Linear Algebra and Its Applications
Sixth Edition, Global Edition
Chapter 1
Linear Equations
in Linear Algebra
Slides by Author
Modified by Fernando H. Calderon
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Section 1.1: Systems of Linear
Equations
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Linear Equation (1 of 3)
• A linear equation in the variables 𝑥1 , … , 𝑥𝑛 is an equation
that can be written in the form
𝑎1 𝑥1 + 𝑎2 𝑥2 + ⋯ + 𝑎𝑛 𝑥𝑛 = 𝑏, (1)
where b and the coefficients 𝑎1 , … , 𝑎𝑛 are real or complex
numbers that are usually known in advance.
• A system of linear equations (or a linear system) is a
collection of one or more linear equations involving the
same variables — say, 𝑥1 , … , 𝑥𝑛 .
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Linear Equation (2 of 3)
• A solution of the system is a list (𝑠1 , 𝑠2 , … , 𝑠𝑛 ) of numbers
that makes each equation a true statement when the values
𝑠1 , … , 𝑠𝑛 are substituted for 𝑥1 , … , 𝑥𝑛 , respectively.
• The set of all possible solutions is called the solution set
of the linear system.
• Two linear systems are called equivalent if they have the
same solution set.
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Linear Equation (3 of 3)
• A system of linear equations has
1. no solution, or
2. exactly one solution, or
3. infinitely many solutions.
• A system of linear equations is said to be consistent if it
has either one solution or infinitely many solutions.
• A system of linear equation is said to be inconsistent if it
has no solution.
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Matrix Notation (1 of 2)
• The essential information of a linear system can be
recorded compactly in a rectangular array called a matrix.
• For the following system of equations,
𝑥1 − 2𝑥2 + 𝑥3 = 0 1 −2 1
2𝑥2 − 8𝑥3 = 8 the matrix 0 2 −8
−4𝑥1 + 5𝑥2 + 9𝑥3 = −9, −4 5 9
is called the coefficient matrix of the system.
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Matrix Notation (2 of 2)
• An augmented matrix of a system consists of the
coefficient matrix with an added column containing the
constants from the right sides of the equations.
• For the given system of equations,
1 −2 1 0
0 2 −8 8
−4 5 9 −9
is called the augmented matrix.
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Matrix Size
• The size of a matrix tells how many rows and columns it
has. If m and n are positive numbers, an m × n matrix
is a rectangular array of numbers with m rows and n
columns. (The number of rows always comes first.)
• The basic strategy for solving a linear system is to
replace one system with an equivalent system (i.e., one
with the same solution set) that is easier to solve.
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Solving System of Equations (1 of 7)
• Example 1: Solve the given system of equations.
𝑥1 − 2𝑥2 + 𝑥3 = 0 −−− − 1
2𝑥2 − 8𝑥3 = 8 −−− − 2
−4𝑥1 + 5𝑥2 + 9𝑥3 = −9 −−− − 3
• Solution: The elimination procedure is shown here with
and without matrix notation, and the results are placed
side by side for comparison.
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Solving System of Equations (2 of 7)
𝑥1 − 2𝑥2 + 𝑥3 = 0 1 −2 1 0
2𝑥2 − 8𝑥3 = 8 0 2 −8 8
−4𝑥1 + 5𝑥2 + 9𝑥3 = −9 −4 5 9 −9
• Keep 𝑥1 in the first equation and eliminate it from the
other equations. To do so, add 4 times equation 1 to
equation 3.
4𝑥1 − 8𝑥2 + 4𝑥3 = 0
−4𝑥1 + 5𝑥2 + 9𝑥3 = −9
−3𝑥2 + 13𝑥3 = −9
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Solving System of Equations (3 of 7)
• The result of this calculation is written in place of the
original third equation.
𝑥1 − 2𝑥2 + 𝑥3 = 0 1 −2 1 0
2𝑥2 − 8𝑥3 = 8 0 2 −8 8
−3𝑥2 + 13𝑥3 0 −3 13 −9
1
• Now, multiply equation 2 by
2
in order to obtain 1 as the
as the coefficient for 𝑥2 .
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Solving System of Equations (4 of 7)
𝑥1 − 2𝑥2 + 𝑥3 = 0 1 −2 1 0
𝑥2 − 4𝑥3 = 4 0 1 −4 4
−3𝑥2 + 13𝑥3 = −9 0 −3 13 −9
• Use the 𝑥2 in equation 2 to eliminate the −3𝑥2 in
equation 3.
3𝑥2 − 12𝑥3 = 12
−3𝑥2 + 13𝑥3 = −9
𝑥3 = 3
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Solving System of Equations (5 of 7)
• The new system has a triangular form.
