**Unformatted text preview: **Systems of Linear Equations
Introduction to Linear Algebra, Math115
Ghazal Geshnizjani —— Reference for some of the slides: Introduction to Linear Algebra for Science and Engineering
by Norman and Wolczuk
Chapter 2 Linear Equations and Their Solutions
Definition (Linear Equation)
A linear equation in n variables x1 , . . . , xn is an equation that can be
written in the form
a1 x1 + a2 x2 + a3 x3 + · · · + an xn = b,
where a1 , a2 , . . . , an ∈ R. Definition (Solution) s1 .. A vector . in Rn is a solution of equation (1) if the equation is
sn
satisfied when we make the substitution
x1 = s1 , x2 = s2 , . . . , xn = sn . (1) A System of Linear Equations A system of linear equations is a collection of finitely many linear
equations.
Example: 3x1 + 2x2 − x3 = 3
2x1
+ x3 = −1 3x2 − 4x3 = 4 is a system of three linear equations in three variables.
How-many solutions can a system of linear equations have? A System of Linear Equations
Example: (
a x1 + b x2 = c
d x1 + e x2 = f a, b, c, d, e, f ∈ R is a system of 2 linear equations in 2 variables.
Each equation is a line in R2 =⇒ Solution should belong to intersection
of the two line.
x2
x2
x2 P nonparallel lines
intersect at a point
ONE solution x1 x1 x1
parallel lines
don’t intersect
NO solution same line
intersect at every
point on the line
INFINITELY many What
about a system of 3 linear equations in 3 variables such as 3x1 + 2x2 − x3 = 3
2x1
+ x3 = −1 3x2 − 4x3 = 4 ? Each equations describes one plane. P1 P2
P3 point of intersection line of intersection P1 P3
P2 P1 P2 P1 P2
P3 P3 solution is the intersection of all the three planes:a point (ONE solution), one line
( INFINITELY many solutions), parallel lines (NO solutions) Consistent Systems and Unique Solutions Theorem Given any system of linear equations, there can be just ONE solution,
INFINITELY many solutions or NO solutions. Terminology
A system is consistent if it has at least one solution.
A system is inconsistent if it does not have any solutions. The Matrix Representation of a System of
Linear Equations
To the linear system
a11 x1
a21 x1 +
+ a12 x2
a22 x2 + ··· +
+ ··· + a1n xn
a2n xn =
=
..
. b1
b2 am1 x1 + am2 x2 + · · · + amn xn = bm
we associate three matrices:
coefficient matrix
constant matrix a11 a12 · · · a1n
b1 a21 a22 · · · a2n b2 A= .
..
.. ~b = .. .. . .
.
am1 am2 · · · amn bm augmented matrix, [A|~b] a11
a21
..
. a12
a22
..
. ···
··· a1n
a2n
..
. am1 am2 · · · amn b1
b2 bm Example
x1 + 3x2 = −1
x1 + x2 = 3 Subtract 1st Eq. from 2nd Eq.
x1 + 1 3 −1
1 1 3 R2 − R1 3x2 = −1
−2x2 = 4 multiply 2nd Eq. by − 12 1 3 −1
0 −2 4 − 12 R2 x1 + 3x2 = −1
x2 = −2 Subtract 3 times 2nd Eq. from 1st Eq. 1 3 −1
0 1 −2 R1 − 3R2
1 0 5
x1
= 5
0 1 −2
x2 = −2
x
5
So we obtain the solution in vector form 1 =
. In point form or
x2
−2
n-tulip form it’s (5,-2).
Example So we can solve the system of equations using the augmented matrix.
x1 + 3x2 = −1
x1 + x2 = 3
We write the augmented matrix
1 3 −1
1 3 −1
1 3 −1
∼
∼
1 1 3
0 −2 4
0 1 −2
R2 −R1
− 12 R2
1 0 5
R1 −3R2
∼
0 1 −2
So we obtain the solution in vector form
x1
5
=
. In point form or
x2
−2 n-tulip form it’s (5,-2).
