- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
2.20 Exercises
Apply the Gauss-Jordan elimination process to each of the following systems. If a solution exists, determine the general solution.
7. $\quad$ \begin{align*} x + y + 2z + 3u + 4v &= 0 \\ 2x + 2y + 7z + 11u + 14v &= 0 \\ 3x + 3y + 6z + 10u + 15v &= 0. \end{align*}
Solution. $\quad$ We recall from Section 18 that the Gauss-Jordan elimination method consists of applying three basic types of operations on a linear system:
$\quad$ (1) Interchanging two equations;
$\quad$ (2) Multiplying all the terms of an equation by a nonzero scalar;
$\quad$ (3) Adding one equation to a multiple of another.
Each time we perform one of these operations on the system, we obtain a new system having exactly the same solutions. Two such systems are called equivalent.
$\quad$ Additionally, we can take the coefficients and values of the system of linear equations and place them in an augmented matrix for ease of computation. This gives us the following: \begin{align*} \begin{bmatrix} \begin{array}{ccccc|c} 1 & 1 & 2 & 3 & 4 & 0 \\ 2 & 2 & 7 & 11 & 14 & 0 \\ 3 & 3 & 6 & 10 & 15 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccccc|c} 1 & 1 & 2 & 3 & 4 & 0 \\ 0 & 0 & 3 & 5 & 6 & 0 \\ 0 & 0 & 0 & 1 & 3 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccccc|c} 1 & 1 & 2 & 3 & 4 & 0 \\ 0 & 0 & 3 & 0 & -9 & 0 \\ 0 & 0 & 0 & 1 & 3 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccccc|c} 1 & 1 & 2 & 3 & 4 & 0 \\ 0 & 0 & 1 & 0 & -3 & 0 \\ 0 & 0 & 0 & 1 & 3 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccccc|c} 1 & 1 & 0 & 3 & 10 & 0 \\ 0 & 0 & 1 & 0 & -3 & 0 \\ 0 & 0 & 0 & 1 & 3 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccccc|c} 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & -3 & 0 \\ 0 & 0 & 0 & 1 & 3 & 0 \end{array} \end{bmatrix} \end{align*} From this we have the following relations \begin{align*} v &= -x - y \\ z &= 3v \\ u &= -3v \end{align*} But since we have three equations in five unknowns, we have two arbitrary variables. If we set $y = t_1$ and $v = t_2,$ where $t_1$ and $t_2$ are arbitrary scalars, we get the following set of solutions: \begin{align*} (x, y, z, u, v) &= (-t_1 - t_2, t_1, 3t_2, -3t_2, t_2) \\ &= t_1(-1, 1, 0, 0, 0) + t_2(-1, 0, 3, -3, 1) \quad \blacksquare \end{align*}