Math Problem Statement

Consider the system of equations: -6a + (k-3)b = 1 and 3ka - 5b = 4. (a) Find the values of the real parameter k such that the system has a unique solution. (b) Find the unique solution in terms of k.

Solution

Let's break down the problem you're working on.

The system of equations is:

−6a+(k−3)b=1-6a + (k - 3)b = 1 3ka−5b=43ka - 5b = 4

Part (a): Find the values of kk for which the system has a unique solution.

A system of linear equations has a unique solution when the determinant of the coefficient matrix is non-zero. The matrix for this system is:

-6 & k-3 \\ 3k & -5 \end{pmatrix}$$ The determinant of this matrix is given by: $$\text{det} = (-6)(-5) - (k-3)(3k)$$ $$\text{det} = 30 - (3k^2 - 9k)$$ $$\text{det} = 30 - 3k^2 + 9k$$ $$\text{det} = -3k^2 + 9k + 30$$ To have a unique solution, the determinant must be non-zero, so: $$-3k^2 + 9k + 30 \neq 0$$ This is a quadratic equation, and you would solve it for $$k$$ to find the values where the system has a unique solution. ### Part (b): Find the unique solution in terms of $$k$$. Once you determine the values of $$k$$ that yield a unique solution, you substitute them back into the system to find the solution for $$a$$ and $$b$$. If you need further steps or clarifications, feel free to ask! --- Would you like more details on solving the determinant or substituting back into the equations? Here are some related questions you might find helpful: 1. How do we solve a quadratic equation? 2. What does it mean for a system of equations to be dependent? 3. How do we determine if a system has no solution or infinitely many solutions? 4. What is Cramer's rule for solving systems of equations? 5. Can a determinant of zero still provide a meaningful solution? **Tip**: Remember, checking the determinant is crucial when determining the nature of solutions in linear systems!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Determinants
Systems of Equations

Formulas

Determinant of 2x2 matrix: ad - bc
Quadratic equation formula ax^2 + bx + c = 0

Theorems

Determinant criterion for unique solutions

Suitable Grade Level

Grades 10-12

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