systems of linear equations, rref
TRANSCRIPT
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Announcements
Ï If anyone has not been able to access the class website please
email me at [email protected]
Ï Two corrections made to Tuesday's lecture slides (Jan 12).
See slides 27, 28 (I had written R3-R1 instead of R3-2R1) and
39 (I didn't make it completely RREF before saving).
Ï If you spot any typos in the slides any time, please bring it to
my attention asap.
Ï Lst day to drop this class with refund is tomorrow.
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De�nition
Echelon form (Row Echelon form, REF): A rectangular matrix is of
Echelon form (Row Echelon Form or REF) if
Ï All nonzero rows are ABOVE any rows with all zeros
Ï Each leading entry of a row in a column is to the RIGHT to
the leading entry of the row above it (results in a STEP like
shape for leading entries)
Ï All entries in a column below the leading entry are zero
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De�nition
Reduced Echelon form (Reduced Row Echelon form, RREF): A
rectangular matrix is of Echelon form (Reduced Row Echelon Form
or RREF) if
Ï The leading entry in each nonzero row is 1
Ï Each leading 1 is the only nonzero element in its column.
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Important
Each matrix is row equivalent to EXACTLY ONE reduced echelon
matrix. In other words, RREF for a matrix is unique.
Note: Once you obtain the echelon form of a matrix
Ï If you do any more row operations, the leading entries will not
change positions
Ï The leading entries are always in the same positions in any
echelon form starting from the same matrix (These become 1
in reduced echelon form, RREF)
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Pivot Position, Pivot Column
De�nition
Given a matrix A, the location in A which corresponds to a leading
1 in the reduced echelon form is called PIVOT POSITION
De�nition
The column in A that contains a pivot position is called a PIVOT
COLUMN
1 0 −3 0
0 1 5 0
0 0 0 1
0 0 0 0
Location in A corresponding to blue circles gives pivot position
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Example, Problem 4 sec 1.2
Row reduce the matrix A below to echelon form and locate the
pivot columns.
1 3 5 7
3 5 7 9
5 7 9 1
Caution!! This is NOT an augmented matrix of any linear system
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Basic row operations as usual1 3 5 7
3 5 7 9
5 7 9 1
. R3-5R1R2-3R1
=⇒1 3 5 7
0 −4 −8 −120 −8 −16 −34
Divide R3 by 2 1 3 5 7
0 −4 −8 −120 −4 −8 −17
Do R3-R2 1 3 5 7
0 −4 −8 −120 0 0 −5
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Basic row operations as usual1 3 5 7
3 5 7 9
5 7 9 1
. R3-5R1R2-3R1
=⇒1 3 5 7
0 −4 −8 −120 −8 −16 −34
Divide R3 by 2
1 3 5 7
0 −4 −8 −120 −4 −8 −17
Do R3-R2 1 3 5 7
0 −4 −8 −120 0 0 −5
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Basic row operations as usual1 3 5 7
3 5 7 9
5 7 9 1
. R3-5R1R2-3R1
=⇒1 3 5 7
0 −4 −8 −120 −8 −16 −34
Divide R3 by 2 1 3 5 7
0 −4 −8 −120 −4 −8 −17
Do R3-R2
1 3 5 7
0 −4 −8 −120 0 0 −5
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Basic row operations as usual1 3 5 7
3 5 7 9
5 7 9 1
. R3-5R1R2-3R1
=⇒1 3 5 7
0 −4 −8 −120 −8 −16 −34
Divide R3 by 2 1 3 5 7
0 −4 −8 −120 −4 −8 −17
Do R3-R2 1 3 5 7
0 −4 −8 −120 0 0 −5
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Make sure we have only 0 above and below the leading 1
Divide R2 by -4 and divide R3 by -5 to get 1 as leading term. This
gives
1 3 5 7
0 1 2 3
0 0 0 1
1 3 5 7
0 1 2 3
0 0 0 1
.R1-3R2
=⇒1 0 −1 −20 1 2 3
0 0 0 1
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Make sure we have only 0 above and below the leading 1
Divide R2 by -4 and divide R3 by -5 to get 1 as leading term. This
