Download Math 308 Solutions Sec. 1.3 Problems(1,3,5,7,8,9,11,12,15,13) page

Math 308 Solutions
Sec. 1.3
Problems(1,3,5,7,8,9,11,12,15,13) page 1
More Problems(16,19,21,23,25,27,28,32)
For 1,3 transform the augmented matrix for the given system to REF and determine n= # unknowns, r= # nonzero rows in row equivalent (R)EF matrix, and n − r, the number of variables which can be assigned arbitrary
values.
2 −1
1
 2
 −2 −2
4
1
1.3.1 System whose augmented matrix is 

2
5
5
 2
 −2 −2 −2 −3 
1  ====>  2 2 −1
1  R + R  2 2 0 5/3 
====> 2 2 −1
1
2



3
2
0 0
1 2/3 ====> 0 0 1 2/3 
R2 + R2 0 0
(1/3)R2
Sol.





. Since (1/2)R1 will transR2 − R2 0 0
6
4  R3 − 2(old)R2  0 0
0
0
0

0 0 0
R2 + R2  0 0 −3 −2  R4 + (old)R2  0 0
0
0
0
0 0 0
form this into (R)REF, there are n = 3 unknowns , r = 2 non − zero rows , (already in echelon form)
and n − r = 3 − 2 = 1 variable can be set arbitrarily. In the present case x 2 will be arbitrary in standard
form of solution. (Other ways to present solution might have x 1 arbitrary.)
1 2
 0 −1 1

1.3.3 System corrsponds to the augmented matrix: 1
2 2 −1 3 


3 1
0 2
1
2 −1
3 ====> 1 2
2 −1
3
2 −1
3
1 2
R1 <====> R2  1 2
 0 1 −1 −1 −2 ====> 0 1 −1 −1 −2 
Sol. −1"new" R2  0 1 −1 −1 −2 






R3 − "new" R1 0 1 −1
1 −1  R3 − R2  0 0
0
2
1  (1/2)R3  0 0
0
1 1/2 

R1 + R3  1
R2 + R3  0

====> 0
2
2 0
7/2  R1 − 2R2  1 0
4 0 13/2 


1 −1 0 −3/2 ====> 0 1 −1 0 −3/2 . So n = 4 variables, r = 3 non-zero rows



0
0 1
1/2 
0 1
1/2 

0 0
in (R)EF, and n − r = 4 − 3 = 1 arbitrary variable in any general solution. (Standard form has x 3 arbitrary.) (Note for purposes of finding n, r, and n − r arbitrary variables, we could have stopped after the
first two sets of row equivalences.)
In 7-18 determine all possiblities for the solution set (from among infinitly many solutions, a unique solution, or
no solution) of the system of linear equations described.)
1.3.7
A homogeneous system of 3 equations in 4 unknowns.
Sol.
Infinitely many solutions. To see why note system is consistent since trivial solution exists. Then at most 3 nonzero rows in (R)EF, as only 3 equations, i.e. r ≤ 3. Hence since n = 4 unknowns, n − r ≥ 4 − 3 = 1, i.e., at least
1 unconstrained variable. Thus infinitely many solutions.
1.3.8
A homogeneous system of 4 equations in 5 unknowns.
Sol.
Infinitely many solutions. To see why note system is consistent since trivial solution exists. Then at most 4 nonzero rows in (R)EF, as only 4 equations, i.e. r ≤ 4. Hence since n = 5 unknowns, n − r ≥ 5 − 4 = 1, i.e., at least
1 unconstrained variable. Thus infinitely many solutions.
1.3.9
A system of 3 equations in 2 unknowns.
Math 308 Solutions
Sec. 1.3
Problems(1,3,5,7,8,9,11,12,15,13) page 2
Sol.
No solution or unique solution or infinitely many. Can not limit possibilities since all we know is r ≤ n = 2. If
inconsistent no solutions. If consistent, and if r = 2, then no arbitrary variables and unique solution. Otherwise,
consistent and r < 2, so then some arbitrary variables and many solutions.
1.3.10
A system of 4 equations in 3 unknowns.
Sol.
No solution or unique solution or infinitely many. Can not limit possibilities since all we know is r ≤ n = 3. If
inconsistent no solutions; if consistent and r = 3 then no arbitrary variables and unique solution; consistent and
r < 3 then some arbitrary variables and many solutions.
1.3.11
A homogeneous system of 3 equations in 2 unknowns.
Sol.
Unique solution or infinitely many. Homogeneous equation always has at least one solution - the trivial solution. If r = 2 in (R)EF then n − r = 2 − 2 = 0, so no arbitrary variables and unique solution; and if r < 2 then
some arbitrary variables and many solutions.
1.3.12
A homogeneous system of 4 equations in 3 unknowns.
Sol.
Unique solution or infinitely many. Homogeneous equation always has at least one solution - the trivial solution. If r = 3 in (R)EF then n − r = 3 − 3 = 0, so no arbitrary variables and unique solution; and if r < 3 then
some arbitrary variables and many solutions.
1.3.13
A system of 2 equations in 3 unknowns that has x 1 = 1, x 2 = 2, and x 3 = −1 as a solution.
Sol.
Infinitely many, because at most 2 non-zero rows in (R)EF as only 2 equations, while at least n − r ≥ 3 − 2 = 1
unconstrained variable.
1.3.17
A homogeneous system of 2 equations in 2 unknowns that has x 1 = 1, x 2 = −1 as a solution.
Sol.
Infinitely many, because the trivial solution solves a homogeneous system and so each point on the line from
(0,0) to (1,-1) solve the system. Or just say since at least two solutions, there must be infinitely many.
In 19,21 determine by inspection whether the given system has nontrivial solutions or only the trivial solution.
1.3.19 with Sol.
2x 1 + 3x 2 −
x3 = 0
has nontrivial solutions since there are only two homogeneous equations
x1 −
x 2 + 2x 3 = 0
(so some solution) but three unknowns, i.e. more equations than unknowns and consistent so at least one
unconstrained variable in the general solution.
1.3.21 with Sol.
x 1 + 2x 2 −
x3 = 0
x 2 + 2x 3 = 0 has only the trivial solution since there are exactly three homogeneous
4x 3 = 0
equations (so at least the trivial solution) in three unknowns and from the given equations (whose matrix is
almost in echelon form) it is clear each variable will be a leading variable, so no non-leading variables, i.e.
no unconstrained variables and thus unique solution.
Math 308 Solutions
Sec. 1.3
Problems(1,3,5,7,8,9,11,12,15,13) page 3
Alt. There are n = 3 unknowns and its clear that there will be r = 3 non-zero rows in the REF for the augmented matrix of this system. Thus n − r = 3 − 3 = 0 unconstrained variables, i.e. unique solutions.
Alt. Clear that working from last equation back up, first x 3 determined uniquely, then plugging in to second
equation, x 2 determined uniquely and then plugging both into the first equation, x 1 is determined uniquely.
x 1 + 2x 2 + x 3 = 0
1.3.23For what value(s) of a does the system have nontrivial solutions? −x 1 −
x2 + x3 = 0
3x 1 + 4x 2 + a = 0
2 1 0  ====>  1
2
1 0
 1



