- 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.
5. $\quad$ \begin{align*} 3x - 2y + 5z &= 1 \\ x + y - 3z &= 2 \\ 6x + y - 4z &= 7 \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}{cccc|c} 3 & -2 & 5 & 1 & 1 \\ 1 & 1 & -3 & 2 & 2\\ 6 & 1 & -4 & 3 & 7 \end{array} \end{bmatrix} \end{align*} First, we rearrange the augmented matrix to place the equation $x + y - 3z + 2u = 2$ in the first row \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 1 & -3 & 2 & 2\\ 3 & -2 & 5 & 1 & 1 \\ 6 & 1 & -4 & 3 & 7 \end{array} \end{bmatrix} \end{align*} Then, we can use this row to make the remaining elements of the first column zero. \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 1 & -3 & 2 & 2\\ 0 & -5 & 14 & -5 & -5 \\ 0 & -5 & 14 & -9 & -5 \end{array} \end{bmatrix} \end{align*} As we can see, subtracting the second row from the third gives us \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 1 & -3 & 2 & 2\\ 0 & -5 & 14 & -5 & -5 \\ 0 & 0 & 0 & 4 & 0 \end{array} \end{bmatrix} \end{align*} From which we deduce that $u = 0.$ Then, dividing the second row by $-5$ and subtracting it from the first gives us \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 0 & -\frac{1}{5} & 1 & 1\\ 0 & 1 & -\frac{14}{5} & 1 & 1 \\ 0 & 0 & 0 & 4 & 0 \end{array} \end{bmatrix} \end{align*} Then, recalling that $u = 0,$ we have $x - \frac{1}{5}z = 1$ and $y - \frac{14}{5}z = 1,$ or in other words \begin{align*} x &= 1 + \frac{1}{5}z \\ y &= 1 + \frac{14}{5}z \end{align*} This gives us the following \begin{align*} (x, y, z, u) &= \left(1 + \frac{1}{5}z, 1 + \frac{14}{5}z, z, 0\right) \end{align*} But in this case $z$ is an arbitrary scalar. Thus, if we set $z = 5t$ with arbitrary $t,$ we have \begin{align*} (x, y, z, u) &= (1, 1, 0, 0) + t(1, 14, 5, 0) \quad \blacksquare \end{align*}