Let a_1 = [1 2 4], a_2 = [1 3 5]. (a) Explain if the following vector b_1 = [2 5
ID: 3038847 • Letter: L
Question
Let a_1 = [1 2 4], a_2 = [1 3 5]. (a) Explain if the following vector b_1 = [2 5 9] is in the plane spanned by a_1, a_2? (b) Explain if the following vector b_2 = [1 2 6] is in the plane spanned by a_1, a_2? Given the augmented matrix of a system of equation [1 2 -1 1 2 4 -2 3 3 6 -3 7 4 8 -4 2 5 10 -5 4] Solve the system of equation. Indicate each step of the necessary elementary row operations. Indicate any pivot column. Indicate any basic and/or free variable. Write the solution set. Write the reduced row echelon form of the matrix. Indicate if no solution, unique solution or infinite solution.Explanation / Answer
2) a) Let A = [a1,a1,b1,b2] =
1
1
2
1
2
3
5
2
4
5
9
6
In order to determine whether b1 and b2 can be expressed as a linear combination of a1, a2 or not, i.e. whether b1 and b2 are in span {a1,a2} or, not, we will reduce A to its RREF as under:
Add -2 times the 1st row to the 2nd row
Add -4 times the 1st row to the 3rd row
Add -1 times the 2nd row to the 3rd row
Multiply the 3rd row by 1/2
Add -1 times the 3rd row to the 1st row
Add -1 times the 2nd row to the 1st row
Then the RREF of A is
1
0
1
0
0
1
1
0
0
0
0
1
Now, it is apparent that b1 = a1+a2 so that b1 is in span {a1,a2}.
b) It is also apparent that b2 cannot be expressed as a linear combination of a1, a2. Hence b2 is not in span {a1,a2}.
3) Let the given matrix be denoted by B = [A|b] , where b = (5,10,-5,4)T and the first 4 columns of B comprise A, the coefficient matrix of the system.Then the given system of linear equations can be expressed as AX = b. To solve this system of linear equations, we will reduce B to its RREF as under:
Add -2 times the 1st row to the 2nd row
Add 1 times the 1st row to the 3rd row
Add -1 times the 1st row to the 4th row
Interchange the 2nd row and the 4th row
Add -2 times the 2nd row to the 1st row
Then the RREF of B is
1
0
-5
8
7
0
1
4
-2
-1
0
0
0
0
0
0
0
0
0
0
Now, if X = (x,y,z,w)T, then the given system of linear equations is equivalent to x -5z+8w= 7 or, x = 7+5z-8w and y +4z-2w = -1 or, y = -1-4z+2w so that X = (7+5z-8w, -1-4z+2w, z,w)T = (7,-1,0,0)T+z(5,-4,1,0)T+ w( -8,2,0,1)T. Thus z and w are the basic variables and x and y are the free variables.We know that a pivot position in a matrix is a position in the matrix whichcorresponds to a row–leading 1 in the RREF of the matrix. Here the 1 in such position in the 1st and the 2nd column indicate pivot columns. The solution set is (x,y,z,w)T = (7,-1,0,0)T+r(5,-4,1,0)T+ t( -8,2,0,1)T where r,t are arbitrary real numbers.Thus, there are infinite solutions.
The RREF of the given augmented matrix is mentioned above as RREF of B.
1
1
2
1
2
3
5
2
4
5
9
6
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