𝑥1 − 2𝑥2 + 𝑥3 = 0 1 −2 1 0
𝑥2 − 4𝑥3 = 4 0 1 −4 4
𝑥3 = 3 0 0 1 3
• Now, you want to eliminate the −2𝑥2 term from equation 1,
but it is more efficient to use the 𝑥3 term in equation 3
first to eliminate the −4𝑥3 and 𝑥3 terms in equations 2
and 1.
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Solving System of Equations (6 of 7)
4𝑥3 = 12 −𝑥3 = −3
𝑥2 − 4𝑥3 = 4 𝑥1 − 2𝑥2 + 𝑥3 = 0
𝑥2 = 16 𝑥1 − 2𝑥2 = −3
• Now, combine the results of these two operations.
𝑥1 − 2𝑥2 = −3 1 −2 0 −3
𝑥2 = 16 0 1 0 16
𝑥3 = 3 0 0 1 3
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Solving System of Equations (7 of 7)
• Move back to the 𝑥2 in equation 2, and use it to eliminate
the − 2𝑥2 above it. Because of the previous work with 𝑥3 ,
there is now no arithmetic involving 𝑥3 terms. Add 2 times
equation 2 to equation 1 and obtain the system:
𝑥1 = 29 1 0 0 29
𝑥2 = 16 0 1 0 16
𝑥3 = 3 0 0 1 3
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Reasonable Answers
• Thus, the only solution of the original system is (29,16,3).
To verify that (29,16,3) is a solution, and hence a
reasonable answer, substitute these values into the left
side of the original system, and compute.
(29) − 2(16) + (3) = 29 − 32 + 3 = 0
2(16) − 8(3) = 32 − 24 = 8
−4(29) + 5(16) + 9(3) = −116 + 80 + 27 = −9
• The results agree with the right side of the original system,
so (29,16,3) is a solution of the system.
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Elementary Row Operations (1 of 2)
• Elementary row operations include the following:
1. (Replacement) Replace one row by the sum of itself
and a multiple of another row.
2. (Interchange) Interchange two rows.
3. (Scaling) Multiply all entries in a row by a nonzero
constant.
• Two matrices are called row equivalent if there is a
sequence of elementary row operations that transforms
one matrix into the other.
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Elementary Row Operations (2 of 2)
• It is important to note that row operations are reversible.
• If the augmented matrices of two linear systems are row
equivalent, then the two systems have the same solution
set.
• Two fundamental questions about a linear system are as
follows:
1. Is the system consistent; that is, does at least one
solution exist?
2. If a solution exists, is it the only one; that is, is the
solution unique?
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Existence And Uniqueness of System
of Equations (1 of 4)
• Example 2: Determine if the following system is consistent.
𝑥2 − 4𝑥3 = 8
2𝑥1 − 3𝑥2 + 2𝑥3 = 1 −−− − 4
5𝑥1 − 8𝑥2 + 7𝑥3 = 1
• Solution: The augmented matrix is
0 1 −4 8
2 −3 2 1
5 −8 7 1
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Existence And Uniqueness of System
of Equations (2 of 4)
• To obtain an 𝑥1 in in the first equation, interchange rows
1 and 2. 2 −3 2 1
0 1 −4 8
5 −8 7 1
• To eliminate the 5𝑥1 term in the third equation, add −5
2
times row 1 to row 3.
2 −3 2 1
0 1 −4 8 −−− − 5
0 −1/2 2 −3/2
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Existence And Uniqueness of System
of Equations (3 of 4)
• Next, use the 𝑥2 term in the second equation to eliminate the −(1
/2)𝑥2
term from the third equation. Add 1 times row 2 to row 3.
2
2 −3 2 1
0 1 −4 8 −−− − 6
0 0 0 5/2
• The augmented matrix is now in triangular form. To interpret it correctly,
go back to equation notation.
2𝑥1 − 3𝑥2 + 2𝑥3 = 1
𝑥2 − 4𝑥3 = 8 −−− − 7
0 = 5/2
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Existence And Uniqueness of System
of Equations (4 of 4)
• The equation 0=
5 is a short form of 0𝑥1 + 0𝑥2 + 0𝑥3 =
5
2 2.
• There are no values of 𝑥1 , 𝑥2 , 𝑥3 that satisfy (7) because
the equation 0=
5 is never true.
2
• Since (7) and (4) have the same solution set, the original
system is inconsistent (i.e., has no solution).
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Section 1.2: Row Reduction and
Echelon Forms
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Echelon Form (1 of 3)
• A rectangular matrix is in echelon form (or row echelon
form) if it has the following three properties:
1. All nonzero rows are above any rows of all zeros.
2. Each leading entry of a row is in a column to the right
of the leading entry of the row above it.
3. All entries in a column below a leading entry are zeros.
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Echelon Form (2 of 3)
• If a matrix in echelon form satisfies the following additional
conditions, then it is in reduced echelon form (or
reduced row echelon form):
4. The leading entry in each nonzero row is 1.
5. Each leading 1 is the only nonzero entry in its column.
• An echelon matrix (respectively, reduced echelon
matrix) is one that is in echelon form (respectively,
reduced echelon form.)
