Download Systems of Linear Equations

Systems of Linear Equations
Linear Expressions
Linear equations represent a good starting point for the study of linear algebra, in that solving them
gives rise to fundamental linear-algebra concepts, such as matrices, elementary matrix operations,
and solution (i.e. vector) spaces.
To begin, a linear expression over real-valued variables x1 , . . . , xn is one of the form
a1 x 1 + · · · + an x n ,
where ai is a real-valued constant called the coefficient
of xi . For example, 3x1 − 2.2x2 + 7x3 and
√
−5x + 2y − 3z + 4w are linear expressions, but 2 x − 3y is not beccause the square-root function
has been applied to variable x, which is not allowed. Linear expressions are also referred to as linear
combinations.
Linear expressions arise quite naturally in science and engineering. For example, Newton’s 2nd law
of motion states that the acceleration a of an object is directly proportional to the sum of the forces
acting on the object:
a ∝ F1 + · · · + Fn .
Also, in chemistry the total mass of a comound can be expressed as a linear combination of the
(number of) base elements that comprise the compound. For example, the compound known as
water has an approximate mass of 2H + 16O units, where H and O are the respective number of
hydrogen and oxygen atoms. And in computer science the activation potential of a biological neuron
n is modeled as a linear combination of the activation potentials of other neurons that are connected
to n.
Thus, linear expressions can arise in practice when attemtping to quantify a phenomenon using
mathematics. This is often referred to as mathematical modeling, or just modeling for short.
Moreover, modeling a quantity as a linear combination of other quantities represents one of the more
basic kinds of math models. A linear model represents a good place to start, and can always be
made more complex (by introducing one or more non-linear functions) if the linear model does not
adequately represent the phenomenon in question.
1
Linear Equations and Inequalities
A linear inequality is one of the form a1 x1 + · · · + an xn ≤ b, where the left side is a linear
expression over variables x1 , . . . , xn , and b is a real-valued constant. Note that ≤ may be replaced
with one of the other three inequality relations: <, ≥, and >. Examples of linear inequalities include
2x1 − 3x3 + 0.51x4 ≤ 7, 3x − 5y > 2, and −4a + b − 9c − 10d ≥ 20. On the other hand, 3x2 + 4y 2 ≤ 16
is not because 3x2 + 4y 2 is not a linear expression (why ?).
A linear equation is one of the form a1 x1 + · · · + an xn = b. Examples of linear equations include
2x1 − 3x3 + 0.51x4 = 7, 3x − 5y = 2, and −4a + b − 9c − 10d = 20. On the other hand, y − sin x = 0
is not a linear equation (why ?).
A linear equation or inequality is an example of a constraint, which restricts the combination of
values that a set of variables may assume. For example, in the equation 2x + 3y = 7, if x is assigned
the value 2, then y is forced to be assigned the value 1. Any other (x, y)-combination with x = 2
is invalid. Constraints arise naturally in problem modeling because every realm of human life is
governed by constraints. For example, if an individual has a total of D dollars to invest in three
different companies, and if xi , i = 1, 2, 3, is the amount she will invest in company i, then necessarily
x1 + x2 + x3 ≤ D.
For the remainder of this lecture and the remaining linear-algebra lectures, we will focus entirely on
linear equations. Problems involving linear inequalities will be treated in a future lecture on linear
programming and optimization.
Given a linear equation a1 x1 + · · · + an xn = b, a solution to the equation is a sequence of numbers
s1 , . . . , sn which, when respectively substituted for the variables, x1 , . . . , xn , satisfies the equation in
the sense that
a1 s 1 + · · · + an s n = b
is a true statement. For example, x = 2, y = 1 is a solution to equation 2x + 3y = 7.
The solution set for an equation is the set of all possible solutions to the equation. For an equation
with n variables, its solution set can be described with the help of n − 1 paramters. For example,
2x + 3y = 7 has two variables and its solution set can be described using a single parameter t. To
do this, set y = t. Then, solving for x in the equation gives
x = (7 − 3t)/2 = 7/2 + (−3/2)t.
