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resuelve mediante el proceso de determinantes

To solve the given equations for the constant \( k \) using determinants, I'll analyze each equation individually. The problems involve determinants of 2x2 matrices set equal to certain values. Let's solve them one by one.

### Problem 1
\[
\text{Given: } \left|\begin{array}{cc} 5k & \frac{k}{3} \\ \frac{k}{3} & 1 \end{array}\right| = 6
\]

The determinant of a 2x2 matrix \(\left|\begin{array}{cc} a & b \\ c & d \end{array}\right|\) is calculated as:
\[
ad - bc
\]
So for this problem:
\[
\text{Determinant} = (5k)(1) - \left(\frac{k}{3}\right)\left(\frac{k}{3}\right) = 5k - \frac{k^2}{9}
\]
Setting the determinant equal to 6:
\[
5k - \frac{k^2}{9} = 6
\]
Multiplying through by 9 to clear the fraction:
\[
45k - k^2 = 54
\]
Rearranging into standard quadratic form:
\[
k^2 - 45k + 54 = 0
\]
This is a quadratic equation, which can be solved using the quadratic formula:
\[
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -45 \), and \( c = 54 \). 

### Problem 2
\[
\text{Given: } \left|\begin{array}{cc} 1 & k \\ k & -4 \end{array}\right| = 2
\]
Determinant:
\[
1(-4) - k^2 = -4 - k^2
\]
Setting the determinant equal to 2:
\[
-4 - k^2 = 2
\]
Solving for \( k^2 \):
\[
k^2 = -6
\]
Since \( k^2 = -6 \) does not have real solutions, the problem might be interpreted differently or complex solutions are expected.

### Problem 3
\[
\text{Given: } \left|\begin{array}{cc} k & 1 \\ 1 & k \end{array}\right| = 3
\]
Determinant:
\[
k(k) - (1)(1) = k^2 - 1
\]
Setting the determinant equal to 3:
\[
k^2 - 1 = 3
\]
\[
k^2 = 4 \quad \Rightarrow \quad k = \pm 2
\]

### Problem 4
\[
\text{Given: } \left|\begin{array}{cc} 2k & \frac{k}{5} \\ \frac{k}{5} & \frac{1}{2} \end{array}\right| = 5
\]
Determinant:
\[
(2k)\left(\frac{1}{2}\right) - \left(\frac{k}{5}\right)\left(\frac{k}{5}\right) = k - \frac{k^2}{25}
\]
Setting the determinant equal to 5:
\[
k - \frac{k^2}{25} = 5
\]
Multiplying by 25:
\[
25k - k^2 = 125 \quad \Rightarrow \quad k^2 - 25k + 125 = 0
\]
This is another quadratic equation to solve using the quadratic formula.

### Problem 5
\[
\text{Given: } \left|\begin{array}{cc} k & 1 \\ 1 & k \end{array}\right| = 0
\]
This is similar to Problem 3, but we set it equal to 0:
\[
k^2 - 1 = 0 \quad \Rightarrow \quad k^2 = 1 \quad \Rightarrow \quad k = \pm 1
\]

### Problem 6
\[
\text{Given: } \left|\begin{array}{cc} 14k & \frac{5k}{2} \\ \frac{5k}{2} & -4 \end{array}\right| = 6
\]
Determinant:
\[
14k(-4) - \left(\frac{5k}{2}\right)\left(\frac{5k}{2}\right) = -56k - \frac{25k^2}{4}
\]
Setting the determinant equal to 6:
\[
-56k - \frac{25k^2}{4} = 6
\]
Multiply by 4 to clear the fraction:
\[
-224k - 25k^2 = 24 \quad \Rightarrow \quad 25k^2 + 224k + 24 = 0
\]
This is a quadratic equation to be solved using the quadratic formula.

Let me know if you'd like further details on solving the quadratic equations or if you have any other questions! Here are five related questions you might explore:

1. What is the determinant of a 3x3 matrix?
2. How do we solve a quadratic equation using the quadratic formula?
3. Can you provide examples of real-life applications where determinants are used?
4. What are the conditions under which a matrix has no inverse?
5. How does changing a single element of a matrix affect its determinant?

**Tip:** Always double-check the signs and calculations when working with determinants and quadratic equations, as small errors can lead to incorrect solutions.

Explore how to solve equations involving determinants with detailed step-by-step solutions. This problem covers determinants of 2x2 matrices and quadratic equations. Suitable for advanced high school students and those preparing for mathematical competitions.

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