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

Answer:

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.