Thus, the solution set for 2x + 3y = 7 is x = 7/2 − 3t/2 and y = t, where t is any real number. The
two equations can be combined into the single vector equation
(x, y) = (7/2, 0) + t(−3/2, 1),
where (7/2, 0) and (−3/2, 1) are two-dimensional vectors, and t is a scalar. To obtain a solution
from this equation, perform the following steps.
Scalar Multiplication Mutiply (−3/2, 1) by scalar t to get (−3t/2, t).
2
Vector Addition Add vectors (7/2, 0) and (−3t/2, t) to get (7/2 − 3t/2, t), which yields x = 7/2 −
3t/2 and y = t.
For example, letting t = 1 gives the solution x = 7/2 − 3/2 = 2 and y = 1.
Example 1. Represent the solution set −2x − 5y + 4z = 10 using two parameters s and t. Write
the solution set as a single vector equation. What values of s and t will generate the solution
x = 4, y = 2, z = 2?
Systems of Linear Equations
A set of linear equations over the same set of variables is called a system of linear equations, or
linear system. Systems of linear equations arise in practice when each constraint has been modeled
with an linear equation. For example,
5x1 − 7x2 + 3x3 = 38
2x1 + 4x2 + 8x3 = 22
represents a system of two equations over the variables x1 , x2 , x3 .
Given a system of linear equations over variables x1 , . . . , xn , a system solution is a sequence of
numbers s1 , . . . , sn which, when respectively substituted for the variables, x1 , . . . , xn , satisfies each of
the equations in the system. For example x1 = 1, x2 = −3, x3 = 4 is a solution to the above system.
Moreover, a system solution set is the set of all system solutions.
Two different systems of equations over a common set of variables are said to be equivalent iff they
both have the same solution set. For example, the system
3
x1 − x2 = 2
x1 + x2 = 4
has the single solution x1 = 3, x2 = 1. This system is equivalent to the following three systems
System1.
5x1 − 5x2 = 10
x1 + x2 = 4
System2.
x1 − x2 = 2
2x1 = 6
System3.
x1 + 0x2 = x1 = 3
0x1 + x2 = x2 = 1
Of the above three equivalent systems, notice that System 3 seems the simplest and most convenient,
as it immediately implies the solution. What is needed is an algorithm that is capable of transforming
a linear system into an equivalent system that has the least number of equations possible, and for
which each equation has the least number of variables possible. This leads to the following definition.
A system of equations is said to be in reduced form iff the following statements hold.
RF1. Each equation is either the tautology 0 = 0, the fallacy 0 = 1, or a conditional equation
of the form x + a1 y1 + · · · + ad yd = b. For a conditional equation, x is called the lead or
dependent variable, while variables y1 , . . . , yd are called independent variables.
RF2. If x is a lead variable for some equation, then it does not appear in any other equation. As a
consequence, no variable can be both dependent and independent, and the number of dependent
variables is equal to the number of conditiional equations.
Without loss of generality, we may assume that each conditional equation has the same set of independent variables (a missing independent variable yi in some equation may be assumed to have
coefficient ai = 0 in that equation). Moreover, we may assume that any variable that does not appear
in any equation is an independent variable. Thus, for a reduced-form system, every variable can be
classified as either independent or dependent.
Notice that if a reduced-form system has a fallacy, then it cannot possibly have a solution, since no
variable assignment can satisfy the fallacy. Such a system is called inconsistent. Otherwise it is
said to be consistent.
4
Now assume the reduced-form system is consistent. If the system has no independent variables, then
every variable is a lead variable for exactly one equation. Moreover, assuming there are n variables
x1 , . . . , xn , it follows that the system has the form x1 = b1 , . . . , xn = bn . In other words, the system
has exactly one solution, and that solution (b1 , . . . , bn ) is immediately obtained from the equations.
Finally, assume the system has d independent variables y1 , . . . , yd . Then let t1 , . . . , td be d free
parameters, and set yi = ti , i = 1, . . . , d. Then any combination of real values assumed by these
parameters leads to a system solution. To see this, note that for each dependent variable x, there
