Problem 28 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Fill in the missing entries by performing the indicated row operations to obtain the row-reduced matrices. \(\left[\begin{array}{rr|r}1 & 2 & 1 \\ 2 & 3 & -1\end{array}\right] \stackrel{R_{2}-2 R_{1}}{\longrightarrow}\left[\begin{array}{ll|l}1 & 2 & 1 \\\ . & . & .\end{array}\right] \frac{-R_{2}}{\longrightarrow}\) \(\left[\begin{array}{ll|l}1 & 2 & 1 \\ . & . & .\end{array}\right] \frac{R_{1}-2 R_{2}}{\longrightarrow}\left[\begin{array}{ll|r}1 & 0 & -5 \\ 0 & 1 & 3\end{array}\right]\)

Step-by-Step Solution

Answer

Performing the indicated row operations, we obtain the row-reduced matrix as follows: \(\left[\begin{array}{rr|r} 1 & 0 & -5 \\ 0 & 1 & 3 \end{array}\right]\)

Step 1: Write Down the Given Matrix

Write down the given matrix as it is: \(\left[\begin{array}{rr|r} 1 & 2 & 1 \\ 2 & 3 & -1 \end{array}\right]\)

Step 2: Perform the First Row Operation (R₂ - 2R₁)

Subtract 2 times the first row (R₁) from the second row (R₂) and replace the second row (R₂) with the result: \(R_{2} = R_{2} - 2R_{1} = (2-2\cdot1, 3-2\cdot2, -1-2\cdot1) = (0, -1, -3)\) The matrix becomes: \(\left[\begin{array}{rr|r} 1 & 2 & 1 \\ 0 & -1 & -3 \end{array}\right]\)

Step 3: Perform the Second Row Operation (-R₂)

Multiply the second row (R₂) by -1: \(R_{2} = -R_{2} = (-0, 1, 3)\) The matrix becomes: \(\left[\begin{array}{rr|r} 1 & 2 & 1 \\ 0 & 1 & 3 \end{array}\right]\)

Step 4: Perform the Third Row Operation (R₁ - 2R₂)

Subtract 2 times the second row (R₂) from the first row (R₁) and replace the first row (R₁) with the result: \(R_{1} = R_{1} - 2R_{2} = (1-2\cdot0, 2 - 2\cdot1, 1-2\cdot3) = (1, 0, -5)\) The matrix becomes: \(\left[\begin{array}{rr|r} 1 & 0 & -5 \\ 0 & 1 & 3 \end{array}\right]\) This is the final row-reduced matrix.

Key Concepts

Matrix Row Reduction

Matrix row reduction is a process used to simplify a matrix into a form that makes it easy to solve systems of linear equations. It involves using a series of specific operations on the rows of the matrix to transform it into a row-reduced form, often referred to as reduced row-echelon form (RREF). This process is a key part of solving linear equations because it allows us to find solutions quickly and efficiently.

When performing matrix row reduction, the goal is to achieve three primary objectives:

• Make leading coefficients in each row equal to 1.
• Ensure that each leading coefficient is the only non-zero entry in its column.
• Position each leading 1 to the right of any leading 1s above it.

Matrix row reduction helps in simplifying complex systems into more manageable forms, thereby making it easier to interpret and solve them.

Elementary Row Operations

Elementary row operations are fundamental operations that can be performed on the rows of a matrix to transform it without changing the solution set of a system of equations. These operations form the basis of processes such as row reduction and Gaussian elimination.

There are three types of elementary row operations:

• Row switching: Swap two rows of the matrix.
• Row multiplication: Multiply each entry of a row by a non-zero scalar.
• Row addition: Add or subtract a multiple of one row from another row.

Using these operations strategically allows us to manipulate matrices into simpler forms, like row-echelon or reduced row-echelon form, thus facilitating easier solutions to the equations represented by the matrix.

Gaussian Elimination

Gaussian elimination is a method for solving systems of linear equations. It uses a sequence of row operations to transform the system's matrix into an upper triangular form, from which the solutions can be easily calculated using back substitution.

The process involves three main phases:

• Transforming the original matrix into an upper triangular matrix using row operations.
• Performing back substitution to find the solutions for the variables.
• Finally, optionally continuing to reduced row-echelon form for exact clarity, although this is not always necessary for finding solutions.

Using Gaussian elimination allows for a systematic approach in solving equations, making it an essential tool in linear algebra. Through careful manipulation of the matrix, it enables clear and concise solutions to potentially complex systems of equations.