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echelon matrix reduced echelon matrix
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Echelon Form (3 of 3)
• Any nonzero matrix may be row reduced (i.e.,
transformed by elementary row operations) into more than
one matrix in echelon form, using different sequences of
row operations. However, the reduced echelon form one
obtains from a matrix is unique.
Theorem 1: Uniqueness of the Reduced Echelon Form
Each matrix is row equivalent to one and only one
reduced echelon matrix.
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Pivot Position (1 of 7)
• If a matrix A is row equivalent to an echelon matrix U, we
call U an echelon form (or row echelon form) of A; if U is
in reduced echelon form, we call U the reduced echelon
form of A.
• A pivot position in a matrix A is a location in A that
corresponds to a leading 1 in the reduced echelon form of
A. A pivot column is a column of A that contains a pivot
position.
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Pivot Position (2 of 7)
• Example 1: Row reduce the matrix A below to echelon
form, and locate the pivot columns of A.
0 −3 −6 4 9
−1 −2 −1 3 1
𝐴=
−2 −3 0 3 −1
1 4 5 −9 −7
• Solution: The top of the leftmost nonzero column is the
first pivot position. A nonzero entry, or pivot, must be
placed in this position.
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Pivot Position (3 of 7)
• Now, interchange rows 1 and 4.
• Create zeros below the pivot, 1, by adding multiples of the
first row to the rows below, and obtain the next matrix.
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Pivot Position (4 of 7)
• Choose 2 in the second row as the next pivot.
• Add −5/2 times row 2 to row 3, and add 3/2 times row
2 to row 4.
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Pivot Position (5 of 7)
1 4 5 −9 −7
0 2 4 −6 −6
0 0 0 0 0
0 0 0 −5 0
• There is no way a leading entry can be created in column
3. But, if we interchange rows 3 and 4, we can produce a
leading entry in column 4.
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Pivot Position (6 of 7)
• The matrix is in echelon form and thus reveals that
columns 1, 2, and 4 of A are pivot columns.
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Pivot Position (7 of 7)
• The pivots in the example are 1, 2 and −5.
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Row Reduction Algorithm (1 of 10)
• Example 2: Apply elementary row operations to transform
the following matrix first into echelon form and then into
reduced echelon form.
0 3 −6 6 4 −5
3 −7 8 −5 8 9
3 −9 12 −9 6 15
• Solution:
• STEP 1: Begin with the leftmost nonzero column. This is a
pivot column. The pivot position is at the top.
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Row Reduction Algorithm (2 of 10)
• STEP 2: Select a nonzero entry in the pivot column as a
pivot. If necessary, interchange rows to move this entry
into the pivot position.
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Row Reduction Algorithm (3 of 10)
• Interchange rows 1 and 3. (Rows 1 and 2 could have also
been interchanged instead.)
• STEP 3: Use row replacement operations to create zeros
in all positions below the pivot.
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Row Reduction Algorithm (4 of 10)
• We could have divided the top row by the pivot, 3, but with two
3s in column 1, it is just as easy to add −1 times row 1 to row 2.
• STEP 4: Cover the row containing the pivot position, and cover
all rows, if any, above it. Apply steps 1–3 to the submatrix that
remains. Repeat the process until there are no more nonzero
rows to modify.
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Row Reduction Algorithm (5 of 10)
• With row 1 covered, step 1 shows that column 2 is the next pivot
column; for step 2, select as a pivot the “top” entry in that
column.
• For step 3, we could insert an optional step of dividing the “top”
row of the submatrix by the pivot, 2. Instead, we add −3/2
times the “top” row to the row below.
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Row Reduction Algorithm (6 of 10)
• This produces the following matrix.
• When we cover the row containing the second pivot position for
step 4, we are left with a new submatrix that has only one row.
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Row Reduction Algorithm (7 of 10)
• Steps 1–3 require no work for this submatrix, and we
have reached an echelon form of the full matrix. We
perform one more step to obtain the reduced echelon
form.
• STEP 5: Beginning with the rightmost pivot and working
upward and to the left, create zeros above each pivot. If a
pivot is not 1, make it 1 by a scaling operation.
• The rightmost pivot is in row 3. Create zeros above it,
adding suitable multiples of row 3 to rows 2 and 1.
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Row Reduction Algorithm (8 of 10)
3 −9 12 −9 0 − 9 ← Row 1 + −6 × row 3
0 2 −4 4 0 − 14 ← Row 2 + −2 × row 3
0 0 0 0 1 4
• The next pivot is in row 2. Scale this row, dividing by the
pivot.
