Question

Find the solution to the given linear system. If the system has infinite solutions, give 2 particular solutions. $$ \begin{array}{c} -3 x_{1}+7 x_{2}=-7 \\ 2 x_{1}-8 x_{2}=8 \end{array} $$

Short answer

The unique solution is \(x_1 = 0\) and \(x_2 = -1\).

Step-by-step solution

Write the System in Matrix Form

The given system of equations can be written in matrix form as \[\begin{bmatrix}-3 & 7 \2 & -8\end{bmatrix}\begin{bmatrix}x_1 \x_2 \end{bmatrix}=\begin{bmatrix}-7 \8 \end{bmatrix}\]This matrix equation represents the original set of linear equations.

Calculate the Determinant

To determine if the system has a unique solution, infinite solutions, or no solution, calculate the determinant of the coefficient matrix:\[d = \begin{vmatrix} -3 & 7 \ 2 & -8 \end{vmatrix} = (-3)(-8) - (7)(2) = 24 - 14 = 10.\]Since the determinant is not zero (\(d eq 0\)), the system has a unique solution.

Solve Using Cramer's Rule

Using Cramer's Rule to find the values of \(x_1\) and \(x_2\), we establish:The determinant of the coefficient matrix, \(D = 10\). Replace the first column with the constants to find \(D_{x_1}\):\[D_{x_1} = \begin{vmatrix} -7 & 7 \ 8 & -8 \end{vmatrix} = (-7)(-8) - (7)(8) = 56 - 56 = 0\]Replace the second column with the constants to find \(D_{x_2}\):\[D_{x_2} = \begin{vmatrix} -3 & -7 \ 2 & 8 \end{vmatrix} = (-3)(8) - (-7)(2) = -24 + 14 = -10\]

Calculate the Solution

Using the formulas from Cramer's rule:\[\x_1 = \frac{D_{x_1}}{D} = \frac{0}{10} = 0\]\[\x_2 = \frac{D_{x_2}}{D} = \frac{-10}{10} = -1\]Thus, the unique solution to the system is \(x_1 = 0\) and \(x_2 = -1\).

Key concepts

Matrix Form

To solve a system of linear equations, one effective way is to express it in matrix form. This organizes the equations clearly and is the first step in applying numerical methods such as Cramer's Rule. A matrix form consists of three key matrices:

• Coefficient Matrix (A): Contains the coefficients of the variables in the linear equations. For example, for the given system:\[\begin{bmatrix} -3 & 7 \ 2 & -8 \end{bmatrix}\]
• Variable Matrix (X): Lists the unknowns vertically. In our case: \[\begin{bmatrix} x_1 \ x_2 \end{bmatrix}\]
• Constant Matrix (B): Holds the constants from the right-hand side of each equation. For instance: \[\begin{bmatrix} -7 \ 8 \end{bmatrix}\]

Combining these, the system of equations can be represented as \(AX = B\), making it easier to manipulate algebraically.

Determinant

The determinant is a special number calculated from a square matrix. In the context of linear systems, the determinant of the coefficient matrix helps us understand the nature of the solutions.

• Non-zero Determinant: If the determinant isn't zero, the system has a unique solution.
• Zero Determinant: A zero determinant indicates the system could have either infinitely many solutions or no solution at all, depending on the relationship of the equations.

Calculating the determinant of a 2x2 matrix like \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]is simple. The formula is \(ad - bc\). For our system, it was found to be 10, confirming that there is a unique solution.

Cramer's Rule

Cramer's Rule provides a systematic way to find the solutions for each variable in a linear system using determinants. It's applicable only when the determinant of the coefficient matrix is non-zero. Here's how it works:

• Calculate the original determinant \(D\) of the coefficient matrix.
• For each variable \(x_i\):
  • Replace the \(i^{th}\) column in the coefficient matrix with the constant matrix.
  • Calculate the new determinant \(D_{x_i}\).
  • Solve for the variable: \(x_i = \frac{D_{x_i}}{D}\).

Using this method on our system, we calculated \(x_1 = 0\) and \(x_2 = -1\), representing the unique solution of the system.

Unique Solution

A unique solution means there is exactly one set of values for the variables that satisfies all the equations in the system. This occurs when:

• The determinant of the coefficient matrix is non-zero.
• The system is consistent, meaning there's no contradiction among the equations.

In our example, the determinant was calculated to be 10, which is non-zero. Hence, the system was determined to have a unique solution of \(x_1 = 0\) and \(x_2 = -1\). This outcome helps avoid the ambiguity of multiple or non-existent solutions, ensuring a clear and definite answer to the problem.