Answer to Question #95817 in Geometry for Max

Question

Show that the vectors a = 3i − j + 4k, b = i − 3j − 2k and c = 4i − 3j + 2k are
linearly independent. Find numbers α, β and γ such that d = 2i + 3j − k can be
expressed in the form d = αa + βb + γc.

Answer 1

Solution: Firstly, show that these vectors are linearly independent. Write our vectors vertical in matrix. Use Gaussian elimination for linearly independent vectors:$\begin{pmatrix} 3 & 1 & 4 \\ -1 & -1 & -3\\ 4 & -2& 2 \end{pmatrix}$

Divide the first row by 3:

$\begin{pmatrix} 1 & 1/3 & 4/3 \\ -1 & -1 & -3\\ 4 & -2& 2 \end{pmatrix}$

Substitute the second and third rows from the first multiplied by -1 and 4:

$\begin{pmatrix} 1 & 1/3 & 4/3 \\ 0 & -8/3 & -5/3\\ 0 & -10/3 & -10/3 \end{pmatrix}$

Divide the second row by 8/3:

$\begin{pmatrix} 1 & 1/3 & 4/3 \\ 0 & -1 & -5/8\\ 0 & -10/3 & -10/3 \end{pmatrix}$

Substitute the third row from second multiplied by 10/3:

$\begin{pmatrix} 1 & 1/3 & 4/3 \\ 0 & -1 & -5/8\\ 0 & 0 & -5/4 \end{pmatrix}$

So we have 3 rows without zero (0) in all three places. So we have linearly independent vectors.

Next task is to find unknown numbers α, β and γ for another vector. Write a supplemented matrix (from previous one) with numbers of another vector.

$\begin{pmatrix} 3 & 1 & 4 && 2 \\ -1 & -1 & -3 && 3\\ 4 & -2& 2 && -1 \end{pmatrix}$

Divide the first row by 3:

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ -1 & -1 & -3 && 3\\ 4 & -2& 2 && -1 \end{pmatrix}$

Substitute the second and the third rows from the first multiplied by 1 and -4

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ 0 & -8/3 & -5/3 && 11/3\\ 0 & -10/3& -10/3 && -11/3 \end{pmatrix}$

Divide the second row by 8/3

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ 0 & -1 & -5/8 && 11/8\\ 0 & -10/3& -10/3 && -11/3 \end{pmatrix}$

Substitute the third row from the second multiplied by 10/3:

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ 0 & -1 & -5/8 && 11/8\\ 0 & 0& -5/4 && -33/4 \end{pmatrix}$

Use Gaussian elimination in another side - from top to bottom and find numbers in the last column when we would have the identity matrix.

Divide the third row by -5/4

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ 0 & -1 & -5/8 && 11/8\\ 0 & 0& 1 && 33/5 \end{pmatrix}$

Divide the second row by -1:

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ 0 & 1 & 5/8 && -11/8\\ 0 & 0& 1 && 33/5 \end{pmatrix}$

Add the third row to second multiplied by -5/8

$\begin{pmatrix} 1 & 1/3 & 4/3 && 2/3 \\ 0 & 1 & 0 && -11/2\\ 0 & 0& 1 && 33/5 \end{pmatrix}$

Add the second row to first multiplied by -1/8

$\begin{pmatrix} 1 & 0 & 4/3 && 5/2 \\ 0 & 1 & 0 && -11/2\\ 0 & 0& 1 && 33/5 \end{pmatrix}$

Add the third row to first multiplied by -4/3

$\begin{pmatrix} 1 & 0 & 0 && -63/10 \\ 0 & 1 & 0 && -11/2\\ 0 & 0& 1 && 33/5 \end{pmatrix}$

We have the identity matrix and have numbers of our vector. Rewrite the system:

α = - 63/10; β = -11/2; γ = 33/5;

or

α = - 6,3; β = -5,5; γ = 6,6;

Answer: Yes, these vectors are linearly independent. Numbers of the last vector are

α = - 6,3; β = -5,5; γ = 6,6