Note: this system is consistent since it has a solution. Row Operations
Types of Elementary Row Operations (EROs)
(1)
(2)
(3) Multiply one row by a non-zero constant. tRi
Interchange two rows. Ri ↔ Rj
Add a multiple of one row to another row. Ri + tRj Terminology
Row reduction: the process of performing elementary row operations
on a matrix to bring it into some simpler form.
Row equivalent matrices: A matrix M is row equivalent to another
matrix N if M can be reduced to N by a sequence of elementary row
operations. Example
2x1 + x2 + 9x3 = 31
x2 + 2x3 = 8
x1
+ 3x3 = 10
We write the augmented matrix
2 1 9 31 0 1 2 8
1 0 3 10 1 0 0 1
∼
R3 −R2
0 0 1 0 3 10 ∼ 0 1 2 8
R3 ↔R1
2 1 9 31 3 10
R1 − 3R3
1 2 8
R2 − 2R3
0
1 3
∼
0 1 0 3 10 ∼ 0 1 2 8 R3 −2R1
0 1 3 11 0 0 1
x1 = 1 1 0 2
=⇒ x2 = 2
0 1 3
x3 = 3 x1
1 So we obtain the solution in vector form x2 = 2 . In point form or
x3
3
n-tulip form it’s (1,2,3). Note: this system is consistent since it has a Remark x1
1
x2 = 2,
x3
3 P1 P2
P3 point of intersection we could have stopped at this step 1 0 3 10 0 1 2 8 0 0 1 3
turn the augmented matrix back to the system :
x1 x2 + 3x3 = 10
+ 2x3 = 8
x3 = 3 substitute x3 = 3 to 1st and 2nd equation to get x2 and x1 ;
This technique is called Back Substitution. Remarks about doing multiple EROs in one step
1
2 No row should be modified more than once in one step. No row should be both modified and be used to modify other rows in
one step. ∼
· · · R2 − 2R1 This is ok.
R3 + 3R1 ··· ··· R1 + R2
∼
R3 + 3R1 This is not ok. 2R1 + 4R2
∼
This is not ok. You have to do 2R1 first then in the next step do R1 + 4R2 . Row Echelon Form We write −1 −2
−1 ∼ R3 −R2 the
5
6
1 −1 0
0 −x1 + 5x2
=
−2x1 + 6x2 + x3 =
−x1
x2 + x3 =
augmented matrix 0 5
∼
−1
1 3 R2 − 2R1 0
1 −2
R3 − R1
0 −R1
1
5 0 5
−4 1 −7 − 14 R2 0
0 0 0
∼
0 5
3
−2 5 0 5
−4 1 −7 −4 1 −7
−5 0
1 − 14
0
0 REF
5
4 1 0
R1 + 5R2 0 1 − 14
∼
0 0 0 RREF − 15
4
7
4
0 REF −5 0 7
4 Row Echelon Form −1 5 0 5 0 −4 1 −7 ,
0
0 0 0 0 5 0 5 0 0 −3 2 ,
0 0 0 1 REF REF 0 2 0 5 −1
0 0 −5 2 0
REF Definition (Row Echelon Form)
A matrix is in row echelon form (REF) if
(1)
when all entries in a row are zeros, this row appears below all rows
that contain a non-zero entry, and
(2)
when two non-zero rows are compared, the first non-zero entry, called
the leading entry(or pivot), in the upper row is to the left of the
leading entry in the lower row.
Note: A matrix can have many REFs! Reduced Row Echelon Form
1 0 54 0 1 −1
4
0 0 0 RREF − 15
4
7
4 0 , 1 0 0 1 0 1 0 2 ,
0 0 1 3 RREF 0 1 2 0 3
0 0 0 1 −4
RREF Definition (Reduced Row Echelon Form)
A matrix is in reduced row echelon form (RREF) if
(1)
it is in row echelon form,
(2)
all leading entries are 1, called a leading 1, and
(3)
in a column with a leading 1, all the other entries are zeros. Theorem For any given matrix A there is a unique matrix in reduced row echelon
form that is row equivalent to A.
