Question

Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (column) vectors in ℝ4 and let A be the 4 ✕ 4 matrix v1 v2 v3 v4 with these vectors as its columns. Then v1, v2, v3, and v4 are linearly dependent if and only if the homogeneous linear system with augmented matrix [A|0] has a nontrivial solution. Consider the following equation. c1 0 0 0 1 + c2 0 0 3 1 + c3 0 4 3 1 + c4 8 4 3 1 = 0 0 0 0 Solve for c1, c2, c3, and c4. If a nontrivial solution exists, state it or state the general solution in terms of the parameter t. (If only the trivial solution exists, enter the trivial solution {c1, c2, c3, c4} = {0, 0, 0, 0}.)

Answer 1

The system of linear equations obtained from the vector combination has only the trivial solution c1 = c2 = c3 = c4 = 0. Therefore, the vectors v1, v2, v3, and v4 are linearly independent as no nontrivial solution exists for these vectors.

Step-by-step explanation:

To determine whether the vectors v1, v2, v3, and v4 are linearly dependent, we need to solve the homogeneous system of equations formed by their combinations with scalars c1, c2, c3, and c4 set equal to a zero vector:

c1(0,0,0,1) + c2(0,0,3,1) + c3(0,4,3,1) + c4(8,4,3,1) = (0,0,0,0)

This gives rise to the following system of linear equations:

• 0*c1 + 0*c2 + 0*c3 + 8*c4 = 0
• 0*c1 + 0*c2 + 4*c3 + 4*c4 = 0
• 0*c1 + 3*c2 + 3*c3 + 3*c4 = 0
• 1*c1 + 1*c2 + 1*c3 + 1*c4 = 0

This system simplifies to the following equations:

• 8*c4 = 0
• 4*c3 + 4*c4 = 0
• 3*c2 + 3*c3 + 3*c4 = 0
• c1 + c2 + c3 + c4 = 0

From the first equation, we immediately get c4 = 0. Substituting c4 = 0 into the second equation, we find that c3 = 0. With c3 = 0, the third equation tells us that c2 = 0 as well. Finally, with c2 = 0, the last equation simplifies to c1 = 0.

The system of equations has only the trivial solution {c1, c2, c3, c4} = {0, 0, 0, 0}, indicating that no nontrivial solution exists, and thus the vectors v1, v2, v3, and v4 are linearly independent.

Answer 2

We have the equation

`c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]`

Then, the augmented matrix of the system is

`\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]`

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

`\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]`

This matrix is in echelon form. Then, now we apply backward substitution:

1.

`8c_4=0\\c_4=0`

2.

`4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0`

3.

`3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0`

4.

`c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0`

Then the system has unique solution that is
`(c_1,c_2c_3,c_4)=(0,0,0,0)` and this imply that the vectors
`v_1,v_2,v_3,v_4` are linear independent.