Problem 22 - Advanced Engineering Mathematics

Question

since \\[ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{l} 0 \\ 0 \end{array}\right) \quad \text { and } \quad\left(\begin{array}{rr} 2 & 1 \\ 1 & -1 \end{array}\right) \mathbf{X}_{p}+\left(\begin{array}{r} -5 \\ 2 \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \end{array}\right) \\] we see that \\[ \mathbf{X}_{p}^{\prime}=\left(\begin{array}{rr} 2 & 1 \\ 1 & -1 \end{array}\right) \mathbf{X}_{p}+\left(\begin{array}{r} -5 \\ 2 \end{array}\right) \\]

Step-by-Step Solution

Answer

The solution is \(\mathbf{X}_{p} = \begin{pmatrix} \frac{7}{3} \\ \frac{1}{3} \end{pmatrix}\).

Step 1: Understand the Given Equations

We are given a system of linear equations where \(\mathbf{X}_{p}^{\prime}\) is a zero vector:\[\mathbf{X}_{p}^{\prime}=\begin{pmatrix} 0 \ 0 \end{pmatrix} \] along with the equation:\[\begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \mathbf{X}_{p} + \begin{pmatrix} -5 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]

Step 2: Solve the Matrix Equation

Our task is to solve the equation:\[\begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \mathbf{X}_{p} = \begin{pmatrix} 5 \ -2 \end{pmatrix} \] by finding \(\mathbf{X}_{p} \).

Step 3: Convert Matrix Equation into Linear Set

Translate the matrix into linear equations:1. \(2x + y = 5\) 2. \(x - y = 2\)

Step 4: Solve the System of Linear Equations

Substitute \(y = x - 2\) from the second equation into the first equation:\[ 2x + (x - 2) = 5 \]\[ 3x - 2 = 5 \]\[ 3x = 7 \]\[ x = \frac{7}{3} \]Substitute back to find \(y\):\[ y = x - 2 = \frac{7}{3} - 2 = \frac{7}{3} - \frac{6}{3} = \frac{1}{3} \]

Step 5: Verify the Solution

Check if \(x = \frac{7}{3}\) and \(y = \frac{1}{3}\) satisfy both equations:1. \(2 \cdot \frac{7}{3} + \frac{1}{3} = \frac{14}{3} + \frac{1}{3} = \frac{15}{3} = 5\), which is correct.2. \(\frac{7}{3} - \frac{1}{3} = \frac{6}{3} = 2\), which is also correct. Therefore, the solution \(\mathbf{X}_{p} = \begin{pmatrix} \frac{7}{3} \ \frac{1}{3} \end{pmatrix}\) is valid.

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations that share a set of variables. The goal is to find values for these variables that satisfy all the given equations simultaneously. In this exercise, we aim to determine values for the variable vector \( \mathbf{X}_p \) composed of \( x \) and \( y \). This is done by analyzing a pair of equations:

• \( 2x + y = 5 \)
• \( x - y = 2 \)

To solve a system like this, it’s common to use methods such as substitution or elimination. These methods help isolate one variable, allowing us to solve for the other. For example, from the equation \( x - y = 2 \), we can express \( y \) as \( x - 2 \) and substitute it into the other equation. This process reduces the complexity of the system and leads us to determine the individual values of the variables involved.

Matrix Equations

Matrix equations simplify linear systems by organizing their coefficients into matrices. In our exercise, the system of equations is represented as a matrix equation. This form uses matrices to express the equations succinctly and is especially helpful for computational solutions.

We have the matrix form:

• \( \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \mathbf{X}_p + \begin{pmatrix} -5 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \).

This representation clarifies the relationships between variables and coefficients. Solving this matrix equation involves transferring constants to the other side, leading to an equation like:

• \( \begin{pmatrix} 2 & 1 \ 1 & -1 \end{pmatrix} \mathbf{X}_p = \begin{pmatrix} 5 \ -2 \end{pmatrix} \).

This prompts further steps to find the variable vector \( \mathbf{X}_p \). Understanding matrix equations allows you to efficiently solve multiple equations together by taking advantage of matrix operations.

Vector Solutions

Vector solutions provide a convenient way to express the results of a system of linear equations. In linear algebra, vectors often represent solutions to systems of equations. A solution vector \( \mathbf{X}_p \) indicates the specific values of the unknowns that satisfy the given equations.

• In this case, our solution vector is \( \mathbf{X}_p = \begin{pmatrix} \frac{7}{3} \ \frac{1}{3} \end{pmatrix} \).

This vector provides the precise values of \( x \) and \( y \) that resolve both equations in our system. Checking these values in each original equation confirms their validity. With \( x = \frac{7}{3} \) and \( y = \frac{1}{3} \), each equation holds true, validating our solution.

The use of vectors is advantageous, particularly when dealing with higher dimensions or complex systems, as it preserves clarity and facilitates calculations.