Note: If a matrix is in RREF, then it is in REF too. Back to our example, We got the RREF:
1 0
0 x1 0
1
0 x2 5
4 − 14 −x1 + 5x2
= 5
−2x1 + 6x2 + x3 = 3
−x1 + x2 + x3 = −2
− 15
4
7
4 0 x3 0 x1 =⇒ x2 5
+
= − 15
4 x3
4
1
7
,
+ (− 4 )x3 =
4
0
=
0 ‘ let x3 = t, t ∈ R =⇒ x1 = − 15
4
7
x2 =
4
x3 =
0 + (− 54 ) t
+ ( 14 )t ,
+
(1)t 15 5
−4
x1
−4
=⇒ x2 = 74 + t 14 , t ∈ R
x3
0
1
x1 , x2 are leading variables and x3 is a free variable. We assign parameters
like t to free variables to solve for leading variables. 3 − 2 = 1 , example continued, −x1 + 5x2
= 5
−2x1 + 6x2 + x3 = 3
−x1 + x2 + x3 = −2 15 5
−4
x1
−4
7 =⇒ x2 =
+ t 14 , t ∈ R
4
x3
0
1 line of intersection P1 P3
P2 System is consistent
and solution
corresponds to a line
in R3 Existence of a parameter in the solution means we have infinitely many
solutions. (one point for each value of parameter.) # of parameters = # of variables − # leading 1s. Another example We get the following RREF:
0 1 2 0 3
0 0 0 1 −4 x1 # of variables= 4,
x2 + 2x3 = 3
x4 = −4 x2 x3 x4 # leading 1s= 2 =⇒ # of parameters = 4 − 2 = 2.
=⇒ x1 = t, x3 = s, x2 = 3 − 2s, x4 = −4, s, t ∈ R 0
1
0
x1
t
0
−2
x2 3 − 2s 3 =⇒ x3 = s = 0 + t 0 + s 1 , t, s ∈ R
−4
0
0
x4
−4
Plane in R4 Rank of a Matrix
We saw # of parameters = # of variables − # leading 1s. Definition (Rank)
The rank of a matrix is the number of leading 1s in its reduced row echelon form
(RREF) and is denoted by rank(A). # of parameters = n − rank(A). −1 5 0 5
1 0 54 − 15
n=3
4
A| ~b = −2 6 1 3 ∼ 0 1 − 14 74 =⇒ rank(A) = 2
−1 1 1 −2
1 parameter
0 0 0
0 h i Remark
rank(A) =# of leading 1s in RREF = the number of leading entries in any REF. Example for Inconsistent System
2x1 + 12x2 − 8x3 = −4
2x1 + 13x2 − 6x3 = −5
−2x1 − 14x2 + 4x3 = 7
We write the augmented matrix ∼
2 12 −8 −4
2
12 −8 −4 2 R2 − R1
0 1
2 −1
13 −6 −5
R3 + R1
0 −2 −4 3
−2 −14 4
7 2 12 −8 −4
2x1 + 12x2 − 8x3 0 1
2 −1
x2 + 2x3
∼
=⇒
R3 +2R2
0 0
0
1
0 P1 P2
P3 P1 P2
P3 = −4
= −1
= 1 System is inconsistent and has no
solutions. Equations describe planes
in R3 that don’t have a common
intersection Theorem
h
i
Suppose that the augmented matrix A ~b of a linear system is row
~
S
c
equivalent
to
, which is in row echelon
form. If some row of
S ~c is of the form 0 0 · · · 0 c with c 6= 0 then the system
is inconsistent.
