Question

. Find the solution of the linear system using Gauss-Jordan elimination

𝑣 + 2𝑤 − 2𝑥 − 𝑦 = 0

2𝑣 + 3𝑤 − 5𝑥 + 𝑦 − 7𝑧 = 0

𝑤 + 𝑥 − 𝑦 + 𝑧 = 0

−𝑣 + 𝑤 + 5𝑥 − 𝑦 = 0

Answer 1

Solution.

$𝑣 + 2𝑤 − 2𝑥 − 𝑦 = 0$

$2𝑣 + 3𝑤 − 5𝑥 + 𝑦 − 7𝑧 = 0$

$𝑤 + 𝑥 − 𝑦 + 𝑧 = 0$

$−𝑣 + 𝑤 + 5𝑥 − 𝑦 = 0$

$\begin{bmatrix} 1 & 2&-2&-1&0&0 \\ 2 & 3&-5&1&-7&0\\ 0&1&1&-1&1&0\\ -1&1&5&-1&0&0 \end{bmatrix}$

Find the pivot in the 1st column in the 1st row

$\begin{bmatrix} 1 & 2&-2&-1&0&0 \\ 2 & 3&-5&1&-7&0\\ 0&1&1&-1&1&0\\ -1&1&5&-1&0&0 \end{bmatrix}$

Eliminate the 1st column

$\begin{bmatrix} 1 & 2&-2&-1&0&0 \\ 0 & -1&-1&3&-7&0\\ 0&1&1&-1&1&0\\ 0&3&3&-2&0&0 \end{bmatrix}$

Find the pivot in the 2nd column in the 2nd row (inversing the sign in the whole row)

$\begin{bmatrix} 1 & 2&-2&-1&0&0 \\ 0 & 1&1&-3&7&0\\ 0&1&1&-1&1&0\\ 0&3&3&-2&0&0 \end{bmatrix}$

Eliminate the 2nd column

$\begin{bmatrix} 1 & 0&-4&5&-14&0 \\ 0 & 1&1&-3&7&0\\ 0&0&0&2&-6&0\\ 0&0&0&7&-21&0 \end{bmatrix}$

Make the pivot in the 4th column by dividing the 3rd row by 2

$\begin{bmatrix} 1 & 0&-4&5&-14&0 \\ 0 & 1&1&-3&7&0\\ 0&0&0&1&-3&0\\ 0&0&0&7&-21&0 \end{bmatrix}$

Eliminate the 4th column

$\begin{bmatrix} 1 & 0&-4&0&1&0 \\ 0 & 1&1&0&-2&0\\ 0&0&0&1&-3&0\\ 0&0&0&0&0&0 \end{bmatrix}$

Solution set:

$v = 4x - z; w = - x + 2z; y = 3z; x, z - free.$

Answer: $v = 4x - z; w = - x + 2z; y = 3z; x, z - free.$