Problem 35 - Precalculus

Question

Perform the indicated row operations on each augmented matrix. $$\left[\begin{array}{rrrr|r} 1 & -2 & 5 & -1 & 2 \\ 0 & 3 & 0 & -1 & -2 \\ 0 & -2 & 1 & -2 & 5 \\ 0 & 0 & 1 & -1 & -6 \end{array}\right] \quad R_{3}+R_{2} \rightarrow R_{2}$$

Step-by-Step Solution

Answer

Replace row 2 with \([0, 1, 1, -3, 3]\) in the augmented matrix.

Step 1: Identify the Matrix and Operation

The problem presents an augmented matrix and instructs us to apply the operation \( R_3 + R_2 \rightarrow R_2 \), which means we add row 3 to row 2 and replace row 2 with the result.

Step 2: Extract the Rows Involved

The current rows are:\- \( R_2 = \left[ 0 \ 3 \ 0 \ -1 \ -2 \right] \)\- \( R_3 = \left[ 0 \ -2 \ 1 \ -2 \ 5 \right] \)\We need to add these rows together.

Step 3: Perform the Addition

Calculate the new entries for each column by adding the corresponding elements of \( R_3 \) and \( R_2 \):\\( 0 + 0 = 0 \)\\( 3 + (-2) = 1 \)\\( 0 + 1 = 1 \)\\( -1 + (-2) = -3 \)\\( -2 + 5 = 3 \)\This gives us the new \( R_2 = \left[ 0 \ 1 \ 1 \ -3 \ 3 \right] \).

Step 4: Update the Matrix

Replace \( R_2 \) in the original matrix with the new values calculated. The augmented matrix now looks like this: \[ \begin{array}{cccc|c} 1 & -2 & 5 & -1 & 2 \ 0 & 1 & 1 & -3 & 3 \ 0 & -2 & 1 & -2 & 5 \ 0 & 0 & 1 & -1 & -6 \end{array} \]

Key Concepts

Augmented Matrix

An augmented matrix is essentially an extension of a typical matrix, specifically one that includes extra columns added to a system of linear equations. These columns represent the constants from the equations' right-hand side. By using this structure, it becomes easier to handle operations and solve systems through matrix methods.

• A typical matrix includes only the coefficients of the variables within the system of equations.
• The augmented part of the matrix contains these constants separated by a vertical bar, which visually distinguishes them from the coefficients.

For instance, if you have a system of equations like:\[\begin{align*}1x - 2y + 5z - 1w &= 2 \0x + 3y + 0z - 1w &= -2 \0x - 2y + 1z - 2w &= 5 \0x + 0y + 1z - 1w &= -6\end{align*}\]The augmented matrix form highlights the coefficients and constants like so:\[\begin{bmatrix}1 & -2 & 5 & -1 & | & 2 \0 & 3 & 0 & -1 & | & -2 \0 & -2 & 1 & -2 & | & 5 \0 & 0 & 1 & -1 & | & -6\end{bmatrix}\]The vertical bar is optional but often used to represent where the equations' constant terms begin. It allows handling complex operations efficiently, laying the groundwork for applying methods such as Gauss-Jordan elimination.

Elementary Row Operations

Elementary row operations are fundamental in manipulating matrices to solve systems of linear equations or find matrix inverses. They consist of three primary types of operations that can be performed on rows of a matrix:

• Swapping two rows within a matrix.
• Multiplying a row by a non-zero scalar.
• Adding a multiple of one row to another row.

In our problem, the operation \( R_3 + R_2 \rightarrow R_2 \) is an example of the third type of operation. By adding row 3 to row 2, we effectively modify the entries of row 2, while keeping row 3 unchanged. This step is crucial for simplifying the matrix or preparing it for solving a system of equations.
These operations are valid because they do not change the set of solutions to the matrix equation. Thus, they serve as effective tools to transform an augmented matrix into an easier form to analyze, such as the reduced row echelon form needed for Gauss-Jordan elimination.

Gauss-Jordan Elimination

Gauss-Jordan elimination is a systematic method to solve a system of linear equations using augmented matrices by reducing them to their simplest form, known as reduced row echelon form (RREF). This process makes it straightforward to solve systems of equations or determine their solvability.

• The main goal is to make every leading coefficient 1 and ensure that all other coefficients in the leading coefficient's column are zero.
• The process begins with the first row and systematically moves down the matrix, applying elementary row operations to achieve RREF.

As the name suggests, this method builds on Gaussian elimination but continues the process to achieve a matrix where each leading variable is isolated.
Consider our updated matrix after applying the row operation:\[\begin{bmatrix}1 & -2 & 5 & -1 & | & 2 \0 & 1 & 1 & -3 & | & 3 \0 & -2 & 1 & -2 & | & 5 \0 & 0 & 1 & -1 & | & -6\end{bmatrix}\]The next steps could involve additional row operations to zero out the x coefficients below the pivot in row 1 and above and below the pivot in rows 3 and 4. This tedious and strategic process eventually reveals the solutions clearly without needing to back-substitute the remaining equations, as is necessary in Gaussian elimination.