Problem 11 - Intermediate Algebra

Question

Solve each of the following systems. If the solution set is \(\varnothing\) or if it contains infinitely many solutions, then so indicate. $$ \left(\begin{array}{rr} x-2 y+z= & -4 \\ 2 x+4 y-3 z= & -1 \\ -3 x-6 y+7 z= & 4 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{49}{20}\), \(y = \frac{3}{8}\), \(z = -\frac{4}{5}\).

Step 1: Write the Augmented Matrix

Start by writing the given system of equations in the form of an augmented matrix. This looks like: \[\left(\begin{array}{ccc|c}1 & -2 & 1 & -4 \2 & 4 & -3 & -1 \-3 & -6 & 7 & 4\end{array}\right)\]

Step 2: Perform Row Operations to Obtain Row Echelon Form

The goal is to simplify the matrix to row echelon form. Start by making the first element of the second row zero using the first row. Subtract 2 times row 1 from row 2: \[R_2 = R_2 - 2R_1 = (0, 8, -5, 7)\]Then make the first element of the third row zero using the first row. Add 3 times row 1 to row 3: \[R_3 = R_3 + 3R_1 = (0, 0, 10, -8)\]So, the matrix becomes:\[\left(\begin{array}{ccc|c}1 & -2 & 1 & -4 \0 & 8 & -5 & 7 \0 & 0 & 10 & -8\end{array}\right)\]

Step 3: Simplify to Reduced Row Echelon Form

Now simplify further. First, divide the third row by 10 to make the leading coefficient 1:\[R_3 = \frac{1}{10}R_3 = (0, 0, 1, -\frac{4}{5})\]Now, adjust the second row to make its third column zero using the third row. Add 5 times the third row to the second row: \[R_2 = R_2 + 5R_3 = (0, 8, 0, 3)\]Lastly, update the first row to make its third column zero using the third row. Subtract the first row by the third row:\[R_1 = R_1 - R_3 = (1, -2, 0, -\frac{16}{5})\]The updated matrix is:\[\left(\begin{array}{ccc|c}1 & -2 & 0 & -\frac{16}{5} \0 & 8 & 0 & 3 \0 & 0 & 1 & -\frac{4}{5}\end{array}\right)\]

Step 4: Solve for Variables

Now, interpret the reduced matrix in terms of variables: 1. From the third row: \(z = -\frac{4}{5}\).2. From the second row: \(8y = 3\), so \(y = \frac{3}{8}\).3. From the first row: \(x - 2y = -\frac{16}{5}\), substitute \(y\) and solve for \(x\): \(x - 2\left(\frac{3}{8}\right) = -\frac{16}{5}\) \(x = -\frac{16}{5} + \frac{3}{4}\)\(x = -\frac{64}{20} + \frac{15}{20}\)\(x = -\frac{49}{20}\)

Step 5: Final Solution

The solution to the system is: \(x = -\frac{49}{20}\), \(y = \frac{3}{8}\), \(z = -\frac{4}{5}\).

Key Concepts

Augmented Matrix

The augmented matrix is a tool used to represent a system of linear equations compactly. This representation organizes coefficients of the variables and the constants into a single matrix. The visual clarity offered by an augmented matrix is crucial when applying matrix operations. For a system like:\[\begin{align*}x - 2y + z &= -4, \2x + 4y - 3z &= -1, \-3x - 6y + 7z &= 4,\end{align*}\]we write it as:\[\left(\begin{array}{ccc|c}1 & -2 & 1 & -4 \2 & 4 & -3 & -1 \-3 & -6 & 7 & 4\end{array}\right)\]This matrix keeps the equation coefficients aligned and the constants right after a separator to denote the equality. Hence, it assists in simplifying solving methods such as Gaussian elimination.

Row Echelon Form

Row echelon form is a specific arrangement of a matrix achieved by performing elementary row operations. This form makes further handling of the system more straightforward. For row echelon form, each leading entry of a row is to the right of the leading entry of the row above it. Besides, any full row of zeros will be at the bottom.
To reach this form, certain operations are performed like adding, subtracting, or swapping rows, as well as multiplying a row by a nonzero scalar. For example, apply:

• Subtracting or adding multiples of rows to make parts of the matrix zero.
• Positioning the zeros below the leading number (or pivot) within a row.

After modifications, the matrix might look like this:\[\left(\begin{array}{ccc|c}1 & -2 & 1 & -4 \0 & 8 & -5 & 7 \0 & 0 & 10 & -8\end{array}\right)\] Here, we have intuitively stepped towards simplifying the system to a form ready for reduced row echelon form.

Reduced Row Echelon Form

Reduced row echelon form (RREF) further refines a matrix to make solving straightforward. In RREF, not only do pivot positions have zeros below them, they also have zeros above them. The pivots themselves must each be equal to one.
Achieving this requires additional row operations like:

• Normalizing rows so that the pivot becomes 1 (by dividing the whole row by the pivot's value).
• Eliminating any existing non-zero entries above the pivots by suitable row manipulations.

For example, post-operation, the augmented matrix can transform into:\[\left(\begin{array}{ccc|c}1 & -2 & 0 & -\frac{16}{5} \0 & 8 & 0 & 3 \0 & 0 & 1 & -\frac{4}{5}\end{array}\right)\]Here, you clearly see the leading ones and how everything is nicely sorted, minimalizing any complexity in solving directly for each variable. This form makes deducing solutions such as \(x = -\frac{49}{20}\), \(y = \frac{3}{8}\), and \(z = -\frac{4}{5}\) very accessible.