is a unique equation of the form x + a1 y1 + · · · + ad yd = b, and this equation yields the following
solution value for x:
x = b − a1 t1 − · · · − ad td .
Therefore, if the reduced system has one or more independent variables, then it has infinitely many
solutions.
The above paragraphs are now summarized in the following theorem.
Theorem 1. Let S be a system of linear equations in reduced form. Then one of the following is
true about S.
1. S has a fallacy and is inconsistent.
2. S has no independent variables and has exactly one solution.
3. S has one or more independent variables and has infinitely many solutions.
The next step is to show that any system of linear equations can be transformed into an equivalent
system in reduced form. Somewhat surprisingly, this can be accomplished by making repeated use
of the following two elementary equation operations.
Constant Times Equation If e is an equation and k is a nonzero constant, then ke is the equation
that results from multiplying both sides of e by k. For example, if e is the equation 3x + 4y = 6,
and k = 1/3, then ke is the equation x + (4/3)y = 2.
Equation Plus Equation If e1 and e2 are equations, then e1 + e2 is the equation that results in
equating the sum of the left sides of e1 and e2 with the sum of the right sides of e1 and e2 . For
example, if e1 is 3x + 4y = 6 and e2 is −2y − z = −4, then e1 + e2 is 3x + 2y − z = 2.
The last two operations are often combined into a single operation of the form ke1 + e2 ; i.e. add a
multiple of e1 to e2 . So from here on we asssume that ke1 + e2 is one of the elementary operations.
When transforming a system of equations S by applying one of the above elementary operations, we
assume that i) ke is replacing e in S, ii) e1 + e2 is replacing e2 in S, and iii) ke1 + e2 is replacing e2
in S. Thus, if the system has m equations, then it will still have m equations after applying one of
5
the elementary operations. Moreover, the next proposition states that the resulting system will be
equivalent to the original.
Proposition 1. Let S0 be a system of linear equations. If S1 is the system that results from applying
a single elementary operation to one or more equations of S0 , then S1 is equivalent to S0 .
Proof of Proposition 1. Any assignment x1 = s1 , . . . , xn = sn that satisfies e ∈ S0 will also
satisfy ke. Conversely, if an assignment satisfies, say ke, then it will also satisfy (1/k)ke = e. Thus,
operation ek does not change the solution set of S0 .
As for e1 + e2 , suppose x1 = s1 , . . . , xn = sn satisfies both e1 and e2 . Then, when substituting the s
values for the variables, the left and right sides of e1 (and those of e2 ) are equal. But then the sum
of the left sides of e1 and e2 will equal the the sum of the right sides of e1 and e2 . In other words,
x1 = s1 , . . . , xn = sn satisfies e1 +e2 . Conversely, if x1 = s1 , . . . , xn = sn satisfies e1 and e1 +e2 , then it
also must satisfy (−1)e1 and, from the previous reasoning, it must also satisfy (e1 + e2 ) + (−1)e1 = e2 .
Therefore, a single elementary operation does not change the solution set, and S0 is equivalent to S1 .
QED
Corollary 1. Let S0 be a system of linear equations. If Sk is the system that results from applying
k consecutive elementary operations to one or more of the equations of S, then Sk is equivalent to
S0 .
Theorem 2. Let S be a system of m linear equations. Then S can be transformed into an equivalent
system in reduced form by applying a sequence of elementary operations.
Proof of Theorem 2. The proof is by induction on m.
Basis Step. If m = 1, then S has a single equation e. If e is a tautology then it must equal 0 = 0,
and S is in reduced form. If e is a fallacy, then it has the form 0 = b, where b 6= 0, then operation
(1/b)e transforms it to 0 = 1. Finally, if e is a conditional equation, then it has at least one variable
x with a nonzero coefficient c. Then (1/c)e transforms e to an equation with lead variable x, while
all other variables appearing in e are independent variables. Thus, this equation is in reduced form.
Inductive Step. Assume the theorem holds for systems with at most m equations. Suppose S is
a system with m + 1 equations. If all of the equations are tautologies and fallacies, then they are
of the form 0 = 0 or may be transformed to 0 = 1 (see basis step). So assume there is at least