3 −9 12 −9 0 −9
1
0 1 −2 2 0 − 7 ← Row scaled by
2
0 0 0 0 1 4
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Row Reduction Algorithm (9 of 10)
• Create a zero in column 2 by adding 9 times row 2 to row 1.
3 0 −6 9 0 − 72 ← Row 1 + 9 × row 2
0 1 −2 2 0 −7
0 0 0 0 1 4
• Finally, scale row 1, dividing by the pivot, 3.
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Row Reduction Algorithm (10 of 10)
1 0 −2 3 0 − 24 1
← Row scaled by
0 1 −2 2 0 −7 3
0 0 0 0 1 4
• This is the reduced echelon form of the original matrix.
• The combination of steps 1–4 is called the forward phase
of the row reduction algorithm. Step 5, which produces the
unique reduced echelon form, is called the backward
phase.
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Solutions of Linear Systems (1 of 4)
• The row reduction algorithm leads to an explicit description
of the solution set of a linear system when the algorithm is
applied to the augmented matrix of the system.
• Suppose that the augmented matrix of a linear system has
been changed into the equivalent reduced echelon form.
1 0 −5 1
0 1 1 4
0 0 0 0
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Solutions of Linear Systems (2 of 4)
• There are 3 variables because the augmented matrix has
four columns. The associated system of equations is
𝑥1 − 5𝑥3 = 1
𝑥2 + 𝑥3 = 4 −− − 1
0=0
• The variables 𝑥1 and 𝑥2 corresponding to pivot columns in
the matrix are called basic variables. The other variable,
𝑥3 , is called a free variable.
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Solutions of Linear Systems (3 of 4)
• Whenever a system is consistent, as in (1), the solution
set can be described explicitly by solving the reduced
system of equations for the basic variables in terms of
the free variables.
• This operation is possible because the reduced echelon
form places each basic variable in one and only one
equation.
• In (1), solve the first and second equations for 𝑥1 and 𝑥2 .
(Ignore the third equation; it offers no restriction on the
variables.)
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Solutions of Linear Systems (4 of 4)
𝑥1 = 1 + 5𝑥3
𝑥2 = 4 − 𝑥3 −− − 2
𝑥3 is free
• The statement “𝑥3 Is free” means that you are free to choose any value for
𝑥3 . Once that is done, the formulas in (2) determine the values for 𝑥1 and 𝑥2 .
For instance, when 𝑥3 = 0, the solution is (1,4,0); when 𝑥3 = 1, the solution
is (6,3,1).
• Each different choice of 𝑥3 determines a (different) solution of the
system, and every solution of the system is determined by a choice of 𝑥3 .
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Reasonable Answers (1 of 2)
• We can now verify that our solution is “reasonable” for a
given matrix. Write down the system of equations
associate with the matrix:
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Reasonable Answers (2 of 2)
• Then substitute in the solution you found for each variable
and verify that the equations on the left add to the correct
amount:
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Parametric Descriptions of Solution
Sets (1 of 2)
• The description in (2) is a parametric description of
solutions sets in which the free variables act as
parameters.
• Solving a system amounts to finding a parametric
description of the solution set or determining that the
solution set is empty.
• Whenever a system is consistent and has free variables,
the solution set has many parametric descriptions.
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Parametric Descriptions of Solution
Sets (2 of 2)
• For instance, in system (1), add 5 times equation 2 to equation 1 and
obtain the following equivalent system.
𝑥1 + 5𝑥2 = 21
𝑥2 + 𝑥3 = 4
• We could treat 𝑥2 as a parameter and solve for 𝑥1 and 𝑥3 in terms of
𝑥2 , and we would have an accurate description of the solution set.
• When a system is inconsistent, the solution set is empty, even when
the system has free variables. In this case, the solution set has no
parametric representation.
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Existence and Uniqueness Theorem
Theorem 2: Existence and Uniqueness Theorem
A linear system is consistent if and only if the rightmost
column of the augmented matrix is not a pivot column—i.e.,
if and only if an echelon form of the augmented matrix has
no row of the form
0 … 0 𝑏 with 𝑏 nonzero.
• If a linear system is consistent, then the solution set
contains either i a unique solution, when there are no
free variables, or ii infinitely many solutions, when there
is at least on free variable.
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Row Reduction to Solve A Linear
System (1 of 2)
• Using Row Reduction to Solve a Linear System
1. Write the augmented matrix of the system.
2. Use the row reduction algorithm to obtain an
equivalent augmented matrix in echelon form.
Decide whether the system is consistent. If there is
no solution, stop; otherwise, go to the next step.
3. Continue row reduction to obtain the reduced
echelon form.
4. Write the system of equations corresponding to the
matrix obtained in step 3.
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Row Reduction to Solve A Linear
System (2 of 2)
5. Rewrite each nonzero equation from step 4 so that its
one basic variable is expressed in terms of any free
variables appearing in the equation.
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