Row space, column space, null space And Rank, Nullity and Rank-Nullity theore...
Matrix Row Echelon Form and Rank (40
1.
2.
3. A row of a matrix is a zero row if it consists entirely of
zeros. A row which is not a zero row is a nonzero row.
A leading entry of a row refers to the leftmost nonzero
entry in a nonzero row
A unit column of a matrix is a column with one
entry a 1 and all other entries 0.
non-zero row 0 2 1 1 0 3
0 0 0 0 0 0
Leading entry 1 6 1 0 2 0
zero row 0 0 0 0 0 0
unit column
4. A matrix is in row-echelon form if and only if:
1. All zero rows occur below all nonzero rows.
2. The leading entry of a nonzero row lies strictly to the
right of the leading entry of any other preceding row.
3. All entries in a column below a leading entry are zeros.
3 0 6 3
3.
2. 0 2 -1 4
0 0 1 2
1. 0 0 0 0
5. A matrix A is in or
if it is in
echelon form and satisfies the following
additional conditions:
1- All the leading entries are 1.
2- Every column containing a leading one is
a .
7. The following matrices are in echelon
form.
2 3 2 1 1 2 1 0
0 1 4 8 0 1 4 4
0 0 0 5 2 0 0 1 3
8. And the following matrices
1 0 0 29 1 0 0 0
0 1 0 16 0 1 0 4
0 0 1 3 0 0 0 0
are in reduced echelon form.
9. Example 3:
not a unit column
1 2 3 A is in row-echelon form.
A fails to be in reduced row-echelon form
A 0 1 7
because the second column contains a
0 0 0 leading 1, but is not a unit column.
leading 1
C fails to be in row-echelon form because
1 2 1 5 the leading 1 in the second row is to the
C 0 0 0 1 right of the leading 1 in the third row. Since
0 0 1 0 C is not in row-echelon form, it cannot be
in reduced row-echelon form.
leading 1s
10. The following matrices are in echelon
form. 0
0 0 0
0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 where
0 0 0
0 0 0 any real number
12. A is one that is in echelon
form.
A is one that is
in reduced echelon form.
13. Each matrix is row equivalent to one
and only one reduced echelon matrix.
14. The rank of the matrix is the number of
nonzero rows of its row echelon form
15. Example 4 .
1 3 2 R1 R 2 1 3
A
2 6 0 0
Thus
r (A ) 1
1 3 2R1 R 2 1 3 R2 1 3 3R 2 R1 1 0
B
2 5 0 1 0 1 0 1
Thus,
r(B) = 2
16. Step 1: Begin with the leftmost nonzero
column say Cr . If a1r =0 and asr 0
interchange R1 by a row Rs.
Step 2: Use elementary row operations to
create zeros in all positions below a1r
-(asr /a1r) R1+Rs s= 2,3, ,m
17. Step 3: go to the next leftmost nonzero
column and repeat steps 1and 2 to the
remaining submatrix. Repeat the process
until there are no more nonzero rows to
modify.
18. Row reduce the matrix below to echelon
form and find the rank of .
0 3 6 4 9
1 2 1 3 1
A
2 3 0 3 1
1 4 5 9 7
27. Assume that A is an invertible matrix of
order n
Step 1: Construct the augmented matrix
B=[A In]
Step 2: Reduce B to its (unique) reduced
echelon form
The resulting equivalent matrix is
-1
n
40. A linear equation in n variables x 1, x 2 , , x n is defined to
be in the form
a1x 1 a2x 2 an x n b
where a1, a2 , , an ,b R .
The variables in a linear equation are called the unknowns.
41. Linear N o t lin e a r
2
x 3y 7 x 3y 7
x1 2 x2 3 x3 x4 7 3x 2 z xz 4
1
y 2 x 3z 1 y sin x 0
x1 x2 xn 1 x1 2x 2 x3 1
42. A finite set of linear equations in the variables x 1 , x 2 , , x n is called
a system of linear equations .
a11x 1 a12x 2 a1n x n b1
a21x 1 a22x 2 a2n x n b2
am1x 1 am 2x 2 amn x n bm
A sequence of numbers s1, s2 , , sn is called a solution of the
system if x1 s1, x2 s2 , , xn sn is a solution of every equation
in the system.
43. Matrix form of a system of linear
equations:
a11 x1 a12 x2 a1n xn b1
a21 x1 a22 x2 a2 n xn b2 m linear equations
am 1 x1 am 2 x2 amn xn bm
a11 a12 a1n x1 b1
a21 a22 a2 n x2 b2 Single matrix equation
Ax b
m n n 1 m 1
am 1 am 2 amn xn bm
A x b
44. A system of equations that has no solution is said to be
inconsistent;
if there is at least one solution of the system, it is said to be
consistent.
The system
x y 4
2x 2 y 6
has no solutions since the equivalent system
x y 4
x y 3
has contradictory equations.
45. Every system of linear equations has either
solutions, exactly solution, or
solutions.
46. If m < n, the system has more than one of
solutions.
If m > n, the system can be reduced to
equivalent system where m n, or it is
inconsistant.
If m = n , the system has either no solution
or one solution.
47. Solution possibilities for two lines
Consider the solutions to the general system of two linear equations
in the two unknowns x and y:
a1 x b1 y c1 (a1 , b1 not both zero)
a2 x b2 y c2 (a2 , b2 not both zero)
The graphs of these equations are lines; call them l1 and l2 .
Then the solutions of the system correspond to the intersections
of the lines.
48. We will study the solution of linear systems under
the following assumptions:
m = n (#of equations = # of unknowns)
b is a non-zero matrix
non-
The system is consistent .
Under these assumptions the system has
either no solution or one solution.
49. For the linear system AX=b , if A is
invertible then the system is consistent
and it has one solution
X = A-1 b
50. Solvin g a syst e m
1-GAUSS ELIMINATION
Reduce the augmented matrix to the echelon
form and use back substitution.
substitution.
2-GAUSS-JORDON ELIMINATION
GAUSS-
Reduce the augmented matrix to the reduced
[A b] [In X]
51. To find the solution and the inverse of
the matrix at the same time reduce the
matrix
[A In b] to the reduced echelon form
to get [I A-1 X]
-1
n