one conditional equation e that has at least one variable x with a nonzero coefficient c. Then (1/c)e
transforms e to an equation e1 with lead variable x. Now let ê be any one of the other equations in
which x appears with nonzero coefficient ĉ. Then applyig operation (−ĉ)e1 + ê results in an equation
for which x does not appear. Thus, we may assume that e1 is the only equation for which x appears.
Now consider the system S1 that results from removing x from each equation other than e1 . If e1 is
removed from this system, then the system only has m equations (none of which depend on x). By
the inductive assumption, this new system can be transformed to an equivalent one, call it S2 , that
is in reduced form.
6
Now add e1 to S2 . If no dependent variable of S2 appears in e1 , then the system is still reduced,
and the theorem is proved. On the other hand, assume that dependent variable y appears in e1
with cofficient c 6= 0. Since y is dependent, there is an equation e2 for which y is the lead variable.
Then operation (−c)e2 + e1 results in an equation for which x is still the lead variable, but no longer
contains variable y. This process can be repeated until all the dependent variables that appeared in
e1 have been removed. QED
The proof of Theorem 2 describes an algorithm called Gauss-Jordan (variable) elimination. The
following summarizes the algorithm.
1. If S has only tautologies and fallacies, then terminate the algorithm.
2. Let e be an equation with at least one variable x that has nonzero coefficient c. Make x the
lead variable via operation e1 = (1/c)e.
3. Remove x from any other equation e2 by adding an appropriate multiple of e1 to e2 .
4. Remove e1 from S and repeat the algorithm on the new system S2 .
5. Add e1 back to the reduced system, and remove from it any dependent variable y by adding an
appropriate multiple of e2 to e1 , where e2 has lead variable y.
Example 2. Use the Gauss-Jordan elimination algorithm and provide a sequence of elementaryequation operations that transforms the linear system
e1 : 2x − 3y = 8
e2 : 5x + 4y = −3
into an equivalent reduced-form system. Find the solution set for the system.
Example 2 Solution. The notation e1 ← 21 e1 means that equation e1 is being replaced by the
equation 21 e1 , with the new equation assuming the label e1 .
e1 ← 12 e1 :
e1 : x − (3/2)y = 4
e2 : 5x + 4y = −3
e2 ← −5e1 + e2 :
e1 : x − (3/2)y = 4
e2 : (23/2)y = −23
e2 ← (2/23)e2 :
e1 : x − (3/2)y = 4
7
e2 : y = −2
e1 ← (3/2)e2 + e1 :
e1 : x = 1
e2 : y = −2
Example 3. Use the Gauss-Jordan elimination algorithm and provide a sequence of elementaryequation operations that transforms the linear system
e1 : 2x + 5y + 7z = 9
e2 : − x + 2y + 4z = −3
e3 : 3x − y − 6z = 6
into an equivalent reduced-form system. Find the solution set for the system.
Example 3 Solution.
8
Example 4. Use the Gauss-Jordan elimination algorithm and provide a sequence of elementaryequation operations that transforms the linear system
e1 : x + y + 3z = 9
e2 : − x + 2y + 4z = −3
into an equivalent reduced-form system. Find the solution set for the system.
Example 4 Solution.
In the last two examples the reader may have noticed that the variables do not play much of a
role, other than providing a label for each coefficient. And that it is the coefficients that are being
multiplied and added during an elementary-equation operation. In light of this, the next step is to
modify the algorithm so that we only have to work with the coefficients and not worry about the
variables.
The first step is to assume a variable ordering, such as x1 , . . . , xn . The next step is to transfer the
coefficients into a matrix, which is a two-dimensional array of numbers. Moreover, assuming m
equations and n variables, the matrix will have m rows and n + 1 columns (the extra column is
needed for the right-side constants). The resulting m × n matrix is called the augmented matrix
for the system of equations. The first n columns are called the variable columns, while the last
column is called the constant column. Moreover, matrix that is formed by removing the constant
column from the augmented matrix is called the coefficient matrix.
9
Example 5. Provide the augmented matrix for the following system of equations.
e1 : 2x + 5y + 7z = 9
e2 : − x + 2y + 4z = −3
e3 : 3x − y − 6z = 6
Example 5 Solution. The following is the augmented matrix for the above system.