Proof: There are no values of x1 , x2 , . . . , xn ∈ R that can satisfy 0 = 1 ,
Note: This indicates that there is a leading entry/pivot in the last column
of augmented matrix. Solving a complex system
j z1 − z2 −
z3
+ (−1 + j) z4 =
−1
− (1 + j)z3 −
2j z4 =
−1 − 3j
2j z1 − 2z2 −
z3
− (1 − 3j) z4 =
j
Solution: j −1 −1
−1 + j
−1
j −1 −1 −1 + j
−1 0 0 −1 − j −2j −1 − 3j 0 0 −1 − j −2j −1 − 3j ∼
2j −2 −1 −1 + 3j
j
R3 − 2R1 0 0
1
1+j
2+j −jR1
1 j
j
1+j
j
1 j j 1+j j
0 0 1 1+j 2+j R2 ↔ R3 0 0
1
1+j 2+j ∼
∼
0 0 (−1 − j) −2j −1 − 3j R3 + (1 + j)R2 0 0 0 0
0 1 j 0 2 1−j
z , z leading
R1 − jR2 0 0 1 1 + j 2 + j =⇒ 1 3
z2 , z4 free,
∼
0 0 0 0
0 z2 =s, z4 =t =⇒ z1
z2
z3
z4 = (1 − j) − js − 2t
= s
= (2 + j) − (1 + j)t
= t z1
1−j
−j
−2
z2 0 1 0 =⇒ z3 = 2 + j + s 0 + t −1 − j , s, t ∈ C
z4
0
0
1 Gaussian Elimination Any matrix can be reduced to row echelon form by using the following
steps:
(1) (2) (3) (4) (5) (6) Consider the first non-zero column of the matrix (i.e. the first column
that does not consist entirely of zero entries).
If necessary, interchange rows so that the top entry in the column is
non-zero. This entry is called a pivot.
Use elementary row operations of type (3) (i.e.Ri + tRj ) to make all
entries below the pivot into zeros.
Consider the submatrix consisting of all columns to the right of the
column you have just worked on and all rows below the row with the
most recently obtained leading entry.
Repeat the procedure for this submatrix to obtain the next pivot with
zeros below it.
Keep repeating the procedure until you have “used up” all rows and
columns of the original matrix. The resulting matrix is in row echelon Examples
1
−6
4
4
, ~y =
, ~u =
, ~v =
. Express ~u and ~v as
2
−12
8
5
linear combination of ~x and ~y (if possible) using the augmented
matrix framework.
123
Let A =
. What is rank of A?
246
Let ~x = Consistent Systems and Unique Solutions
2x1
2x1
−2x1 12x2
13x2
14x2 +
+
− −
−
+ 8x3
6x3
4x3 =
=
= −4
−5
7 2
∼ 0
0 =⇒ h A ~b i 12
1
0 −8
2
0 −4
−1 1 Theorem
h
Suppose that the augmented matrix A
S ~c , which is in row echelon form. ~b i of a linear system is row equivalent to (1) The
system is inconsistent
if and only if some row of
0 0 · · · 0 c with c 6= 0. (2) If the system is consistent, there are two possibilities:
(a) (b) −1 0
0 S ~c is of the form The number of pivots in S is equal to the number of variables, and the
system has a unique solution.
The number of pivots in S is less than the number of variables, and the
system has infinitely many solutions. 5
−4
0 0
1
0 5
−7 =⇒
0 n=3
rank(A) = 2
1 parameter Proof: exercise for you. System Rank Theorem 2 1 9 31
1 0 0 1
A| ~b = 0 1 2 8 ∼ 0 1 0 2 1 0 3 10
0 0 1 3
h
i
rank(A) = 3, rank( A ~b ) = 3, 3 = 3
System is consistent and has a unique solution. 5
−1 5 0
5
1 0
− 15
h
i
4
4
7 3 ∼ 0 1 − 14
A| ~b = −2 6 1
4
−1 1 1 −2
0 0
0
0
h
i
rank(A) = 2, rank( A ~b ) = 2, 2 = 2
System is consistent and has infinite solutions. 2
12
−8 −4
2 12 −8 −4
h
i
13
−6 −5 ∼ 0 1
2
−1 A ~b ∼ 2
−2 −14
4
7
0 0
0
1
h
i
rank(A) = 2, rank( A ~b ) = 3, 2 < 3
System is inconsistent and has no solutions. h P1 P2
P3 point of intersection line of intersection P1 P1 P2
P3 P3
P2 P1 P2
P3 i System Rank Theorem (part 1)
Theorem
Let
(1) (2) h be a system of m linear equations in n variables.
h
The system is consistent if and only if rank(A) = rank
A
A ~b i ~b i . If the system is consistent, then the number of parameters ( or number of free
variables) in the general solution is the number of variables minus the rank of the
matrix:
# of parameters = n − rank(A). h
i
Proof (1):First n columns in RREF of A ~b form the RREF for A.
