- 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.
3. $\quad$ \begin{align*} 3x + 2y + z &= 1 \\ 5x + 3y + 3z &= 2 \\ 7x + 4y + 5z &= 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}{ccc|c} 3 & 2 & 1 & 1 \\ 5 & 3 & 3 & 2 \\ 7 & 4 & 5 & 3 \end{array} \end{bmatrix} \end{align*} First, rearranging the rows of the matrix, we get \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 7 & 4 & 5 & 3 \\ 5 & 3 & 3 & 2 \\ 3 & 2 & 1 & 1 \\ \end{array} \end{bmatrix} \end{align*} Then, we subtract two times the third row from the first row \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 3 & 1 \\ 5 & 3 & 3 & 2 \\ 3 & 2 & 1 & 1 \\ \end{array} \end{bmatrix} \end{align*} We can now use this row to make the other entries of the first column zero \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 3 & 1 \\ 0 & 3 & -12 & -3 \\ 0 & 2 & -8 & -2 \\ \end{array} \end{bmatrix} \end{align*} Now, we can do the same to the second column \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 3 & 1 \\ 0 & 1 & -4 & -1 \\ 0 & 2 & -8 & -2 \\ \end{array} \end{bmatrix} \end{align*} \begin{align*} \begin{bmatrix} \begin{array}{ccc|c} 1 & 0 & 3 & 1 \\ 0 & 1 & -4 & -1 \\ 0 & 0 & 0 & 0 \\ \end{array} \end{bmatrix} \end{align*} As we can see, $x + 3z = 1$ and $y - 4z = -1.$ Rearranging terms, we find that $x = 1 - 3z$ and $y = 4z - 1.$ Then, if we set $z = t,$ where $t$ is an arbitrary scalar, we find that $x = 1 - 3t, y = -1 + 4t,$ and $z = t.$ Put another way, the general solution is given by \begin{align*} (x, y, z) &= (1, -1, 0) + t(-3, 4, 1) \quad \blacksquare \end{align*}