- 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.
1. $\quad$ \begin{align*} x + y + 3z &= 5 \\ 2x - y + 4z &= 11 \\ -y + z &= 3. \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} 1 & 1 & 3 & 5 \\ 2 & -1 & 4 & 11 \\ 0 & -1 & 1 & 3 \end{array} \end{bmatrix} \end{align*} Adding the third equation to the first \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 4 & 8 \\ 2 & -1 & 4 & 11 \\ 0 & -1 & 1 & 3 \end{array} \end{bmatrix} \end{align*} Subtracting the first equation from the second \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 4 & 8 \\ 1 & -1 & 0 & 3 \\ 0 & -1 & 1 & 3 \end{array} \end{bmatrix} \end{align*} Subtracting the second equation from the first \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 0 & 1 & 4 & 5 \\ 1 & -1 & 0 & 3 \\ 0 & -1 & 1 & 3 \end{array} \end{bmatrix} \end{align*} Adding the first equation to the third and dividing the result by $5$ \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 0 & 1 & 4 & 5 \\ 1 & -1 & 0 & 3 \\ 0 & 0 & 1 & \frac{8}{5} \end{array} \end{bmatrix} \end{align*} Subtracting four times the third equation from the first \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 0 & 1 & 0 & \frac{-7}{5} \\ 1 & -1 & 0 & 3 \\ 0 & 0 & 1 & \frac{8}{5} \end{array} \end{bmatrix} \end{align*} Finally, adding the first equation to the second and rearranging rows gives us \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 0 & \frac{8}{5} \\ 0 & 1 & 0 & \frac{-7}{5} \\ 0 & 0 & 1 & \frac{8}{5} \end{array} \end{bmatrix} \end{align*} This gives us the solution to the linear system: \begin{align*} x &= \frac{8}{5}, \quad y = \frac{-7}{5}, \quad z = \frac{8}{5} \quad \blacksquare \end{align*}