Solving Linear Systems: True or False?

In summary: For (2), there is no known way to do it, so you would get an error. For (3), you would need to find the determinant of the matrix and use the rule of Sarrus. For (4), I believe you are doing this for practice, and this is not a homework question.
  • #1
System
42
0
Hi

Answer as T or F:

1) Every linear system consisting of 3 equations in 4 unknowns has infinitely many solutions.
2) If A and B are 3 x 3 matrices , then det(AB - A (B^T) ) = 0
3) If A and B are n x n matrices, nonsingular matrices and AB=BA, then A(B^-1) = (B^-1)A
4) If A is a singular n x n matrix, then Aadj(A)=0

For (1):
I think its true
since # of columns > # of rows
so we will have recall a parameter
and this means we will a infinitely many solutions

For (2):
I do not know how to do it =(

For (3):
I got the answer, its true
but how ?

For (4):
I completely stopped here :/

Any help please?


this is not for my homework
I swear
am solving these for fun
 
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  • #2
A hint for the second one: factor out A and use the rule det(XY)=det(X)det(Y). Then, can you conclude something about det(B - B^T)? What are the diagonal elements of that matrix? What is the general element of that matrix? Use the rule of Sarrus to calculate the determinant.
 
  • #3
A hint for the third one: multiply the left side of the equation AB = BA by B^-1.
 
  • #4
A hint for the fourth one: take an element cij from the matrix C = Aadj(A) and write down what it equals. Can you conclude something from that?
 
  • #5
Correct me if I'm wrong, but for (1), I believe the answer is False. While you are correct most of the time, you have to consider the situation when the system might have 0 solutions. In general - if r is the rank of the matrix - r < n and r < m implies the system will have 0 or an infinite amount of solutions. Thus, not every system described will have an infinite amount of solutions.
 
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FAQ: Solving Linear Systems: True or False?

Is it possible to have a system of linear equations with no solution?

Yes, it is possible to have a system of linear equations with no solution. This is known as an inconsistent system, and it means that the equations are contradictory and do not have a common solution.

Can a system of linear equations have more than one solution?

Yes, it is possible for a system of linear equations to have more than one solution. This is known as a dependent system, and it means that the equations are not independent and have multiple solutions that satisfy all the equations.

Are there any shortcuts or tricks for solving linear systems?

Yes, there are a few shortcuts and tricks that can be used to solve linear systems. These include using the elimination method, substitution method, and graphing method. However, the best approach may vary depending on the specific system of equations.

Can a linear system have an infinite number of solutions?

Yes, it is possible for a linear system to have an infinite number of solutions. This is known as an inconsistent system, and it means that the equations are not independent and have an infinite number of solutions that satisfy all the equations.

Can a linear system have no solution or an infinite number of solutions?

No, a linear system cannot have both no solution and an infinite number of solutions. It can only have one of these outcomes. A system with no solution means that the equations are contradictory, while a system with an infinite number of solutions means that the equations are not independent and have multiple solutions.

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