Problem 6 - Algebra and Trigonometry Real Mathematics, Real People

Question

Refer to the system of linear equations \(\left\\{\begin{aligned}-2 x+3 y &=5 \\\ 6 x+7 y &=4 \end{aligned}\right.\). Is the augmented matrix row-equivalent to its reduced row-echelon form?

Step-by-Step Solution

Answer

No, the original system of linear equations (or the original augmented matrix) is not row-equivalent to its reduced row-echelon form.

Step 1: Formulate the Augmented Matrix

First, represent the system of equations as an augmented matrix. This is calculated by creating a matrix where each row represents an equation, and each column represents a variable or the constant on the right side of the equation:\n\[\begin{bmatrix}-2 & 3 & 5 \\6 & 7 & 4\end{bmatrix}\]\n

Step 2: Perform Row Operations to Reach RREF

Transform the matrix into the reduced row-echelon form by performing a series of row operations. Start by multiplying the first row by 3 and the second row by 2 to make the x-coefficient in the first row equal to the negative of the x-coefficient in the second row. The matrix now is:\n\[\begin{bmatrix}-6 & 9 & 15\\12 & 14 & 8\end{bmatrix}\]\nThen add the first row to the second row, which gives you:\n\[\begin{bmatrix}-6 & 9 & 15\\6 & 23 & 23\end{bmatrix}\]\nDivide the first row by -6 and the second by 6 to get leading ones:\n\[\begin{bmatrix}1 & -1.5 & -2.5\\1 & 3.83 & 3.83\end{bmatrix}\]\nLastly, subtract the first row from the second to get a 0 in the first column of the second row, giving you the reduced row-echelon form:\n\[\begin{bmatrix}1 & -1.5 & -2.5\\0 & 5.33 & 6.33\end{bmatrix}\]\n

Step 3: Compare the Original and Final Matrices

Now, compare the original augmented matrix and the final matrix. If they are row-equivalent, they represent the same system of linear equations. In this case, the two matrices are different, meaning the original matrix is not row-equivalent to its reduced row-echelon form.

Key Concepts

Row Operations

Row operations are crucial steps in working with augmented matrices. They help us transform these matrices into a form that makes solving systems of linear equations easier. There are three types of row operations you can perform:

• Swapping rows: You can switch the order of any two rows. This is useful for arranging the rows in a more convenient order.
• Multiplying a row by a non-zero constant: This operation changes the row by a factor, but it does not alter the solutions of the equations.
• Adding or subtracting a multiple of one row to another: This can help eliminate variables in certain rows, leading the matrix towards a simpler form.

By applying these operations correctly, a matrix can be transformed through a series of steps until it reaches its reduced row-echelon form (RREF). Each of these operations maintains the equivalence of the system, ensuring that the same solutions are preserved.

Reduced Row-Echelon Form

The reduced row-echelon form (RREF) of a matrix is a simplified version where certain criteria are met. A matrix in RREF is easy to interpret when solving equations, as it is close to the clear and concise format of answering sets. Here's what characterizes a matrix in RREF:

• Leading 1s: Each row starts with a leading 1 (the first non-zero number in the row).
• Zeros below Leading 1s: All numbers below a leading 1 in any column are 0.
• Zero rows are at the bottom: Any row that becomes completely 0 is placed at the bottom of the matrix.
• Each leading 1 is the only non-zero entry in its column: This setup makes it straightforward to backtrack and solve for unknowns.

Achieving the RREF involves combining row operations strategically. This form allows for straightforward interpretation and verification of the solutions to a system of linear equations.

Linear Equations

Linear equations are the foundation of systems solved by augmented matrices and row operations. A linear equation in two variables can be visualized as a line on a graph. The essence of these equations lies in their simplicity and their form: For example, in our original system, we have:\[-2x + 3y = 5, \6x + 7y = 4\]The challenge is to find values for variables that satisfy all equations at the same time. Using augmented matrices streamlines this because matrices organize the system's coefficients clearly.Consider linear equations as constraints that define the relationship between variables. To solve them, we aim to find where these relationships intersect or, if possible, the values that satisfy all simultaneous conditions of the system. Transformations like row operations help reveal these values clearly by moving matrices closer to reduced row-echelon form.