Problem 26 - College Algebra with Modeling and Visualization

Question

Perform each row operation on the given matrix by completing the matrix at the right. $$ \left[\begin{array}{rrr|r} 1 & -2 & 3 & 6 \\ 2 & 1 & 4 & 5 \\ -3 & 5 & 3 & 2 \end{array}\right] \begin{array}{r} R_{2}-2 R_{1} \rightarrow \\ R_{3}+3 R_{1} \rightarrow \end{array}\left[\begin{array}{rrr|r} 1 & -2 & 3 & 6 \\ & & & \end{array}\right] $$

Step-by-Step Solution

Answer

The transformed matrix is \( \begin{bmatrix} 1 & -2 & 3 & 6 \\ 0 & 5 & -2 & -7 \\ 0 & -1 & 12 & 20 \end{bmatrix} \).

Step 1: Understand the Operations

We need to perform two row operations. The first is \( R_2 - 2R_1 \) which means from row 2, we will subtract 2 times each element of row 1. The second is \( R_3 + 3R_1 \) which means to each element of row 3 we will add 3 times the corresponding element of row 1.

Step 2: Perform First Row Operation (R2)

Take row 2 \([2, 1, 4, 5]\) and subtract 2 times row 1 \([1, -2, 3, 6]\). This yields: - First column: \(2 - 2(1) = 2 - 2 = 0\) - Second column: \(1 - 2(-2) = 1 + 4 = 5\) - Third column: \(4 - 2(3) = 4 - 6 = -2\) - Fourth column: \(5 - 2(6) = 5 - 12 = -7\) The new row 2 is \([0, 5, -2, -7]\).

Step 3: Perform Second Row Operation (R3)

Take row 3 \([-3, 5, 3, 2]\) and add 3 times row 1 \([1, -2, 3, 6]\). This yields: - First column: \(-3 + 3(1) = -3 + 3 = 0\) - Second column: \(5 + 3(-2) = 5 - 6 = -1\) - Third column: \(3 + 3(3) = 3 + 9 = 12\) - Fourth column: \(2 + 3(6) = 2 + 18 = 20\) The new row 3 is \([0, -1, 12, 20]\).

Step 4: Write the Final Matrix

The transformed matrix after performing the operations is: \[\begin{bmatrix}1 & -2 & 3 & 6 \0 & 5 & -2 & -7 \0 & -1 & 12 & 20 \end{bmatrix}\] Each row now contains the result of the specified operations.

Key Concepts

Elementary Row Operations

Elementary row operations are the fundamental tools used to modify matrices in linear algebra. They are essential when solving systems of equations using matrix methods. These operations help us transform matrices into simpler forms, often making it easier to interpret or solve. The three types of elementary row operations are:

• **Row swapping**: Interchange two rows in a matrix. This can help in bringing a pivotal element to the right position.
• **Scalar multiplication of a row**: Multiply every element in a row by a non-zero scalar to make calculations straightforward.
• **Row addition (or subtraction)**: Add or subtract a multiple of one row to another row. This is often used to eliminate elements and simplify the matrix.

In our exercise, we specifically focused on the row addition/subtraction operations. These involve scaling one row and adding it to another to eliminate variables and simplify the system of equations represented by the matrix. For instance, subtracting twice the first row from the second (as shown in the solution), modifies each element, ultimately reducing the complexity of the matrix system.

Augmented Matrix

An augmented matrix is a powerful tool in the representation of systems of linear equations. It combines the coefficient matrix of the variables with the constants from the right side of the equations into one matrix. This matrix typically has an additional column, marked with a vertical line (|), signifying the presence of constants.
The structure of an augmented matrix makes it convenient for applying techniques like row operations, as changes can be conducted in an organized fashion. For example, our exercise begins with a matrix:\[\left[\begin{array}{rrr|r}1 & -2 & 3 & 6 \2 & 1 & 4 & 5 \-3 & 5 & 3 & 2 \end{array}\right]\]Here, the first three columns represent coefficients of variables in equations, while the last column represents constants. The clear separation helps execute row transformations that aim to find solutions for variables systematically.

Gaussian Elimination

Gaussian elimination is a systematic method used to solve systems of linear equations. It involves using elementary row operations to bring a matrix into a reduced row echelon form (or row-reduced echelon form), from which the solutions can be easily deduced. The process usually involves the following steps:

• Identify a pivot position; a non-zero leading entry in a row of the matrix.
• Use row operations to create zeros below the pivot position.
• Move to the next column and repeat the process, working down and to the right.

With our exercise, Gaussian elimination has been applied by transforming the given augmented matrix step-by-step. After performing row operations on the original matrix, a simpler matrix is obtained:\[\begin{bmatrix}1 & -2 & 3 & 6 \0 & 5 & -2 & -7 \0 & -1 & 12 & 20\end{bmatrix}\]This stage helps identify simpler relationships between variables depicted in the matrix, setting the stage for further operations until solutions are revealed, characteristic of the Gaussian elimination method.