gives 1 3 5 7
0 1 2 3
0 0 0 1
1 3 5 7
0 1 2 3
0 0 0 1
.R1-3R2
=⇒1 0 −1 −20 1 2 3
0 0 0 1
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Make sure we have only 0 above and below the leading 1
Divide R2 by -4 and divide R3 by -5 to get 1 as leading term. This
gives 1 3 5 7
0 1 2 3
0 0 0 1
1 3 5 7
0 1 2 3
0 0 0 1
.R1-3R2
=⇒1 0 −1 −20 1 2 3
0 0 0 1
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Make sure we have only 0 above and below the leading 1
1 0 −1 −2
0 1 2 3
0 0 0 1
. R1+2R3
=⇒1 0 −1 0
0 1 2 3
0 0 0 1
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Make sure we have only 0 above and below the leading 1
1 0 −1 −2
0 1 2 3
0 0 0 1
. R1+2R3
=⇒1 0 −1 0
0 1 2 3
0 0 0 1
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Make sure we have only 0 above and below the leading 1
1 0 −1 0
0 1 2 3
0 0 0 1
.R2-3R3
1 0 −1 0
0 1 2 0
0 0 0 1
Pivot positions in blue circle
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Make sure we have only 0 above and below the leading 1
1 0 −1 0
0 1 2 3
0 0 0 1
.R2-3R3
1 0 −1 0
0 1 2 0
0 0 0 1
Pivot positions in blue circle
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Answer
Ï We have pivot positions at the blue circles
Ï We need to �nd the pivot columns
Ï These are the corresponding columns in the original matrix,
that is Column 1, Column 2 and Column 4 in A (not the
columns in RREF)
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Answer
Ï We have pivot positions at the blue circles
Ï We need to �nd the pivot columns
Ï These are the corresponding columns in the original matrix,
that is Column 1, Column 2 and Column 4 in A (not the
columns in RREF)
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Solutions of Linear systemsThe row reduction steps gives solution of linear system when
applied to augmented matrix of the linear system
Suppose the following matrix is the RREF of the augmented matrix
of a linear system. 1 0 8 5
0 1 4 7
0 0 0 0
.
What is the corresponding system of equations? (Note that this
system has 3 variables since the augmented matrix has 4 columns)
x + 8z = 5
y + 4z = 7
0 = 0
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Solutions of Linear systemsThe row reduction steps gives solution of linear system when
applied to augmented matrix of the linear system
Suppose the following matrix is the RREF of the augmented matrix
of a linear system. 1 0 8 5
0 1 4 7
0 0 0 0
.
What is the corresponding system of equations? (Note that this
system has 3 variables since the augmented matrix has 4 columns)
x + 8z = 5
y + 4z = 7
0 = 0
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Basic Variables, Free variables
Ï Here x and y correspond to the pivot columns of the matrix.
They are called BASIC variables
Ï The remaining variable z which corresponds to the non-pivot
column is called a FREE variable
Ï For a consistent system, we can express the solution of the
system by solving for the basic variables in terms of free
variables
Ï This is possible because RREF makes sure that each basic
variable lies in one and only one equation.
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Solution to the system is thus...
x = 5 − 8z
y = 7 − 4z
z free
Ï When we say that z is free, we mean that we can assign any
value for z . Depending on your choice of z , x and y will get
�xed.
Ï For each choice of z , there is a di�erent solution set.
Ï Every solution of the system is determined by choice of z
We thus have a GENERAL solution of the system.
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Existence and Uniqueness theorem
For a linear system to be consistent,
Ï The rightmost column of an augmented matrix MUST NOT
be a pivot column to avoid the "0=non-zero" situation.
