Linear Algebra & Linear Systems

Question

Consider the linear system

x1 + x2 - bx3 + (b-a)x4 = b-2

x1 + ax2 - bx3 = 1

2x1 + 2x2 + ax4 = 0

x1 + x2 -bx3 = 0

for some real numbers a, b = R

(a) Determine the values of a and b such that the system has no solution.

(b) Determine the values of a and b such that the system has only one solution. Write down the unique solution.

(c) Determine the values of a and b such that the system has infinitely many solutions with 1 parameter in the general solution. Write down the general solution.

(d) Is it possible for the system to have infinitely many solutions with 2 parameters in the general solution? Explain your answer.

I have some difficulty in understanding linear systems, I did tried my best to get the above system into a rref form but I could not understand which variables to shift upwards and which to move down, I understand that for a no solution the left side should generally be all 0 and there’s a value on the right side.

Answer 1

(a) No solution:

For a system to have no solution, the RREF should have a leading 1 in a row followed by all zeros below and a non-zero constant on the right-hand side. This represents a situation where one variable is completely independent (no other variable affects it), but the equation doesn't hold true for any value of that variable.

Here, we can manipulate the system without reducing it to RREF:

1. Observe that the last equation (x1 + x2 - bx3 = 0) doesn't involve x4. This means we can solve for x1 and x2 in terms of x3 and x4 from the first three equations and substitute them back into the last equation.
2. Combining the first and fourth equations, we get (2a-b)x4 = b-4. Since we want the system to have no solution, set (2a-b) = 0 and b-4 ≠ 0. This leads to a = b/2 and b ≠ 4.

(b) One solution:

For one unique solution, the RREF should have a leading 1 in each row, and the last column shouldn't have any leading 1s (indicating all variables are dependent).

Here, we can again avoid RREF:

1. The equations already suggest a dependence between all variables.
2. Looking at the last equation, if x3 = 0 (which is possible), then x1 and x2 can be any values that satisfy the first equation (x1 + x2 = b-2).
3. Therefore, the system has one unique solution only when x3 = 0 and any values for x1 and x2 that satisfy x1 + x2 = b-2.

(c) Infinitely many solutions with 1 parameter:

For infinitely many solutions with 1 parameter, the RREF should have two leading 1s followed by a row with a leading non-zero number below, and the last two columns should be dependent (meaning one variable can be expressed in terms of the other).

Without RREF, we can observe:

1. We can eliminate x3 by adding the first and third equations. This gives 3x1 + 3x2 + (a+b)x4 = b-2.
2. If a+b = 3, then x3 becomes dependent on x1, x2, and x4. With one variable dependent, the system becomes dependent, and we can solve for one variable in terms of another.

Therefore, the system has infinitely many solutions with 1 parameter when a + b = 3. We can rewrite the equations in terms of x4 (assuming x4 ≠ 0):

1. x1 = (b-2 - 3x4)/(3+a)
2. x2 = (b-2 - ax4)/(3+a)
3. x3 = (x1 + x2 - b)/(-b) (substitute the x1 and x2 values from above)

(d) Infinitely many solutions with 2 parameters:

This is not possible in this system because RREF would require three leading 1s in the first three rows and all zeros below, which is impossible with four variables involved. This situation would indicate no dependence between any variables, contradicting the system having infinitely many solutions with two parameters