- 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.
8. $\quad$ \begin{align*} x - 2y + z + 2u &= -2 \\ 2x + 3y - z - 5u &= 9 \\ 4x - y + z - u &= 5 \\ 5x - 3y + 2z + u &= 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}{cccc|c} 1 & -2 & 1 & 2 & -2 \\ 2 & 3 & -1 & -5 & 9 \\ 4 & -1 & 1 & -1 & 5 \\ 5 & -3 & 2 & 1 & 3 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & -2 & 1 & 2 & -2 \\ 0 & 7 & -3 & -9 & 13 \\ 0 & 7 & -3 & -9 & 13 \\ 0 & 7 & -3 & -9 & 13 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & -2 & 1 & 2 & -2 \\ 0 & 7 & -3 & -9 & 13 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 3 & -6 & 3 & 6 & -6 \\ 0 & 7 & -3 & -9 & 13 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 3 & 1 & 0 & -3 & 7 \\ 0 & 7 & -3 & -9 & 13 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} Then, we have the equations \begin{align*} 3x &= 7 + 3u - y \\ 3z &= 7y - 9u - 13 \end{align*} Then, if we set $y = 3t_1$ for arbitrary $t_1$ and $u = t_2$ for arbitrary $t_2,$ the solution takes the form \begin{align*} (x, y, z, u) &= \left(\frac{7}{3} + t_2 - t_1, 3t_1, 7t_1 - 9t_2 - \frac{13}{3}, t_2\right) \\ &= \left(\frac{7}{3}, 0, -\frac{13}{3}, 0\right) + t_1(-1, 3, 7, 0) + t_2(-1, 0, -9, 1) \end{align*} If instead, we zero out the remaining element of the second column, we get the following: \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 7 & -14 & 7 & 14 & -14 \\ 0 & 14 & -6 & -18 & 26 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 7 & 0 & 1 & -4 & 12 \\ 0 & 7 & -3 & -9 & 13 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} Then, taking $x$ and $y$ in terms of $z$ and $u$ \begin{align*} 7x &= 4u - z + 12 \\ 7y &= 9u + 3z + 13 \end{align*} If we set $z = 7t_1, u = 7t_2$ for arbitrary $t_1, t_2,$ we have \begin{align*} (x, y, z, u) &= (4t_2 - t_1 + \frac{12}{7}, 9t_2 + 3t_1 + \frac{13}{7}, 7t_1, 7t_2) \\ &= \left(\frac{12}{7}, \frac{13}{7}, 0, 0\right) + t_1(-1, 3, 7, 0) + t_2(4, 9, 0, 7) \end{align*} If we set $x$ and $y$ to $1,$ we get the solution going through the point $(1, 1, 1, -1)$ giving us the back-of-book solution: \begin{align*} (x, y, z, u) &= (4t_2 - t_1 + \frac{12}{7}, 9t_2 + 3t_1 + \frac{13}{7}, 7t_1, 7t_2) \\ &= \left(1, 1, 1, -1\right) + t_1(-1, 3, 7, 0) + t_2(4, 9, 0, 7) \quad \blacksquare \end{align*}