- 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.
4. $\quad$ \begin{align*} 3x + 2y + z &= 1 \\ 5x + 3y + 3z &= 2 \\ 7x + 4y + 5z &= 3 \\ x + y - z &= 0. \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}{ccc|c} 3 & 2 & 1 & 1 \\ 5 & 3 & 3 & 2 \\ 7 & 4 & 5 & 3 \\ 1 & 1 & -1 & 0 \end{array} \end{bmatrix} \end{align*} First, we rearrange the augmented matrix to place the equation $x + y - z = 0$ in the first row \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 1 & -1 & 0 \\ 3 & 2 & 1 & 1 \\ 5 & 3 & 3 & 2 \\ 7 & 4 & 5 & 3 \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}{ccc|c} 1 & 1 & -1 & 0 \\ 0 & -1 & 4 & 1 \\ 0 & -2 & 8 & 2 \\ 0 & -3 & 12 & 3 \end{array} \end{bmatrix} \end{align*} Then, we can use the $-1$ in row $2$ column $2$ to make the remaining elements of the second column zero, multiplying the same row by $-1$ to give us \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 3 & 1 \\ 0 & 1 & -4 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \end{bmatrix} \end{align*} But as we saw in Exercise 3, this system of equations is satisfied by \begin{align*} (x, y, z) &= (1, -1, 0) + t(-3, 4, 1) \quad \blacksquare \end{align*}