2
5
7
 −1 2
4
3 −1 −6

9
−3 
6
Example 6. Provide the augmented matrix for the following system of equations. Assume the
variable ordering x, y, z, w.
x − 3z = −2
5y + 8w = −2
2x + 7w = −4
6y − 5z = 0
Example 6 Solution.
We now re-visit elementary equation operations, and provide analogous definitions and results for
augmented matrices. To begin, the augmented-matrix analog of a reduced-form system of equations
is a matrix in reduced row-echelon form. An augmented matrix M for a system of equations is in
reduced row-echelon form iff the following statements hold.
REF1. Each row of M either consists of all zeros (tautology), all zeros except a 1 in the constant column
(fallacy), or has a leading 1 in a variable column thar is preceded by all zeros (conditional).
REF2. If a column of M has a leading 1 from some conditional row, then all othe entries in that column
are zero.
10
REF3. The rows of M are ordered top to bottom as follows: conditional rows, followed by fallacies,
followed by tautologies. Moreover, if r1 and r2 are two conditional rows, then r1 occurs before
r2 iff its leading 1 occurs in an earlier column than r2 ’s leading 1.
Example 7. Explain why, of the following matrices, only M4 is in reduced row-echelon form.


3
−3 
7

1 0 0

1 3 0
M2 =
0 0 0

3
−3 
0


3
0 
6
1 0 0

0 2 0
M1 =
0 0 1
0 1 4

0 0 0
M3 =
1 0 2

0 1 0 2

0 0 1 −4
M4 =
0 0 0 0

7
9 
0
Notice that for an augmented matrix in reduced row-echelon form, a variable column that has a
leading 1 coincides with a dependent variable of the corresponding system, while a variable column
with no leading 1 corresponds with an independent variable. With this observation, the following
theorem is equivalent to Theorem 1, but now makes use of the augmented matrix and reduced
row-echelon concepts.
Theorem 3. Let S be a system of linear equations whose augmented matrix M is in reduced
row-echelon form. Then one of the following is true about S.
1. M has a fallacy row and S is inconsistent.
11
2. Every variable column of M has a leading 1, and so S has exactly one solution.
3. M has at least one variable column with no leading 1, and so S has infinitely many solutions.
We now describe a set of elementary row operations that are analogous to the elementary equation
operations. The only difference is that we must now include an operation for swapping two rows of
a matrix, since reduced row-echelon form requires a specific ordering of the rows.
Constant Times Row If r is a row and k is a nonzero constant, then kr is the row whose i th
entry is equal to kai , where ai is the i th entry of r, for i = 1, . . . , n + 1.
Row Plus Row If r1 and r2 are rows, then r1 + r2 is the row whose i th entry is ai + bi , where ai
and bi are the respective i th entries of r1 and r2 , for i = 1, . . . , n + 1.
Swap Two Rows If r1 and r2 are two rows of a matrix, then this operation swaps the positions of
r1 and r2 .
The first two operations are often combined into a single operation of the form kr1 + r2 ; i.e. add a
multiple of r1 to r2 . So from here on we asssume that kr1 + r2 is one of the elementary operations.
When transforming an augmented matrix M by applying one of the above elementary operations,
we assume that i) kr is replacing r in M , ii) r1 + r2 is replacing r2 in M , iii) kr1 + r2 is replacing r2
in M , and iv) r1 ↔ r2 is swapping rows r1 and r2 of M . Thus, if M has m rows, then it will still
have m rows after applying one of the elementary operations. The next proposition is analogous to
Proposition 1 and Corollary 1, and is stated without proof.
Proposition 2. Let M0 be the augmented matrix for a system of linear equations S. If Mk is the
matrix that results from applying k consecutive elementary row operations to M0 , then Mk is the
augmented matrix of a linear system that is equivalent to S.
The following theorem is analogous to Theorem 2, and is stated without proof.
Theorem 4. Let M be the augmented matrix for a system of linear equations. Then M can
be transformed into a matrix that is in reduced row-echelon form by applying to it a sequence of
elementary row operations.
From the point of view of augmented matrices, the Gauss-Jordan elimination algorithm may now be
viewed as an algorithm for transforming an augmented matrix into one that is in reduced row-echelon
form.
Example 8. Use the Gauss-Jordan elimination
reduced row-echelon form.

