Where is the Mistake? Solving a System of Equations

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In summary, the mistake is in the operation $R_3 \to (a-2)R_3 -3R_2$, which results in the third row disappearing and causing an underfined system.
  • #1
Yankel
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Hello

I have a system of equations with the question "for which values of a the system has: a single solution, no solution and infinite number of solutions". In addition, I have some solution, and I need to find the mistake in the solution, I need some help with it...

So, for which values of a, the next system:

2x+y=1
4x+ay+z=0
3y+az=2

has a single solution, no solution and infinite number of solutions ?

View attachment 434

if a=2, the matrix has proportional rows (r2 and r3), and we get infinite number of solutions. However, if we set a=2 in the original system, there is a single solution.

Where is the mistake ?

Thanks !
 

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  • #2
You have it there in your graphic.

a = -1

This results in the third row disappearing, making an underfined system and infinitely many solutions.

a = 3

This results in an inappropriate result indicating no solution.

a = Anything Else

Unique Solution.
 
  • #3
yes, but put a=2, you get

2 1 0 1
0 0 1 -2
0 0 -3 6

now multiply the 3rd row by (1/3) and add the 2nd row to it, you get:

2 1 0 1
0 0 1 -2
0 0 0 0

which is not a single solution...something is wrong here...
 
  • #4
Yankel said:
yes, but put a=2, you get

2 1 0 1
0 0 1 -2
0 0 -3 6

now multiply the 3rd row by (1/3) and add the 2nd row to it, you get:

2 1 0 1
0 0 1 -2
0 0 0 0

which is not a single solution...something is wrong here...
The operation $R_3 \to (a-2)R_3 -3R_2$ is the culprit. When $a=2$, that has the effect of killing the third row and replacing it by a multiple of the second row. So it is hardly surprising that rows 2 and 3 are then proportional. This is not so in the original system, it is something that you have introduced by performing a dodgy operation.

In fact, operations of the form $R_i\to pR_i+qR_j$ are best avoided unless $p=1.$ What you should have done instead of $R_3 \to (a-2)R_3 -3R_2$ is $R_2\to \frac1{a-2}R_2.$ When $a=2$ that gives a division by zero, which should have sounded an alarm bell and warned you to treat that case separately.
 
  • #5
Understood, thank you !
 

FAQ: Where is the Mistake? Solving a System of Equations

What is a system of equations?

A system of equations is a set of equations that contain multiple variables and must be solved simultaneously. This means that all of the equations in the system must be satisfied by the same set of values for the variables.

How do I know if there is a mistake in solving a system of equations?

If there is a mistake in solving a system of equations, the solution will not satisfy all of the equations in the system. This can be checked by plugging the solution into each equation and seeing if it results in a true statement.

Can a system of equations have more than one solution?

Yes, a system of equations can have more than one solution. This is known as an infinite number of solutions, where the equations are satisfied by any value for the variables.

What are the different methods for solving a system of equations?

There are several methods for solving a system of equations, including substitution, elimination, and graphing. Each method has its own advantages and may be more suitable for certain types of systems.

How can I avoid making mistakes when solving a system of equations?

To avoid mistakes when solving a system of equations, it is important to double check your work and make sure your solution satisfies all of the equations in the system. It can also be helpful to use multiple methods to solve the system and compare the results to ensure accuracy.

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