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Math100
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No.Math100 said:Homework Statement
For what value(s) of k is the linear system with augmented matrix inconsistent?
Homework Equations
None.
The Attempt at a Solution
I know that inconsistent means no solution. So do I set k=2k and solve for k?View attachment 219506
So what value of k would make it so that this system of equations doesn't have a solution? One suggestion has already been given.Math100 said:k+2=1
1+2k=1
These are the world's easiest equations to deal with. Why are you having trouble?Math100 said:k+2=1
1+2k=1
The augmented matrix in the OP represents the system of equationMath100 said:k+2=1
1+2k=1
Zero determinant does not necessarily imply that the system is inconsistent.NFuller said:The most direct way to declare that a system is inconsistent is by looking for the value of ##k## which causes the determinant of the matrix to be zero. Since this is a fairly simple system however, you could easily pick a value of ##k## which causes one of the rows to be a multiple of the other.
What did you get for k?Math100 said:Got it!
Thank you so much!ehild said:Zero determinant does not necessarily imply that the system is inconsistent.
An inconsistent linear system is a set of linear equations that does not have a solution. This means that there is no set of values that can make all of the equations true at the same time.
A linear system is inconsistent if it has no solution, which means that the equations are contradictory or impossible to satisfy. This can be determined by graphing the equations and seeing if they intersect at one point, which would indicate a unique solution, or if they are parallel lines, which would indicate no solution.
K is a constant that is used in the equation to represent a specific value that is unknown. In an inconsistent linear system, k can be any value, but it is usually used to represent a value that would make the equations contradictory, resulting in no solution.
No, an inconsistent linear system does not have any solutions. This means that there is no set of values that can make all of the equations true at the same time. If there were multiple solutions, it would indicate a consistent linear system.
Solving for k in an inconsistent linear system can be used to model real-life situations such as balancing chemical equations, determining the break-even point in business, and analyzing market trends. It can also be used in engineering to find the optimal design for a structure by considering different values for k.