2
3
 −5 1
1 −2
algorithm to transform the following matrix into

1 −9
4
10 
5
15
12
Find the solution set for the corresponding system.
Example 8 Solution.
r1 ↔ r3 :
5r1 + r2 :

1 −2 5
 −5 1 4
2
3 1

15
10 
−9


15
85 
−9
1 −2 5
 0 −9 29
2 3 1
−2r1 + r3 :

1 −2 5
 0 −9 29
0 7 −9

15
85 
− 39
(−1/9)r2 :
−7r2 + r3 :

1 −2
5
 0 1 −29/9
0 7
−9

15
− 85/9 
− 39

1 −2
5
 0 1 −29/9
0 0 122/9

15
− 85/9 
244/9


15
− 85/9 
2
9/122r3 :
1 −2
5
 0 1 −29/9
0 0
1
29/9r3 + r2 :

1 −2 5
 0 1 0
0 0 1
2r2 + r1 :

1 0 5
 0 1 0
0 0 1
13

15
−3 
2

9
−3 
2
−5r3 + r1 :

1 0 0
 0 1 0
0 0 1

−1
−3 
2
Homogeneous Systems of Equations
A homogeneous linear equation is one of the form a1 x1 + · · · + an xn = 0. In other words, a
homogeneous equation has a constant term b = 0. A homogeneous system of linear equations is
one in which each equation is homogeneous. Notice that a homogeneous system can never possess a
fallacy, because the right side of each equation is always zero. Moreover, a homogeneous system will
always possess the trivial solution x1 = x2 = · · · = 0, since this assignment makes the left side of
each equation evaluate to zero. From this we may conclude that a homogeneous system of equations
will either have one (trivial) solution, or infinitely many solutions, depending on whether or not its
reduced form has independent variables.
Another observation about a homogeneous system is that the last column of its augmented matrix
always consists of zeros. For this reason the last column can be omitted, and one can work exclusively
with the systems coefficient matrix.
Example 9. Use Gauss-Jordan elimination to find the solution set to the following homogeneous
system.
2x1 + 2x2 − x3 + x5 = 0
−x1 − 2x2 + 2x3 − 3x4 + x5 = 0
x1 + x2 − 2x3 − x5 = 0
x3 + x4 + x 5 = 0
Example 9 Solution.
14
Exercises
1. Which of the following are linear equations?
a. x1 + 2x1 x2 + 4x3 = 2
√
b. 2x1 − 3x2 + 4x3 = k, where k > 0 is a positive constant.
c. x1 + 1/x2 + 4x3 = 8
d. x1 = 2x2
2. Find the solution set for 6x − 7y + z = 3
3. Provide the augmented matrix for the following system of linear equations.
x1 − 2x2 = 0
3x1 + 4x2 = −1
2x1 − x2 = 3
4. Provide the augmented matrix for the following system of linear equations.
x1 + x2 = 1
2x2 − x3 + x5 = 2
2x3 + x4 = 3
5. Find a linear system of equations that corresponds with the following augmented matrix.


