 1 1
0 0 0
has 2 pivot positions.
(b) Since the 3 × 3 matrix A has only 2 pivot positions, Ax = 0 has nontrivial solutions, or equiva- lently, the column vectors of A are linearly dependent.
(c) For each
b1 b2 b3
∈ R3, since
1 −2 −1 b1
−2 2 0 b2
4 −1 3 b3
R2→ R2+ 2R1
−→
R3→ R3− 4R1
1 −2 −1 b1 0 −2 −2 b2+ 2b1 0 7 7 b3− 4b1
R2→ −1 2R2
−→
R3→ R3− 3R1
1 −2 −1 b1 0 1 1 −b2
2 − b1 0 7 7 b3− 4b1
R3→ R3− 7R2
−→
1 −2 −1 b1
0 1 1 −b2 2 − b1 0 0 0 b3+ 3b1+7
2b2
the span of column vectors of A is
Span(A) = {
b1 b2 b3
∈ R3| 3b1+7
2b2+ b3= 0}.
