- 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.
6. $\quad$ \begin{align*} x + y - 3z + u &= 5 \\ 2x - y + z - 2u &= 2 \\ 7x + y - 7z + 3u &= 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 & 1 & -3 & 1 & 5 \\ 2 & -1 & 1 & -2 & 2 \\ 7 & 1 & -7 & 3 & 3 \end{array} \end{bmatrix} \end{align*} First, we use the first row to make the remaining elements of the first column zero, and then divide the third row by $-2$ \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 1 & -3 & 1 & 5 \\ 0 & -3 & 7 & -4 & -8 \\ 0 & 3 & -7 & 2 & 16 \end{array} \end{bmatrix} \end{align*} Now, we use the second element of the second row to make the remaining elements of the second column zero \begin{align*} \begin{bmatrix} \begin{array}{cccc|c} 1 & 1 & -3 & 1 & 5 \\ 0 & -3 & 7 & -4 & -8 \\ 0 & 0 & 0 & -2 & 8 \end{array} \end{bmatrix} \end{align*} From this, we find that $u = -4.$ Substituting this into the remaining equations and dividing the second row by $-3$ gives us the updated matrix \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 1 & -3 & 9 \\ 0 & 1 & \frac{-7}{3} & 8 \\ 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} Subtracting the second row from the first \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & \frac{-2}{3} & 1 \\ 0 & 1 & \frac{-7}{3} & 8 \\ 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} As we can see, $x - \frac{2}{3}z = 1$ and $y - \frac{7}{3}z = 8.$ Combining these results, with $u = -4,$ we have \begin{align*} (x, y, z, u) &= \left(1 + \frac{2}{3}z, 8 + \frac{7}{3}z, z, -4\right) \end{align*} But in this relation, the value of $z$ is arbitrary. Thus, if we set $z = 3t$ for arbitrary $t,$ we have \begin{align*} (x, y, z, u) &= (1, 8, 0, -4) + t(2, 7, 3, 0) \quad \blacksquare \end{align*}