Mathematics-3
Section (2)
Faculty of Information Technology
Egyptian E-Learning University
Fall 2023-2024
9. Row echelon form, and reduced row echelon form
A matrix is in row echelon form if its entries satisfy the
following conditions
1. The first nonzero entry in each row is a 1 (called a leading 1).
2. Each leading 1 comes in a column to the right of the leading 1s in
rows above it.
3. All rows of all 0s come at the bottom of the matrix.
A matrix in row echelon form is in reduced row echelon form
when every column that has a leading 1 has zeros in every position
above and below its leading 1.
Exercise 7:
Row reduce the following matrix to reduced echelon form and circle the pivot
position:
1 2 4 8
𝐴= 2 4 6 8
3 6 9 12
Solution:
1 2 4 8
𝐴= 2 4 6 8
3 6 9 12
1 2 4 8
0 0 −2 −8
0 0 −3 −12
1 2 4 8
0 0 1 4
0 0 14
1 2 4 8
0 0 1 4 , then the pivots position are in column 1 & 3
0 0 0 0
Exercise 8:
Determine which matrices are in reduced echelon form and which are only in
echelon form?
1 0 0 0
• 0 1 0 0
0 0 1 1
Solution:
the matrix is in echelon form but and reduced echelon form, because every pivot
at its column has zero above and below it.
Exercise 8:
Determine which matrices are in reduced echelon form and which are only in
echelon form?
1 0 1 0
• 0 1 1 0
0 0 0 1
Solution:
the matrix is in reduced echelon form, because each leading 1 is the only nonzero
entry in its column, and the leading 1’s form is a staircase pattern.
Exercise 8:
Determine which matrices are in reduced echelon form and which are only in
echelon form?
1 0 1 1
• 0 1 1 1
0 0 0 0
Solution:
the matrix is in reduced echelon form.
Exercise 8:
Determine which matrices are in reduced echelon form and which are only in
echelon form?
0 0 0 0
• 1 2 0 0
0 0 1 0
0 0 0 1
Solution:
the matrix is not echelon form, because the first row is zero.
Exercise :
Find the inverse of the following matrix using Elementary row
operations:
1 0 1
𝐴= 2 5 12
−9 1 1
7 1 5
− −
40 40 40
11 1 5
𝐴−1 = − −
4 4 20
47 1 5
−
40 40 40
Exercise :
Row reduce the following matrix to reduced echelon form and circle the pivot
position:
1 2 4 5
𝐴= 2 4 5 4
4 5 4 2
1 0 0 1
ans= 0 1 0 −2
0 0 1 2
Inverse of a Square Matrix
If 𝐴−1 = 𝐴𝑇 , then 𝐴 is called an orthogonal matrix.
Inverse of a Square Matrix
Exercise 1
Solution
Solution
Exercise 2:
Find the inverse of the following matrix:
1 4 1
𝐴= 1 3 2
−1 2 7
Solution:
1 4 1 ⋮ 1 0 0
𝐴⋮𝐼 = 1 3 2 ⋮ 0 1 0
−1 2 7 ⋮ 0 0 1
Using the elementary row
1 4 1 ⋮ 1 0 0
0 1 −1 ⋮ 1 −1 0
0 6 8 ⋮ 1 0 1
1 0 5 ⋮ −3 4 0
0 1 −1 ⋮ 1 −1 0
0 0 14 ⋮ −5 6 1
1 0 5 ⋮ −3 4 0
0 1 −1 ⋮ 1 −1 0
0 0 1 ⋮ −5/14 6/14 1/14
Solution:
1 0 5 ⋮ −3 4 0
0 1 −1 ⋮ 1 −1 0
0 0 1 ⋮ −5/14 6/14 1/14
1 0 0 ⋮ −17/14 26/14 −5/14
0 1 0 ⋮ 9/14 −8/14 1/14
0 0 1 ⋮ −5/14 6/14 1/14
17 26 5
− −
14 14 14
−1 9 8 1
𝐴 = −
14 14 14
5 6 1
−
14 14 14
Exercise 3
Solution
Solution
At this point we would like to change the 0 in the (3,3) position of this matrix to a 1 without
changing the zeros in the (3,1) and (3, 2) positions. But there is no way to accomplish this,
because no matter what multiple of rows 1 and/or 2 we add to row 3, we can’t change the third
zero in row 3 without changing the first or second zero as well.
Thus we conclude that this matrix has no inverse or non-invertible.
Exercise 6:
Find the inverse of the following matrix:
1 5 2
𝐴= 0 −1 2
0 0 1
Solution:
1 5 2 ⋮ 1 0 0
𝐴 ⋮ 𝐼 = 0 −1 2 ⋮ 0 1 0
0 0 1 ⋮ 0 0 1
Using the elementary row
1 5 0 ⋮ 1 0 −2
0 −1 0 ⋮ 0 1 −2
0 0 1 ⋮ 0 0 1
1 0 0 ⋮ 1 5 −12
0 −1 0 ⋮ 0 1 −2
0 0 1 ⋮ 0 0 1
1 0 0 ⋮ 1 5 −12 1 5 −12
0 1 0 ⋮ 0 −1 2 𝐴−1 = 0 −1 2
0 0 1 ⋮ 0 0 1 0 0 1
