Download Linear System - Linear Algebra - Solved Exam and more Exams Linear Algebra in PDF only on Docsity!
MT210 MIDTERM 1 ANSWERS
İLKER S. YÜCE
MARCH 15, 2011
SURNAME, NAME:
QUESTION 1. SYSTEMS OF LINEAR EQUATIONS
Determine the values of k such that the linear system
9 x 1 + kx 2 = 9
kx 1 + x 2 = − 3
is consistent.
ANSWER
We apply row-reduction algorithm to the augmented matrix corresponding to the system given above:
Assume that k ̸ = 0, then we get
[
9 k 9
k 1 − 3
]
( −k/ 9) R 1 + R 2 ÏR 2 //
[
9 k 9
k^2 9
− 3 − k
]
By Theorem 2, we know that the system above is consistent if and only if 1 −
k 2
9
We need to examine the case k = 0. If k = 0, then we have 9 x 1 = 9 or x 1 = 1 and x 2 = − 3. So, the
system is consistent. Note that if k = − 3 the given system is still consistent. Finally, we conclude that the
system above is consistent if and only if k ̸ = 3.
QUESTION 2. ROW REDUCTION AND ECHELON FORMS
Determine when the augmented matrix below represents a consistent linear system.
1 − 1 2 1 a
− 1 3 1 1 b
3 − 5 5 1 c
2 − 2 4 2 d
ANSWER
We apply row-reduction algorithm to the augmented matrix corresponding to the system given above:
1 − 1 2 1 a
− 1 3 1 1 b
3 − 5 5 1 c
2 − 2 4 2 d
1 − 1 2 1 a
0 2 3 2 a + b
0 0 2 0 b + c − 2 a
0 0 0 0 d − 2 a
By Theorem 2, we know that the system above is consistent if and only if d − 2 a = 0.
QUESTION 4. THE MATRIX EQUATION Ax=b
A. Solve the matrix equation A x = b where
A =
(^) , b =
ANSWER
We apply row-reduction algorithm to the augmented matrix corresponding to the system given above:
We have
G.S. =
x 1 = 1 − 3 x 3
x 2 = x 3
x 3 is free.
B. Is it possible to solve A x = b for any given b =
b 1
b 2
b 3
(^) where A is the matrix given in part A? Explain.
ANSWER
The coefficient matrix A has only 2 pivot positions. Therefore, it is NOT possible to solve Ax=b for any
given b.
C. Describe the set of all b =
b 1
b 2
b 3
(^) for which A x = b does have a solution.
We apply row-reduction algorithm to the augmented matrix corresponding to the system given above:
1 2 1 b 11
1 3 0 b 2
− 1 − 4 1 b 3
1 2 1 b 1
0 1 − 1 b 2 − b 1
0 0 0 −b 1 + 2 b 2 + b 3
We conclude that A x = b does have a solution if and only if −b 1 + 2 b 2 + b 3 = 0.
QUESTION 5. SOLUTION SETS OF LINEAR SYSTEMS
Consider the linear system A x = b where
A =
(^) , b =
A. Solve the linear system.
B. Write the general solution in parametric-vector form.
C. Give a particular solution p.
D. Write the solution set for the homogeneous equation A x = 0.
ANSWER
A. Note that there are 5 variables, x 1 , x 2 , x 3 , x 4 , x 5 , and 3 equations. We reduce the augmented matrix
corresponding to the given system. We get
[ A b] =
Variables x 1 and x 2 are BASIC and variables x 3 , x 4 , x 5 are. The general solution is
G.S. =
x 1 = − 7 − 2 x 3 − 2 x 4 − 2 x 5
x 2 = − 10 − 4 x 3 − 4 x 4 − 4 x 5
x 3 is free
x 4 is free
x 5 is free.
B.
G.S. =
x 1
x 2
x 3
x 4
x 5
: x 3 , x 4 , x 5 ∈ R
C. The particular solution is given as p =
D. The parametric vector form of homogeneous part of the general solution set is
vh =
x 3
: x 3 , x 4 , x 5 ∈ R
