Solving a system of linear equations with one unknown value

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SUMMARY

The discussion focuses on determining the values of 'a' in a system of linear equations that dictate the conditions for no solutions, a unique solution, or infinitely many solutions. The equations provided are: 3x1 + 6x2 = -6, 3x1 + 9x2 - 6x3 = -12, and x1 + x2 + ax3 = 1. It is established that a unique solution exists when the determinant of the coefficient matrix is non-zero, specifically when 'a' is not equal to 2. The conditions for no solutions occur when 'a' equals 2, while infinitely many solutions are not possible in this scenario.

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  • Familiarity with determinants and their role in determining the nature of solutions.
  • Knowledge of matrix representation of linear equations.
  • Ability to manipulate algebraic expressions and inequalities.
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  • Learn how to calculate the determinant of a matrix in linear algebra.
  • Study the conditions for unique, infinite, and no solutions in systems of linear equations.
  • Explore the concept of row echelon form and its application in solving linear systems.
  • Investigate the implications of parameterized solutions in linear algebra.
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Students studying linear algebra, educators teaching systems of equations, and anyone seeking to understand the conditions for solutions in linear systems.

exqiron
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Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values.

3x1+6x2 = −6
3x1+9x2−6x3 = −12
x1+x2+ax3 = 1No Solutions: ?
Unique Solution: ?
Infinitely Many Solutions: ?

all i could conclude from this was
1 1 a 1
0 1 -2 -2
0 0 (2-a) -1
don't know what to do next i have tried different questions, don't understand :/
 
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exqiron said:
Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', 'a = ', or 'a ≠', then specify a value or comma-separated list of values.

3x1+6x2 = −6
3x1+9x2−6x3 = −12
x1+x2+ax3 = 1No Solutions: ?
Unique Solution: ?
Infinitely Many Solutions: ?

all i could conclude from this was
1 1 a 1
0 1 -2 -2
0 0 (2-a) -1
don't know what to do next i have tried different questions, don't understand :/

To start with, you can find where you are going to have unique solutions by evaluating the determinant of your coefficient matrix. For all values of a that give a nonzero determinant, you will have unique solutions.
 
Prove It said:
To start with, you can find where you are going to have unique solutions by evaluating the determinant of your coefficient matrix. For all values of a that give a nonzero determinant, you will have unique solutions.

so in other words if i write this is this correct?
No solutions: when a = 2
Unique solution: when a not equal to 2
Infinite Many solutions: Never
 

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