3 0 −1
1
 −2 1 4
−3 
0 1 3
3
6. Use induction to prove that if two linear equations e1 and e2 over n variables have the same
solution set, then there exists a k 6= 0 for which e1 = ke2 . Hint: use induction on the number
of variables n.
7. Solve the following linear system using Gauss-Jordan elimination.
x1 + x2 + 2x3 = 8
−x1 − 2x2 + x3 = 1
3x1 − 7x2 + 4x3 = 10
8. Solve the following linear system using Gauss-Jordan elimination.
x − y + 2z − w = −1
2x + y − 2z − 2w = −2
−x + 2y − 4z + w = 1
3x − 3w = −3
15
9. Solve the following linear system using Gauss-Jordan elimination.
2x + 2y + 2z = 0
−2x + 5y + 2z = 0
−7x + 7y + z = 0
10. Solve the following linear system using Gauss-Jordan elimination.
2x1 − 3x2 = −2
2x1 + x2 = 1
3x1 + 2x2 = 1
11. Solve the following linear system using Gauss-Jordan elimination.
3x1 + 2x2 − x3 = −15
5x1 + 3x2 + 2x3 = 0
3x1 + x2 + 3x3 = 11
11x1 + 7x2 = −30
12. Solve the following linear system using Gauss-Jordan elimination.
4x1 − 8x2 = 12
3x1 − 6x2 = 9
−2x1 + 4x2 = −6
13. Find a linear equation over constants a, b, and c in order for the system to have a solution.
x1 + x2 + 2x3 = a
x1 + x3 = b
2x1 + x2 + 3x3 = c
14. Prove that if ad − bc 6= 0, then the reduced row-echelon form of
a b
c d
is
1 0
0 1
.
15. Let A be a 3 × 3 matrix. How many different possible reduced-row echelon forms can A have?
16
Exercise Solutions
1. Which of the following are linear equations?
a. No, because 2x1 x2 is a nonlinear term.
b. Yes, since k is not a variable.
c. No, because 1/x2 is nonlinear.
d. Yes, subtract 2x2 from both sides in order to put it into standard form.
2. Find the solution set for 6x − 7y + z = 3. Set y = s, z = t, where s and t can be any real
numbers. Then x = 1/2 + 7s/6 − t/6. In vector notation, this reads
(x, y, z) = (1/2, 0, 0) + s(7/6, 1, 0) + t(1/6, 0, 1).
3. The augmented matrix is

1 −2
 3 4
2 −1

0
− 1 .
3
4. The augmented matrix is

1 2 0 0 0
 0 2 −1 0 1
0 0 2 1 0

1
2 .
3
5. The following linear system of equations corresponds with the augmented matrix.
3x − z = 1
−2x + y + 4z = −3
y + 3z = 3
6. Basis step: n = 1. Assume ax = b and cx = d have the same solution set, where a, b 6= 0.
Then, we must have b/a = d/c, which implies d = (bc)/a. Then multiplying ax = b by c/a
yields, cx = (bc)/a = d, and thus the second equation is a multiple of the first.
Inductive step: assume that if two equations over n variables have the same solution set,
then they are multiples of each other. Let e1 and e2 be equations over n + 1 ≥ 2 variables, say
x1 , . . . , xn , xn+1 , and assume e1 and e2 have the same solution set. Suppose e1 has the form
a1 x1 + · · · + an xn + an+1 xn+1 = b and e2 has the form c1 x1 + · · · + cn xn + cn+1 xn+1 = d. Since
n ≥ 2, we may assume that xn+1 is an independent variable in the solution, meaning that we
may set xn+1 = t, where t is an arbitrary real number. Then, since e1 and e2 have the same
solution set over variables x1 , . . . , xn+1 , for any value of t, the equations
a1 x1 + · · · + an xn = (b − an+1 t)
and
c1 x1 + · · · + cn xn = (d − cn+1 t)
17
must have the same solution set over variables x1 , . . . , xn . But by the inductive assumption,
this implies that there is a constant k 6= 0 for which the first equation is k times the second.
But this means that a1 = kc1 , . . . , an = kcn , and that
(b − an+1 t) = k(d − cn+1 t).
In particular, for t = 0, we have b = kd. Then setting t = 1 yields
b − an+1 = k(d − cn+1 ) = kd − kcn+1 ,
which implies an+1 = kcn+1 . Therefore e1 = ke2 , and the theorem is proved.
7. The augmented matrix for the linear system system is


