- 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
10. $\quad$(a) $\quad$ Determine all the solutions of the system \begin{align*} 5x + 2y - 6z + 2u &= -1 \\ x - y + z - u &= -2 \end{align*} (b) $\quad$ Detemine all solutions of the system \begin{align*} 5x + 2y - 6z + 2u &= -1 \\ x - y + z - u &= -2 \\ x + y + z &= 6 \end{align*}
Solution. $\quad$ (a) $\quad$ The system \begin{align*} 5x + 2y - 6z + 2u &= -1 \\ x - y + z - u &= -2 \end{align*} gives us the augmented matrix \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 5 & 2 & -6 & 2 & -1 \\ 1 & -1 & 1 & -1 & -2 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 0 & 7 & -11 & 7 & 9 \\ 1 & -1 & 1 & -1 & -2 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 0 & 1 & -\frac{11}{7} & 1 & \frac{9}{7} \\ 1 & 0 & -\frac{4}{7} & 0 & -\frac{5}{7} \end{array} \end{bmatrix} \end{align*} This gives us $x$ and $y$ in terms of $z$ and $u$ as follows \begin{align*} x &= \frac{1}{7}(4z - 5) \qquad y = \frac{1}{7}(11z - 7u + 9) \end{align*} Setting $z = 7t_1 + 3$ and $u = t_2$ for arbitrary $t_1$ and $t_2,$ we have \begin{align*} (x, y, z, u) &= (4t_1 + 1, 11t_1 - t_2 + 6, 7t_1 + 3, t_2) \\ &= ( 1, 6, 3, 0) + t_1(4, 11, 7, 0) + t_2(0, -1, 0, 1) \end{align*} Note: The back-of-book solution contains a typo in the coefficient of $t_2$ for $y,$ stating $y = 6 + 3t_1 + 0t_2.$
(b) $\quad$ The system \begin{align*} 5x + 2y - 6z + 2u &= -1 \\ x - y + z - u &= -2 \\ x + y + z &= 6 \end{align*} gives the augmented matrix \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 5 & 2 & -6 & 2 & -1 \\ 1 & -1 & 1 & -1 & -2 \\ 1 & 1 & 1 & 0 & 6 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 7 & 0 & -4 & 0 & -5 \\ 1 & -1 & 1 & -1 & -2 \\ 1 & 1 & 1 & 0 & 6 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 7 & 0 & -4 & 0 & -5 \\ 1 & -1 & 1 & -1 & -2 \\ 0 & 2 & 0 & 1 & 8 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 11 & 0 & 0 & -2 & 3 \\ 0 & 1 & 0 & \frac{1}{2} & 4 \\ 0 & 0 & 11 & -\frac{7}{2} & 19 \\ \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 0 & 0 & -\frac{2}{11} & \frac{3}{11} \\ 0 & 1 & 0 & \frac{1}{2} & 4 \\ 0 & 0 & 1 & -\frac{7}{22} & \frac{19}{11} \\ \end{array} \end{bmatrix} \end{align*} From this, we get the equivalent system of equations in terms of the variable $u:$ \begin{align*} x &= \frac{1}{11}(3 + 2u) \\ y &= 4 - \frac{1}{2}u \\ z &= \frac{1}{22}(38 + 7u) \end{align*} If we set $u = 22t,$ for arbitrary $t,$ we get the following general solution \begin{align*} (x, y, z, u) &= \left(\frac{3}{11}, 4, \frac{19}{11}, 0 \right) + t(4, -11, 7, 22) \quad \blacksquare \end{align*}