Consider the following system of equations x_1 - x_3 = 2 3x_1 + x_2 + x_3 = 1 -x
ID: 3142980 • Letter: C
Question
Consider the following system of equations x_1 - x_3 = 2 3x_1 + x_2 + x_3 = 1 -x_1 + x_2 + 2x_3 = 3 (a) Write the system of equations as Ax = b, and identify A, x, and b. (b) Use elementary row operations to reduce A to row echelon form. (c) Determine if there is a solution to Ax = 0 other than x = 0. If there is a solution, express it as a linear combination of the columns of A. (d) Determine if the columns of A are linearly independent. (e) Find the rank (A). (f) If possible, solve for x, otherwise explain why it can't be found.Explanation / Answer
Part (a)
The given system of equations can be written in matrix form as Ax = b, where
A = [ 1 0 - 1]
[ 3 1 1]
[- 1 1 2]
x = [x1]
[x2]
[x3]
and b = [2]
[1]
[3]
ANSWER
Part (b)
Echelon Form of A is shown in bold in the last 3 rows of the table below.
The other rows show the steps.
A
I
Elementary operations
1
0
- 1
1
0
0
(1)
3
1
1
0
1
0
(2)
- 1
1
2
0
0
1
(3)
1
0
- 1
1
0
0
(4) = (1)
0
1
4
- 3
1
0
(5) = (2) - 3(4)
0
1
1
1
0
1
(6) = (3) + (4)
1
0
- 1
1
0
0
(7) = (4)
0
1
4
- 3
1
0
(8) = (5)
0
0
3
- 4
1
- 1
(9) = (5) - (6)
1
0
0
- 1/3
1/3
- 1/3
(10) = (7) + (12)
0
1
0
7/3
- 1/3
4/3
(11) = (8) - 4(12)
0
0
1
- 4/3
1/3
- 1/3
(12) = (9)/3
ANSWER
Part (c)
Since A has 3 columns and the Echelon form of A has 3 columns none of which is a zero
column, the columns of A are linearly independent. ANSWER
Part (d)
Since A has 3 columns and the columns of A are linearly independent,
rank of A = 3 ANSWER
Part (e)
The Echelon form of A represents A- 1, the inverse of A.
The solution of Ax = b is obtained by pre-multiplying b by A- 1.
By actual pre-multiplication, x1 = - 4/3, x2 = - 25/3 and x3 = - 10/3 ANSWER
A
I
Elementary operations
1
0
- 1
1
0
0
(1)
3
1
1
0
1
0
(2)
- 1
1
2
0
0
1
(3)
1
0
- 1
1
0
0
(4) = (1)
0
1
4
- 3
1
0
(5) = (2) - 3(4)
0
1
1
1
0
1
(6) = (3) + (4)
1
0
- 1
1
0
0
(7) = (4)
0
1
4
- 3
1
0
(8) = (5)
0
0
3
- 4
1
- 1
(9) = (5) - (6)
1
0
0
- 1/3
1/3
- 1/3
(10) = (7) + (12)
0
1
0
7/3
- 1/3
4/3
(11) = (8) - 4(12)
0
0
1
- 4/3
1/3
- 1/3
(12) = (9)/3