1
1 2 8
 −1 −2 1 1  .
3 −7 4 10
Transform into reduced-row echelon form.
r1 + r2 , −3r1 + r3 :


8
9 .
− 14


8
− 9 .
− 104
1 1
2
 0 −1 3
0 −10 −2
−r2 , 10r2 + r3 :
1 1 2
 0 1 −3
0 0 −32
−1/32r3 :
−r2 + r1 :

1 1 2
 0 1 −3
0 0 1

8
− 9 .
13/4


17
− 9 .
13/4
1 0 5
 0 1 −3
0 0 1
3r3 + r2 , −5r3 + r1 :

1 0 0
 0 1 0
0 0 1

3/4
3/4  .
13/4
Solution: x1 = 3/4, x2 = 3/4, x3 = 13/4.
8. The augmented matrix for the linear system system is

1 −1 2 −1
 2
1 −2 −2

 −1 2 −4 1
3
0
0 −3
18

−1
−2 
.
1 
−3
Transform into reduced-row echelon form.
−2r1 + r2 , r1 + r3 , −3r1 + r4 :

1 −1 2 −1
 0 3 −6 0

 0 1 −2 0
0 3 −6 0

−1
0 
.
0 
0


−1
0 
.
0 
0
r2 ↔ r3 , −3r2 + r3 , −3r2 + r4 :
1 −1 2 −1
 0 1 −2 0

 0 0
0
0
0 0
0
0
r2 + r1 :

1
 0

 0
0
0 0 −1
1 −2 0
0 0
0
0 0
0

−1
0 
.
0 
0
Solution: x1 = t − 1, x2 = 2s, x3 = s, x4 = t.
9. Solution: x = −3t, y = −4t, z = 7t.
10. No solution. The system is inconsistent.
11. x1 = −4, x2 = 2, x3 = 7.
12. x1 = 3 + 2t, x2 = t.
13. The augmented matrix for the linear system system is


1 1 2 a
 1 0 1 b .
2 1 3 c
Transform into reduced-row echelon form.
−r1 + r2 , −2r1 + r3 :


a
b − a .
c − 2a


a
a − b .
c−a−b
1 1
2
 0 −1 −1
0 −1 −1
−r2 , r2 + r3 :
1 1 2
 0 1 1
0 0 0
Thus, the system will be consistent provided row 3 does not represent a fallacy; i.e. provided
−a − b + c = 0.
19
14. Since ad − bc 6= 0, matrix
a b
c d
cannot have a row or column of zeros. So without loss of generality, assume a 6= 0. If c = 0,
then it must be the case that d 6= 0. Then the row operations (1/d)e2 , −be2 + e1 , and (1/a)e1
produce the matrix
1 0
.
0 1
On the other hand, if c 6= 0, then (1/a)e1 and −ce1 + e2 transforms the first column into
1
.
0
Moreover, the second entry of row 2 is now δ = −cb/a + d 6= 0, since ad − bc 6= 0. Thus, (1/δ)e2
and (−b/a)e2 + e1 transforms the second column into
0
.
1
15. All rows equal zero: 1. The last two rows equal zero: 3 (why?). The last row equals zero: 3
(why?). Total: 7.
20