Consider for every real number a the linear system of equations:
{ x +( a + 1 )y + a^2 z = a^3
{ ( 1 - a )x +( 1 - 2a )y = a^3
{ x +( a + 1 )y + az = a^2
b)Find the values of a for which the system has no solution, infinitely many solutions, and a unique solution
Judging from what wolframalpha gets at
http://www.wolframalpha.com/input/?i=solve+x+%2B%28+a+%2B+1+%29y+%2B+a^2+z+%3D+a^3%2C+%28+1+-+a+%29x+%2B%28+1+-+2a+%29y+%3D+a^3%2C+x+%2B%28+a+%2B+1+%29y+%2B+az+%3D+a^2
I suspect a typo.
note first that if a = 1, the first and third rows are identical (infinite number of solutions)
then
calculate the determinant of the coefficient matrix. Find its zeros.
One will be a = 0 Look at second equation when a = 0, x+y = 0 and first equation gives x+y = 0 and third equation gives x+y = 0 so any old x = -y will do for any old z
To find the values of "a" for which the system of linear equations has no solution, infinitely many solutions, and a unique solution, we need to analyze the properties of the system.
Let's start by writing the system of equations in matrix form:
| 1 (a+1) a^2 | | x | | a^3 |
| 1-a 1-2a 0 | | y | = | a^3 |
| 1 (a+1) a | | z | | a^2 |
We can use Gaussian elimination or a similar method to solve the system. However, let's simplify the system first to analyze its properties.
Equation 2 can be simplified by multiplying it by (a+1):
(1-a)(x) + (1-2a)(y) = a^3
(-ax + x - ay) + (-2ay + 2y) = a^3
(-ax - 2ay + x + 2y) = a^3
Now let's compare it with Equation 1:
x + (a+1)y + a^2z = a^3
We can see that Equations 1 and 2 are equivalent when:
-a = 1 and -2a = 0
From the second equation, we get a = 0. Substituting this value in the first equation, we find a contradiction (-0 ≠ 1). Therefore, these equations are inconsistent. The system has no solution when a = 0.
To determine if the system has infinitely many solutions or a unique solution, we need to consider its determinant.
The determinant of the coefficient matrix is:
D = | 1 (a+1) a^2 |
| 1-a 1-2a 0 |
| 1 (a+1) a |
Expanding this determinant using cofactor expansion along the second row, we get:
D = (1-2a) ( (1)(a) - ((a+1)(1-a)) ) - (a+1) ( (1-a)(a) - (1)(a^2) )
= (1-2a) (a - a^2 - (a+1)(1-a)) - (a+1)(a - a^2 - a^2)
= (1-2a) (a - a^2 - a + a^2 + 1 - a) - (a+1)(a - 2a^2)
= (1-2a) (0) - (a+1)(a - 2a^2)
= - (a+1)(a - 2a^2)
If D = 0, the system has infinitely many solutions.
If D ≠ 0, the system has a unique solution.
Let's set the determinant equal to zero and solve for "a":
-(a+1)(a - 2a^2) = 0
This equation is satisfied when a = -1 (the first factor is zero) and a = 0 (the second factor is zero).
Therefore, the system of linear equations has no solution when a = 0. It has infinitely many solutions when a = -1. For any other value of "a", the system has a unique solution.