Problem 9 - College Algebra Essentials

Question

Solve each system. $$\left\\{\begin{array}{l} 3 x+2 y-3 z=-2 \\ 2 x-5 y+2 z=-2 \\ 4 x-3 y+4 z=10 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=-32/3\), \(y=28/3\), and \(z=7\).

Step 1: Write the system as an augmented matrix

The system of equations can be written as an augmented matrix: \[ \left[ \begin{array}{ccc|c} 3 & 2 & -3 & -2 \\ 2 & -5 & 2 & -2 \\ 4 & -3 & 4 & 10 \end{array} \right] \]. Each column corresponds to the coefficients of the variables \(x\), \(y\), \(z\) respectively, and the column after the line corresponds to the constants on the right side of the equations.

Step 2: Go to Row-Echelon form using Gaussian Elimination

Divide the first row by 3 to make the first leading coefficient a 1. Multiply the first row by -2 and add to the second row to eliminate \(x\) from the second equation. Also, multiply the first row by -4 and add to the third row to eliminate \(x\) from the third equation. The augmented matrix is now: \[\left[\begin{array}{ccc|c} 1 & 2/3 & -1 & -2/3 \\ 0 & -9/3 & 4 & 0 \\ 0 & -5 & 8 & 14 \end{array}\right]\]. Next, divide the second row by -3, then multiply it by 5 and add to the third row. Resultant matrix will be: \[\left[\begin{array}{ccc|c} 1 & 2/3 & -1 & -2/3 \\ 0 & 1 & -4/3 & 0 \\ 0 & 0 & 2 & 14 \end{array}\right]\]. Lastly, to get \(z\), divide the last row by 2, the matrix becomes: \[\left[\begin{array}{ccc|c} 1 & 2/3 & -1 & -2/3 \\ 0 & 1 & -4/3 & 0 \\ 0 & 0 & 1 & 7 \end{array}\right]\]

Step 3: Solve for \(z\), \(y\), and \(x\) by back substitution

The system of equations corresponding to the final matrix is: \[\left\{\begin{array}{l} x+2/3y-z=-2/3 \\ y-4/3z=0 \\ z=7 \end{array}\right.\] From equation (3), we get \(z=7\). Substituting \(z=7\) into equation (2), we get \(y=28/3\). Substituting these values into equation (1), we get \(x=-2/3-2(28/3) + 7=-32/3\).