- 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.
2. $\quad$ \begin{align*} 3x + 2y + z &= 1 \\ 5x + 3y + 3z &= 2 \\ x + y - z &= 1. \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}{ccc|c} 3 & 2 & 1 & 1 \\ 5 & 3 & 3 & 2 \\ 1 & 1 & -1 & 1 \end{array} \end{bmatrix} \end{align*} Adding the third row to the first \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 4 & 3 & 0 & 2 \\ 5 & 3 & 3 & 2 \\ 1 & 1 & -1 & 1 \end{array} \end{bmatrix} \end{align*} Subtracting the first row from the second \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 4 & 3 & 0 & 2 \\ 1 & 0 & 3 & 0 \\ 1 & 1 & -1 & 1 \end{array} \end{bmatrix} \end{align*} Subtracting the second equation from the third \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 4 & 3 & 0 & 2 \\ 1 & 0 & 3 & 0 \\ 0 & 1 & -4 & 1 \end{array} \end{bmatrix} \end{align*} Subtracting three times the third equation from the first \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 4 & 0 & 12 & -1 \\ 1 & 0 & 3 & 0 \\ 0 & 1 & -4 & 1 \end{array} \end{bmatrix} \end{align*} But as we can see, if we subtract four times the second equation from the first equation, we see that \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 0 & 0 & 0 & -1 \\ 1 & 0 & 3 & 0 \\ 0 & 1 & -4 & 1 \end{array} \end{bmatrix} \end{align*} Or in other words, that $0 = -1,$ giving a contradiction. Thus, the system has no solution.