- 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
9. $\quad$ Prove that the system \begin{align*} x + y + 2z &= 2 \\ 2x - y + 3z &= 2 \\ 5x - y + az &= 6 \end{align*} has a unique solution if $a \neq 8.$ Find all solutions when $a = 8.$
Proof. $\quad$ The above system of equations gives the following augmented matrix \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 2 & -1 & 3 & 2 \\ 5 & -1 & a & 6 \end{array} \end{bmatrix} \end{align*} Which can be modified as follows \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 0 & -3 & -1 & -2 \\ 0 & -6 & (a - 10) & -4 \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 3 & 0 & 5 & 4 \\ 0 & -3 & -1 & -2 \\ 0 & 0 & (8 - a) & 0 \end{array} \end{bmatrix} \end{align*} As we can see, if $a \neq 8,$ we can add $-5/(8 - a)$ times the third row to the first and $1/(8 - a)$ times the third row to the second to get \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 3 & 0 & 0 & 4 \\ 0 & -3 & 0 & -2 \\ 0 & 0 & (8 - a) & 0 \end{array} \end{bmatrix} \end{align*} This gives us the following system of equations \begin{align*} 3x &= 4, \quad 3y = 2, \quad (8 - a)z = 0 \end{align*} which has the unique solution $(x, y, z) = \left(\frac{4}{3}, \frac{2}{3}, 0\right).$
$\quad$ If $a = 8,$ then the previously computed augmented matrix is as follows \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 3 & 0 & 5 & 4 \\ 0 & 3 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} corresponding to the system of equations \begin{align*} 3x + 5z &= 4 \\ 3y + z &= 2 \end{align*} If we set $z = 3t$ for arbitrary $t,$ we have the set of solutions \begin{align*} (x, y, z) &= \left(\frac{4}{3}, \frac{2}{3}, 0\right) + t \left(-5, -1, 3\right) \end{align*} (Note: since $t$ is arbitrary, we can equivalently set $z = -3t$ to give the back-of-book solution) \begin{align*} (x, y, z) &= \left(\frac{4}{3}, \frac{2}{3}, 0\right) + t \left(5, 1, -3\right) \quad \blacksquare \end{align*}