Problem 31 - Algebra and Trigonometry Real Mathematics, Real People

Question

(a) perform the row operations to solve the augmented matrix, (b) write and solve the system of linear equations represented by the augmented matrix, and (c) compare the two solution methods. Which do you prefer? $$\left[\begin{array}{rrrr} -3 & 4 & \vdots & 22 \\ 6 & -4 & \vdots & -28 \end{array}\right]$$ (i) Add \(R_{2}\) to \(R_{1}\) (ii) Add -2 times \(R_{1}\) to \(R_{2}\) (iii) Multiply \(R_{2}\) by \(-\frac{1}{4}\) (iv) Multiply \(R_{1}\) by \(\frac{1}{3}\)

Step-by-Step Solution

Answer

The solution to the matrix and the system of equations is \(x = -2, y = 4\). Both methods yield the same result so one can prefer either based on their individual comfort.

Step 1: Perform Row Operations on the Augmented Matrix

It's given that the first operation is to add \(R_{2}\) to \(R_{1}\). This gives a new matrix as: \[\left[\begin{array}{cc|c} 3 & 0 & -6 \\ 6 & -4 & -28 \end{array}\right]\] Next, add -2 times \(R_{1}\) to \(R_{2}\). The resulting matrix is: \[\left[\begin{array}{cc|c} 3 & 0 & -6 \\ 0 & -4 & -16 \end{array}\right]\] Now, multiply \(R_{2}\) by \(-\frac{1}{4}\). This produces: \[\left[\begin{array}{cc|c} 3 & 0 & -6 \\ 0 & 1 & 4 \end{array}\right]\] Lastly, multiply \(R_{1}\) by \(\frac{1}{3}\), giving: \[\left[\begin{array}{cc|c} 1 & 0 & -2 \\ 0 & 1 & 4 \end{array}\right]\] This is the row echelon form of the original augmented matrix.

Step 2: Write and Solve the System of Equations

The resulting matrix from step 1 corresponds to the system of equations: \[\begin{cases} x = -2 \\ y = 4 \end{cases}\] Thus, the solution to this system of equations is \(x = -2, y = 4\).

Step 3: Compare the Solutions

We get the same solution from row reduction of the augmented matrix as well as the system of equations. Hence, the methods are equivalent. As to which method one prefers will depend on individual understanding and comfort.

Key Concepts

Row Operations

To solve linear algebra problems involving matrices, one of the fundamental processes we use is row operations. These operations allow us to manipulate the rows of a matrix to simplify its form, making it easier to identify solutions to the system of equations it represents. Three primary row operations can be used during this process:

• Row swapping: exchanging two rows.
• Row multiplying: multiplying a row by a non-zero scalar.
• Row addition: adding the multiples of one row to another.

Each operation is designed to transform the matrix into an equivalent one where the solution set does not change, but the matrix becomes simpler to work with. For example, in the given exercise, operations such as adding one row to another and multiplying a row by a scalar are utilized to reach the simplified row echelon form.

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Finding a solution to this system means determining the values of the variables that satisfy all equations simultaneously. In matrix terms, these systems can be represented using an augmented matrix, which combines the coefficients of the variables and the constants from the right-hand side of the equations into one rectangular array. The goal is to convert this matrix to a form where the correspondence between the matrix and the system of equations is clear, and the solutions can be read off directly. For example, the row echelon form obtained through row operations in the exercise directly resulted in a simple set of equations, with each equation corresponding to a matrix row.

Row Echelon Form

To solve matrix equations efficiently, we strive to reach what is called row echelon form (REF). A matrix is in REF when it meets certain criteria:

• All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes.
• The leading coefficient (also known as the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it.
• All entries in a column below a leading coefficient are zeros.

The advantage of achieving the REF is that it reflects a system of equations that can be solved using back-substitution, starting from the last row. Our exercise demonstrates the process of transforming an augmented matrix into its REF, which reveals the solutions to the associated system of equations.

Matrix Row Reduction

The process of matrix row reduction involves using row operations to achieve a simplified form of a matrix—ideally, the row echelon form or even a further simplified version called the reduced row echelon form. Row reduction is useful because it allows us to solve for the variables more directly, especially when dealing with systems of equations. The step-by-step solution in our exercise involves matrix row reduction, which simplifies the augmented matrix in a systematic manner until the solutions are apparent. By eliminating the variables incrementally, we arrive at a point where the variables can be readily solved, as seen when the augmented matrix was reduced to a form with clear and immediate solutions for the variables.