We use proof by contraposition.
System is inconsistent ⇐⇒ RREF has some row 0 0 · · · 0 c with c 6= 0.
h
i
⇐⇒ rank(A) < rank
.
A ~b
Proof (2): for you More Examples
(for you)
h A ~b i x + 3x2 = −1
= 1
x1 + x2 = 3
=⇒ 1 3 −1
1 1 3
∼ 1 0 5
0 1 −2 h
i
rank(A) = 2 = rank( A ~b ) consistent X
total # of variables = 2, # of leading 1’s in RREF = 2 =⇒ # of free
variables = 2 − 2 = 0 =⇒ unique solution. X
x
5
solution in vector form 1 =
.
x2
−2 2 1 2 0 3
0 0 0 1 −4 rank(A) = 2, m = 2, n = 4, rank(A) ≤ min{2, 4} Theorem
h
i
Let A ~b be a system of m linear equations in n variables. Then
rank(A) ≤ min{ m, n}.
Proof: There can be at most one leading entry in each row of any REF of
A =⇒ rank(A) ≤ m
There can be at most one leading entry in each column of any REF of
A =⇒ rank(A) ≤ n.
=⇒ rank(A) ≤ min{ m, n} 2
A| ~b = 0
1
h
rank(A) = 3, rank( A 9 31
1 0 0 1
2 8 ∼ 0 1 0 2 3 10
0 0 1 3
i
~b ) = 3, 3 = 3
System is consistent. If we change ~b, it will still be consistent!
rank(A) = 3, m = 3 (m for number of rows/equations) h P1 P2
P3 point of intersection 1
1
0 1
−1 5 0
5
3 ∼ 0
A| ~b = −2 6 1
−1 1 1 −2
0
h
i
rank(A) = 2, rank( A ~b ) = 2, 2 = 2
System is consistent but if we change ~b it may 0
1
0 line of intersection P1 i h P3
P2 rank(A) = 2, m = 3 , i 5
4 − 15
4 0 0 − 14 7
4 become inconsistent! 3 6= 2 System-Rank Theorem (part 2)
(3) Let be a system of m linear equations in n variables. The system is
consistent for all ~b ∈ Rm if and only if rank(A) = m.
h A ~b i Remark: rank(A) = m doesn’t mean the solution is always unique. 2x1 + x2 +
− 2x3
x3 + x4 =
= 3
−4 2
0 1
0 2
−1 0
1 3
−4 , m = 2, n = 4 Definition
Let h A ~b i be a system of m linear equations in n variables. The system is underdetermined if n > m, that is it has more variables than
equations.
The system is overderdetermined if n < m, that is it has more equations than
variables. Remark: rank(A) ≤ m =⇒ n − rank(A) ≥ n − m =⇒ # of free variables ≥ n − m
So if we have a consistent underdetermined system (n > m) then it has infinitely many
solutions.
Remark: An overderdetermined system is often inconsistent. Definition (Homogeneous Linear Equations)
A linear equation is homogeneous if the right-hand side (constant
term) is zero.
A system of equations is homogeneous if all of its equations are
homogeneous.
Augmented Matrix is h A ~0 i Remark
The zero vector ~0 is always a solution of any homogeneous system. This is
called the trivial solution.
Example: 1 1 1 0
0 3 −1 0 x1 +
∼
1
3 R2 x2 + x3 = 0
3x2 − x3 = 0
1 1 1 0
1 0 43
R1 − R2
∼
0 1 − 13 0
0 1 − 13 0
0 We often omit zero column and since only need to row reduce, the
coefficient matrix (A).
4
1
1 0 43 0
=⇒ x3 = t, x1 = − t, x2 = t, t ∈ R
0 1 − 13 0
3
3 4 4
−3t
−3
x1
1 =⇒ x2 = 3 t = t 13 , t ∈ R, line through origin
x3
t
1
Remarks:
you can multiply the direction vector by a number (3 here )to simplify
it.