Sol. Form the augmented matrix and attempt to get to (R)EF: −1 −1 1 0 R2 + R1 0
1
2 0




4 a 0  R3 − 3R1  0 −2 a − 3 0 
 3
1 0
1 2

====> 0 1
2 0 . A homogeneous system is always consistent and has nontrivial solutions when


R3 + 2R2 0 0 a + 1 0


there is at least one unconstrained(free) variable. So there must be at most 2 nonzero rows in EF, i.e. the
3rd row must be all 0’s. So a = −1 is the only value for which nontrivial solutions exist.
1.3.25Let B be a (4 × 3) matrix in reduced echelon form.
(a) If B has three nonzero rows, then determine the form of B.
(b) Suppose that a system of 4 linear equations in 2 unknowns has augmented matrix A where A is
(4 × 3) matrix row equivalent to B. Demonstrate that the system of equations is inconsistent.
Sol. (a)
4 rows but 3 non-zero rows, so last row is zero. 3 non-zero rows and only 3 columns, so each column must
have a leading 1. Hence row 1 must have leading 1 in column 1 (otherwise not two more columns to the
right for the leading 1’s in rows 2 and 3). Move down - row 2 must have leading 1 in col 2 and row 3 must
1 0 0
0 1 0
havleading 1 in column 3. Thus the form of B (due to REF assumption) must be: B = 

0 0 1
0 0 0
Sol. (b)
The equation corrsponding to the third row in B is 0x 1 + 0x 2 + 0x 3 = 1, or 0 = 1, which is impossible to
solve.
1.3.32The equation of a circle has the form ax 2 + ay 2 + bx + cy + d = 0. Hence a circle is determined by three
points. Find the equation of the circle through the points (1,1), (2,1) and (3,2).
a(12 + 12 ) + b ⋅ 1 + c ⋅ 1 + d = 0
Sol. Plug the 3 (x, y) points into the equation: a(22 + 12 ) + b ⋅ 2 + c ⋅ 1 + d = 0. The augmented matrix for this
a(32 + 22 ) + b ⋅ 3 + c ⋅ 2 + d = 0
system with reduction to REF is:
1/2
1/2
1/2 0  ==>
 2 1 1 1 0 (1/2)R1  1 1/2 1/2 1/2 0  ==>  1
 5 2 1 1 0
 5


2
1
1 0 R2 − 5R1 0 −1/2 −3/2 −3/2 0 (−2R2






3
2
1 0  R3 − 13R1  0 −7/2 −9/2 −11/2 0 
 13 3 2 1 0 
 13
1/2
1/2
1/2 0 
1
 1 1/2 1/2 1/2 0 
 1 1/2 1/2 1/2 0  R1 − 1/2R3
0



1
3
3 0
==>
0
1
3
3 0 ==>  0
1
3
3 0  R2 − 3R3






0
6
5 0 (1/6)R3  0
0
1 5/6 0  ==>
 0 −7/2 −9/2 −11/2 0  R3 + 7/2R2  0
 1 1/2 0 1/12 0  R1 − 1/2R2  1 0 0 −1/6 0 
0
1 0
1/2 0  ==>  0 1 0
1/2 0 . So from the corresponding equations the coeffi



0 1
5/6 0 
5/6 0 
0
0 0 1
Math 308 Solutions
Sec. 1.3
Problems(1,3,5,7,8,9,11,12,15,13) page 4
cients are a = (1/6)d, b = −(1/2)d, c = − (5/6)d Or, with the choice d = 6, the circle has the equation:
x 2 + y 2 − 3x − 5y + 6 = 0.