Ï If the linear system is consistent, (i)the solution is either
unique (no free variables) or (ii) in�nitely many solutions (at
least one free variable)
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=1 1 0 1
0 0 1 1
0 0 0 0
RREF
B =1 1 0 0
0 1 1 0
0 0 1 1
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=1 1 0 1
0 0 1 1
0 0 0 0
RREF
B =1 1 0 0
0 1 1 0
0 0 1 1
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=1 1 0 1
0 0 1 1
0 0 0 0
RREF
B =1 1 0 0
0 1 1 0
0 0 1 1
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=1 1 0 1
0 0 1 1
0 0 0 0
RREF
B =1 1 0 0
0 1 1 0
0 0 1 1
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=
1 0 0 0
1 1 0 0
0 1 1 0
0 0 1 1
Neither (The leading element in a row must be to the right of the
leading element in the row above, no step pattern)
B =
0 1 1 1 1
0 0 2 2 2
0 0 0 0 3
0 0 0 0 0
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=
1 0 0 0
1 1 0 0
0 1 1 0
0 0 1 1
Neither (The leading element in a row must be to the right of the
leading element in the row above, no step pattern)
B =
0 1 1 1 1
0 0 2 2 2
0 0 0 0 3
0 0 0 0 0
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=
1 0 0 0
1 1 0 0
0 1 1 0
0 0 1 1
Neither (The leading element in a row must be to the right of the
leading element in the row above, no step pattern)
B =
0 1 1 1 1
0 0 2 2 2
0 0 0 0 3
0 0 0 0 0
REF
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Problem 2, section 1.2
Which of the following matrices are REF and which others are in
RREF?
A=
1 0 0 0
1 1 0 0
0 1 1 0
0 0 1 1
Neither (The leading element in a row must be to the right of the
leading element in the row above, no step pattern)
B =
0 1 1 1 1
0 0 2 2 2
0 0 0 0 3
0 0 0 0 0
REF
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Problem 12, section 1.2
Find the general solutions of the systems whose augmented matrix
is given below
A= 1 −7 0 6 5
0 0 1 −2 −3−1 7 −4 2 7
Do R3+R1 to get new R3
B =1 −7 0 6 5
0 0 1 −2 −30 0 −4 8 12
Divide last row by -4
B =1 −7 0 6 5
0 0 1 −2 −30 0 1 −2 −3
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Problem 12, section 1.2
Find the general solutions of the systems whose augmented matrix
is given below
A= 1 −7 0 6 5
0 0 1 −2 −3−1 7 −4 2 7
Do R3+R1 to get new R3
B =1 −7 0 6 5
0 0 1 −2 −30 0 −4 8 12
Divide last row by -4
B =1 −7 0 6 5
0 0 1 −2 −30 0 1 −2 −3
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Problem 12, section 1.2
Find the general solutions of the systems whose augmented matrix
is given below
A= 1 −7 0 6 5
0 0 1 −2 −3−1 7 −4 2 7
Do R3+R1 to get new R3
B =1 −7 0 6 5
0 0 1 −2 −30 0 −4 8 12
Divide last row by -4
B =1 −7 0 6 5
0 0 1 −2 −30 0 1 −2 −3
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Problem 12, section 1.2
Do R3-R2 to get new R3
B =1 −7 0 6 5
0 0 1 −2 −30 0 0 0 0
Our system of equations thus is
x1 − 7x2 + 6x4 = 5
x3 − 2x4 = −3
0 = 0
Column 1 and Column 3 have the pivot positions. The
corresponding variables x1 and x3 are basic variables. So, x2 and x4are free variables.
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Problem 12, section 1.2
To get the general solution, express the basic variables in terms of
the free variables. This gives
x1 = 5 − 6x4 + 7x2
x3 = −3 + 2x4
x2 and x4 free
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Problem 20, section 1.2
Choose h and k such that the system has (a) no solution (b)
exactly one solution and (c) in�nitely many solutions
x1 + 3x2 = 2
3x1 + hx2 = k
Solution: As usual, start with the augmented matrix
1 3 2
3 h k
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Problem 20, section 1.2
1 3 2
3 h k
. R2-3R1
[1 3 2
0 h−9 k −6
]
Ï If h= 9 AND k = 6, we have 0=0 which means there is a free
variable. So in�nite number of solutions for this choice.
Ï If h= 9 and k 6= 6 we have "0=non-zero" situation. So
inconsistent for this choice of h and k .
Ï If h 6= 9 and for any value of k we have exactly one solution.