This system was underdetermined, so it is not a surprise that has
infinitely many solution.
Solution to the homogeneous system corresponded to a subspace of
R3 Theorem Let S be the set of solutions to a homogeneous system of m linear
equation and n variables. Then S is a subspace of Rn .
We call S solution space of the homogeneous system.
Proof: Since system has n variables, S ∈ Rn
~
Since system is homogeneous 0 ∈ S =⇒ non-empty
z1
y1 .. .. Let ~z = . and ~y = . be vectors in S. Then for any arbitrary
zn
yn
equation in the system:
a1 z1 + · · · + an zn = 0 and a1 y1 + · · · + an yn = 0
=⇒ a1 (z1 + y1 ) + · · · + an (zn + yn ) = 0 =⇒ ~z + ~y ∈ S =⇒ closed
underaddition.
also ∀k ∈ R we have
k(a1 y1 + · · · + an yn ) = k0 =⇒ a1 (ky1 ) + · · · + an (kyn ) = 0 =⇒ k~y ∈ S
so it’s closed under scalar multiplication =⇒ S is a sub space of Rn . now lets solve an non-homogeneous system with same coefficient matrix:
x1 +
x + x2 + x3 = 1
x2 + x3 = 0
−→ 1
3x2 − x3 = 3
3x2 − x3 = 0 1 1 1 1
0 3 −1 3 ∼
1
3 R2 1 1 1
0 1 − 13 1
1 R1 − R2
∼ 1 0 43
0 1 − 13 0
1 4
1
=⇒ x3 = t, x1 = − t, x2 = 1 + t, t ∈ R
3
3 4 4
−3t
−3
x1
0
1 =⇒ x2 = 1 + 3 t = 1 + t 13 , t ∈ R
x3
0
t
1 4
−3
x1
solution to associated homogeneous was: x2 = t 13 , t ∈ R
x3
1 0
1 is a particular solution to the non-homogeneous system.
0 Relation between non-homogeneous and its associated homogeneous
system
The complete solution to a consistent non-homogeneous system is a
particular solution to the system plus the complete solution to
corresponding homogeneous system. 4
−3
x1
0
x2 = 1 + t 1 , t ∈ R
3
x3
0
1 Example: Solve the following system and find a basis for solution space.
4x1 − 2x2 + 3x3 + 5x4 = 0
8x1 − 4x2 + 6x3 + 11x4 = 0
−4x1 + 2x2 − 3x3 − 7x4 = 0
Solution: 4 −2 3
5
∼
4 −2 3 5 8 −4 6 11 R2 − 2R1 0 0 0 1 ∼
−4 2 −3 −7
R3 + R1
0 0 0 −2
R3 + 2R2 4 −2 3 5
4 −2 3 0 0 0 0 1 R1 − 5R2 0 0 0 1 ∼
0 0 0 0
0 0 0 0 continued... 1
4 R1 ∼ 1 − 12 0 0
0 0 3
4 0
0 1 0 0 =⇒ x1 , x4 leading, x2 , x3 are free.
we take x2 = s, x3 = t =⇒ x1 = 12 s − 34 t, we also have x4 = 0 1
1 3
3 x1
−4
2s − 4t
2
x2 s 1 0 = s + t s, t ∈ R
=⇒ x3 = t 0 1 x4
0
0
0
Solution
is plane
through origin, a subspace in R4 and
space 1
−3 2 4 1 0 Span , 0 1 0
0 1 3 −4 2 1 0 Also B = , is linearly independent, so together it implies B
0
1 0
0
is a basis for solution space. Example
Consider: x1 + x2 + x3 x4 + 4x5 + 5x6 = 0
+ 2x5 + 3x6 = 0 homogenous system of equations. the augmented matrix:
1 1 1 0 4 5
0 0 0 1 2 3
Already RREF so we have x2 = t1 , x3 = t2 , x5 = t3 , x6 = t4
− 3t4 =⇒ x1 =−t1 −t2 −4t3 −5t4 , x4 = −2t 3
x1
−1
−1
−4
−5
x2 1
0
0
0 x3 0
1
0
0 =⇒ = t1 + t2 + t3 + t4 −3where
x4 0
0
−2 x5 0
0
1
0
x6
0
0
0
1
t1 , t2 , t3 , t4 ∈ R −1
−1
−4
−5 1 0 0 0 0
1
0
0
, , , =⇒ B = spanning set for solution 0 0 −2 −3 0 0 1 0 0
0
0
1
space.
