Alg1131W10T1 - Gaussian elimination and back substitution
Sunday, 13 May 2018
11:28 PM
Created with Microsoft OneNote 2016.
![Untitled picture.png Machine generated alternative text:
leading row of a matrix is
leading entry of a leading row
leading column of a matrix is
O
O
O
O
O
O
a row which staffs with a zero
a row which is followed by other rows
a row which is above a row of zeros
a row which is not all zeros.
the left-most non-zero entry
the left-most zero entry
the right-most non-zero entry
the right-most zero entry.
a column which contains a leading
entry
a column which contains all leading
O
entries
a column which is followed by other
O
columns
a column which contains a non-zero
O
entry
Untitled picture.png Machine generated alternative text:
An augmented matrix (A I b) has been reduced to row-echelon form (U I y). This means that [select all that apply]
any all-zero rows are at the bottom of the matrix
any all-zero columns are at the right of the matrix
in every leading row; the leading entry is further to the right than the leading entry in any higher row
there is a solution to the corresponding system
all entries below a leading entry are zero.
Untitled picture.png Machine generated alternative text:
Let's solve the following system of linear equations using the matrix approach:
In matrix form this is represented by the augmented matrix (A I b) where
is the matrix of coefficients and
b
is the right-hand side vector.
Note:
• The Maple syntax for the matrix
• The Maple syntax for a vector is<l , 2 ,
If we row reduce the matrix (A I b) we may get to the row-echelon form
x —y+z=O
2m — y + 3z
x — 2y + 2z
1
o
0
a
1
2
o
b
c
where a
4
Now we may work backwards to find that z
O
3
and c
O
6
then y
—1
1
o
O
and finally x
-2](Alg1131W10T1%20-%20Gaussian%20elimination%20and%20back%20substitution_files/image003.png)