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Problem 20, section 1.2
1 3 2
3 h k
. R2-3R1
[1 3 2
0 h−9 k −6
]
Ï If h= 9 AND k = 6, we have 0=0 which means there is a free
variable. So in�nite number of solutions for this choice.
Ï If h= 9 and k 6= 6 we have "0=non-zero" situation. So
inconsistent for this choice of h and k .
Ï If h 6= 9 and for any value of k we have exactly one solution.
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Section 1.3, Vector Equations
Ï A vector is nothing but a list of numbers (for the time being).
Ï In the 2 dimensional x −y plane, this will look like u=[1
2
]or
v=[5
−2]or w=
[.1
.52
]Ï Such a matrix with only 1 column is called a column vector or
just vector
Ï Adding (or subtracting) 2 vectors is easy, just subtract the
corresponding entries. So if u=[1
4
]and v=
[6
−2]then
u+v=
[1+6
4−2
]=
[7
2
]Ï You can multiply a vector with a number, it scales the vector
accordingly. So if u=[1
4
]then 5u=
[5
20
]
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Some Geometry
Ï The column vector =[a
b
]identi�es the point (a,b) on the x−y
plane
Ï The vector u=[3
2
]is an arrow from the origin (0,0) to (3,2)
y
x(0, 0)
(3, 2)
(-2, 1)
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Some Geometry
Ï The column vector =[a
b
]identi�es the point (a,b) on the x−y
plane
Ï The vector u=[3
2
]is an arrow from the origin (0,0) to (3,2)
y
x(0, 0)
(3, 2)
(-2, 1)
![Page 44: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/44.jpg)
Some Geometry
Ï The column vector =[a
b
]identi�es the point (a,b) on the x−y
plane
Ï The vector u=[3
2
]is an arrow from the origin (0,0) to (3,2)
y
x(0, 0)
(3, 2)
(-2, 1)
![Page 45: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/45.jpg)
Some Geometry
Ï The column vector =[a
b
]identi�es the point (a,b) on the x−y
plane
Ï The vector u=[3
2
]is an arrow from the origin (0,0) to (3,2)
y
x(0, 0)
(3, 2)
(-2, 1)
![Page 46: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/46.jpg)
Some Geometry
Ï The column vector =[a
b
]identi�es the point (a,b) on the x−y
plane
Ï The vector u=[3
2
]is an arrow from the origin (0,0) to (3,2)
y
x(0, 0)
(3, 2)
(-2, 1)
![Page 47: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/47.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and v
y
x0
u
v
u+v
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Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 49: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/49.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 50: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/50.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 51: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/51.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 52: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/52.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 53: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/53.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 54: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/54.jpg)
Parallelogram rule for vector addition
If u and v are points in a plane, then u+v is the fourth vertex of a
parallelogram whose other vertices are u, 0 and vy
x0
u
v
u+v
![Page 55: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/55.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.
y
x0
u
0.5u
2u
![Page 56: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/56.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.y
x0
u
0.5u
2u
![Page 57: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/57.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.y
x0
u
0.5u
2u
![Page 58: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/58.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.y
x0
u
0.5u
2u
![Page 59: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/59.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.y
x0
u
0.5u
2u
![Page 60: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/60.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.y
x0
u
0.5u
2u
![Page 61: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/61.jpg)
Scalar Multiplication
Multiplying a vector by a number scales it correspondingly.y
x0
u
0.5u
2u
![Page 62: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/62.jpg)
Vectors in 3-D
Vectors in the x −y −z plane is a 3X1 column matrix. Here u=234
is illustrated.
y
x
z
0
u
![Page 63: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/63.jpg)
Vectors in 3-D
Vectors in the x −y −z plane is a 3X1 column matrix. Here u=234
is illustrated.
y
x
z
0
u
![Page 64: Systems of Linear Equations, RREF](https://reader036.vdocuments.net/reader036/viewer/2022062319/55628971d8b42a950c8b4609/html5/thumbnails/64.jpg)
Vectors in 3-D
Vectors in the x −y −z plane is a 3X1 column matrix. Here u=234
is illustrated.
y
x
z
0
u