Question

State for which values of \(k\) the given system will have exactly 1 solution, infinite solutions, or no solution. $$ \begin{array}{l} x_{1}+2 x_{2}=1 \\ x_{1}+3 x_{2}=k \end{array} $$

Short answer

The system has a unique solution for all values of \( k \).

Step-by-step solution

Determine the System Matrix and Augmented Matrix

First, write the system of linear equations in matrix form. The system's coefficient matrix \( A \) and augmented matrix \( [A|b] \) are given by:\[A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}, \quad [A|b] = \begin{bmatrix} 1 & 2 & 1 \ 1 & 3 & k \end{bmatrix}\]

Compute the Determinant of the Coefficient Matrix

To find when the system has a unique solution, we first find the determinant of matrix \( A \). The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by \( ad - bc \). For our matrix:\[\text{det}(A) = (1)(3) - (1)(2) = 3 - 2 = 1\]

Analyze the Determinant

Since the determinant \( \text{det}(A) eq 0 \), the coefficient matrix is invertible. This implies that the system of equations has a unique solution for any value of \( k \), as the row lines will never be parallel and won't coincide regardless of the last column.

Key concepts

Determinant

When we solve a system of linear equations using matrices, the determinant plays a crucial role. For a 2x2 matrix, the determinant is calculated as follows: if you have a matrix \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \] its determinant is \( ad - bc \). The determinant tells us about the properties of the matrix.

• If the determinant is non-zero, the matrix is invertible, and we have a unique solution.
• If the determinant is zero, the matrix is not invertible. This might mean either no solutions or infinitely many solutions, depending on other conditions.

In this particular exercise, the determinant of our coefficient matrix is 1, which is non-zero, so the matrix is invertible. Thus, the system can have a unique solution.

System of Equations

A system of equations consists of multiple equations working together. These might be linear or non-linear, but in this exercise, we're dealing with two linear equations:

• \(x_{1} + 2x_{2} = 1\)
• \(x_{1} + 3x_{2} = k\)

To find solutions, we consider these equations at the same time. The methods to solve them can vary from substitution, elimination, to using matrices like in this exercise. In general, our goal is to find the values for the variables that satisfy all equations simultaneously. This particular system can be represented in matrix form, which provides a structured way to analyze its solutions.

Matrix Form

Matrices offer a compact way to represent and manipulate systems of equations. Our given system of equations can be written in matrix form as follows:Coefficient matrix \( A \): \[A = \begin{bmatrix} 1 & 2 \ 1 & 3 \end{bmatrix}\]Augmented matrix \([A|b]\): \[[A|b] = \begin{bmatrix} 1 & 2 & 1 \ 1 & 3 & k \end{bmatrix}\]The coefficient matrix \( A \) includes only the coefficients of the variables, while the augmented matrix \([A|b]\) includes the constants from the right side of the equations.Matrices simplify operations such as adding or subtracting equations and determining the solutions using resources like determinants or row reduction.

Unique Solution

In linear algebra, a unique solution means there's exactly one set of values for the variables that satisfies all equations in the system. Finding a unique solution often requires conditions such as a non-zero determinant. For the exercise's system of equations, since the determinant of the coefficient matrix \( A \) is 1 (non-zero), we can conclude:

• The system has a unique solution for all values of \( k \).

With a unique solution, the equations intersect at a single point, providing a clear, exact solution rather than multiple or none at all. This property helps differentiate between systems that are solvable and those that are not or have infinite possibilities.