91Ó°ÊÓ

Understanding how to represent a system of linear equations as a matrix is the first step in solving it through matrix methods. In this exercise, we convert the given system of equations into its matrix form. Each equation's coefficients are placed into rows of a matrix, known as the coefficient matrix.
This is done as follows:

• The coefficients of the variables from each equation form the rows: for the equation \(x + y + kz = 1\), the row is \([1, 1, k]\).
• The augmented matrix includes the constant terms, forming a fourth column: \([1, 1, k \vert 1], [1, k, 1 \vert 1], [k, 1, 1 \vert -2]\).

This matrix representation allows us to apply linear algebra techniques to the problem, simplifying calculations such as computing determinants or finding solutions by back-substitution.

The determinant of a matrix is a special number that helps us understand certain properties of the matrix, particularly in finding solutions to systems. In a system of linear equations, the determinant of the coefficient matrix tells us if the system has a unique solution (when it's non-zero) or if more investigation is needed (when it is zero).
For our coefficient matrix:\[\begin{vmatrix}1 & 1 & k \1 & k & 1 \k & 1 & 1 \end{vmatrix}\]the determinant is calculated by expanding along a row or column.

• For example, expanding using the first row results in: \(k - 1 - (1 - k^2) + k(1 - k)\).
• Simplifying yields \(2k^2 - 2k - 2\).

The determinant needs to be evaluated to determine the case for unique, no, or infinitely many solutions.

In the scenario where the determinant of a matrix is zero, we set it equal to zero and solve the resulting quadratic equation. This helps identify specific values of \(k\) that result in either no solutions or infinitely many solutions. The quadratic equation derived is:
\[k^2 - k - 1 = 0\]
We solve it using the quadratic formula:

• The formula is \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-1\), \(c=-1\).
• Plugging in the values gives: \(k = \frac{1 \pm \sqrt{5}}{2}\).

These \(k\) values are pivotal, leading to conditions for either no solutions or infinitely many solutions, determined by further matrix evaluation.

A system of linear equations has a unique solution when the determinant of its coefficient matrix is non-zero. If the determinant is non-zero, it means the matrix can be inverted, and the system of equations has one definitive solution.
For this exercise, expect unique solutions when \(k\) does not match the critical values found through the quadratic formula \(\left(\frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}\right)\).
Take, for instance, \(k = 0\). Substituting \(k = 0\) into the original system results in solving:

• \(x + y = 1\)
• \(x + z = 1\)
• \(y + z = -2\)

Solving this consistent and robust set yields a unique solution: \(x = \frac{3}{2}, y = -\frac{1}{2}, z = -\frac{1}{2}\). This assures us that by choosing suitable \(k\), unique solutions are achievable.