Every vector in B has a 1 for one of its entries that others have zeros
there.
=⇒ B is linearly independent =⇒ B is a Basis for Solution space. Theorem
Let [A|~b] be the augmented matrix for a consistent system of m equations
in n variables. If rank(A) = k < n, then the solution of the system has
the form
~x = ~b + t1~v1 + · · · + tn−k ~vn−k
where ~b ∈ Rn , t1 , . . . , tn−k ∈ R and {~v1 , . . . , ~vn−k } is linearly independent
in Rn .
In particular the solution set is an (n − k)−flat in Rn . Applications of Systems of Linear Equations:
Chemical Reactions
Example: Hydrogen molecules (H2 ) and Oxygen molecules (O2 ) can
combine to produce water (H2 O)
( )H2 + ( )O2 −→ ( )H2 O
balance chemical reaction: the same number of atoms of each type before
and after the chemical reaction.
2H2 + 1O2 −→ 2H2 O
inspection becomes difficult for more complex molecules! Example, the chemical reaction photosynthesis
Plants combine carbon dioxide (CO2 ) and water (H2 O) to produce glucose
(C6 H12 O6 ) and oxygen (O2 ):
( )CO2 + ( )H2 O −→ ( )C6 H12 O6 + ( )O2
New method:
x1 CO2 + x2 H2 O −→ x3 C6 H12 O6 + x4 O2
number of atoms of each type before and after the reaction:
C:
x1 = 6x3
O : 2x1 + x2 = 6x3 + 2x4
H:
2x2 = 12x3 =⇒ x1
2x1 + x2
2x2 − 6x3
= 0
− 6x3 − 2x4 = 0
− 12x3
= 0 1
=⇒ 2
0 1 0
0 1 0
0 x1
2x1 + 0
1
2 −6
−6
−12 0
−2
0 0
1
0 −6
6
−12 0
−2
2 0
1
0 0
0
6 −1
−1
−1 x2
2x2 0
0
0 −
−
− 6x3
6x3
12x3 − 2x4 =
=
= 0
0
0 0
∼
1 0 −6
0 0
∼
0 R2 −2R1 0 1
6 −2 0 1R
0
0 1 −6
0 0
R3 −R2
2 3 0
∼
1 0 −6
0 0
R1 +R3 0 1
0 6 −2 0 R2 −R3
0
− 12 R3
0 0
6 −1 0
∼ ∼
1 0 0
−1
0 0 1 0
−1
0 1R
0 0 1 −1/6 0
6 3 =⇒ x4 = t =⇒ x1 = t, x2 = t, x3 = t/6, t ∈ R
infinitely many solutions! Cannot have a fractional or negative number of molecules!
=⇒ x1 , x2 , x3 and x4 nonnegative integers.
=⇒ t is positve multiple of 6. Take the simplest (or smallest) solution, t = 6.
=⇒ x1 = x2 = x4 = 6 and x3 = 1 . Thus,
6CO2 + 6H2 O −→ C6 H12 O6 + 6O2 Example, the chemical reaction The fermentation of sugar
Balance this chemical reaction: C6 H12 O6 −→ CO2 + C2 H5 OH
where C6 H12 O6 is glucose, CO2 is carbon dioxide and C2 H5 OH is ethanol.
Solution:
x1 C6 H12 O6 −→ x2 CO2 + x3 C2 H5 OH =⇒ x2 + 2x3
2x2 + x3 =⇒
6x3 6 −1 −2 0
∼ 6 −2 −1 0 R2 −R1 12
0 −6 0
R3 −2R1 1R
1
6
0 −3 0
6 1 0 −1
1 0 −R2 0
∼
0
0
0
0 0
C:
O:...

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