Problem 23 - Intermediate Algebra

Question

Perform each of the following elementary row operations on the augmented matrix \(\left[\begin{array}{rrrr}3 & 6 & -9 & 0 \\ 1 & 5 & -2 & 1 \\ -2 & 2 & -2 & 5\end{array}\right]\) $$ -R_{1}+R_{2} $$

Step-by-Step Solution

Answer

The resulting matrix is \(\begin{bmatrix} 3 & 6 & -9 & 0 \\ -2 & -1 & 7 & 1 \\ -2 & 2 & -2 & 5 \end{bmatrix}\).

Step 1: Identify the Matrix

The given augmented matrix is: \[\begin{bmatrix}3 & 6 & -9 & 0 \1 & 5 & -2 & 1 \-2 & 2 & -2 & 5\end{bmatrix}\]We will be performing the row operation \(-R_1 + R_2 \rightarrow R_2\).

Step 2: Perform the Operation -R1 + R2

To perform the operation \(-R_1 + R_2 \rightarrow R_2\), we first take the negative of the first row \([-3, -6, 9, 0]\) and add it to the second row element-wise. This operation is applied to each element of the rows as follows:1. First element: \(1 + (-3) = -2\)2. Second element: \(5 + (-6) = -1\)3. Third element: \(-2 + 9 = 7\)4. Fourth element: \(1 + 0 = 1\)This gives us the new second row \([-2, -1, 7, 1]\).

Step 3: Write the Resulting Matrix

Replace the old second row with the new second row in the matrix. The resulting matrix after the row operation is:\[\begin{bmatrix}3 & 6 & -9 & 0 \-2 & -1 & 7 & 1 \-2 & 2 & -2 & 5\end{bmatrix}\]

Key Concepts

Understanding Augmented Matrices

An augmented matrix is a convenient way of representing a system of linear equations. It combines the coefficients of the variables and the constants from each equation into one compact matrix form.
This allows us to apply matrix operations to solve the system more easily. Imagine a system of equations, like the one in the original exercise. The first three columns of the augmented matrix represent the coefficients of three variables, while the last column contains the constants of each equation.

• For example, in the matrix \[ \begin{bmatrix} 3 & 6 & -9 & | & 0 \ 1 & 5 & -2 & | & 1 \ -2 & 2 & -2 & | & 5 \end{bmatrix} \]the first row corresponds to the equation \(3x + 6y - 9z = 0\),indicating that the coefficients for \(x\), \(y\), and \(z\) are 3, 6, and -9, respectively, and the constant is 0.

This format simplifies the process of applying operations to solve for the variable values.

The Magic of Matrix Transformation

Matrix transformation involves changing a matrix from one state to another using specific operations, often to solve linear equations. An example is transforming an augmented matrix to row-echelon form or reduced row-echelon form to make it easier to determine the solution of a system of equations.
By employing operations like row swapping, multiplication, or addition, we modify the matrix while maintaining its fundamental properties. This helps to isolate variables and reach solutions systematically. Remember, the aim in matrix transformations is often simplification. We apply transformations that yield a convenient form, such as identity matrix, where non-zero entries of diagonals indicate variable solution points.

Breaking Down Row Operation Steps

Row operations are tools we use to manipulate and simplify matrices. In our exercise, the specific operation used is row addition/subtraction. The step \(-R_1 + R_2 \rightarrow R_2\)means to add the negative of the first row to the second row to form a new second row. Each element of these rows was modified individually in the given problem. Let's briefly look at each part of this step:

• Negate the First Row: Change the sign of each element in the first row to get \([-3, -6, 9, 0]\).
• Add Element-wise: Combine this negated row with the second row, i.e., add \([-3, -6, 9, 0]\) to each corresponding element of the second row \([1, 5, -2, 1]\).
• This results in the new second row: \([-2, -1, 7, 1]\).

These steps help simplify the matrix by reducing one row, making it closer to a form where solution extraction is straightforward.