Question

Determine if the columns of the matrix form a linearly independent set. Justify your answer. -2 -1 01 0 - 1 3 1 1 -6 2 1 - 12 Select the correct choice below and fill in the answer box within your choice. (Type an integer or simplified fraction for each matrix element.) O A. If A is the given matrix, then the augmented matrix represents the equation Ax = 0. The reduced echelon form of this matrix indicates that Ax = 0 has more than one solution. Therefore, the columns of A form a linearly independent set. OB. If A is the given matrix, then the augmented matrix represents the equation Ax = 0. The reduced echelon form of this matrix indicates that Ax = 0 has more than one solution. Therefore, the columns of A do not form a linearly independent set. O C. If A is the given matrix, then the augmented matrix represents the equation Ax = 0. The reduced echelon form of this matrix indicates that Ax = 0 has only the trivial solution. Therefore, the columns of A do not form a linearly independent set. OD. If A is the given matrix, then the augmented matrix represents the equation Ax = 0. The reduced echelon form of this matrix indicates that Ax = 0 has only the trivial solution. Therefore, the columns of A form a linearly independent set.

Answer 1

Answer:

Data given:

Given matriz, A= -2 -1 0 -1 1 1 2 1 0 3 -6 -12

We need to find if the columns of the matrix form a linearly independent set.

Now, in order to find if the columns of the matrix form a linearly independent set, we need to solve the equation given by -

Ar = 0

We can write the augmented matrix for the above system as -

Augmented Matriz, M = -2 0 1 2 -1 -1 1 1 0 0 3 0 -6 jo -12 0

Now, let us reduce the above matrix to the reduced row-echelon form as -

1-2 -1 0 -1 M = 0 0 3 0 -6 0 -12 10 2 1

On performing (interchanging row1 and row3), we get -

R1 + R2

1 0 I = |-2 |2 1 -1 -1 1 -6 3 0 -12 0 0 0 0|

On performing , we get -

R3 R3 +2R1

11 -6 10 0 -1 3 M = lo 1 -12 10 2 1 -12 0

On performing , we get -

R4 → R4 – 2R1

1 1 -6_0 10 -1_3_01 AI-101-120 ܬ ܚ ܢ ܗ 10 -1 _0

On performing , we get -

R2+(-1) R2

ſi 0 1 1 -6 -3 10 M = lo 1 -12 10 0 -1 0 0

On performing , we get -

R1 + R1 - R

10 -3 0 0 1 -3 M = 0 1 -12 10 0 -1 0 0

On performing , we get -

R3 R3 - R

1 0 0 1 -3 -3 10 10 M= lo 0-90 0 -1 0 0

On performing , we get -

R + R +R-

Ti 0 0 1 -3 107 -3 jo M = lo 0-90 0 0 -3 0

On performing , we get -

R3 + 7-9) R3

M 1 0 -3 10 0 1 -3 0 = lo 0 1 0 0 0-30

On performing , we get -

R1 + R1 +3R3

M 1 0 0 1 = lo 0 0 0 0 -3 1 -3 0 0 0 0

On performing , we get -

R2 + R2 +3R3

M Ti 0 0 1 = lo 0 0 0 0 0 1 -3 0 0 0 0

On performing ​​​​​​​ , we get -

R4 R4 +3R3

0 0 0 | 이 100 이 0 [ 0] 0 0 0 ] = IN

The above matrix is in the reduced row-echelon form, and the corresponding system is given by -

11 = 0 12 = 0 23 = 0

Thus, the equation has only the trivial solution.

Ar = 0

If A is the given matrix, then the augmented matrix
「-2 -1 0 -1 | 1 1 | 2 1 0 0 3 0 -6 0 -12 lol
represents the equation
Ar = 0
.

The reduced row-echelon form of this matrix indicates that Ax=0 has only the trivial solution.

Therefore, the columns of A form a linearly dependent set.

Hence, the correct option is (D) .