Consider the following 2x2 system of linear equations,
x + y = 3
x + (a^2 - 8)y = a
If the system have a unique solution, then a takes any value except a=-3 and what?
To find the values of a for which the system of linear equations has a unique solution, we can solve the system by using the method of elimination or substitution.
Let's solve the system using the method of elimination:
Given system of equations:
1) x + y = 3
2) x + (a^2 - 8)y = a
To eliminate x, we can subtract equation 1 from equation 2:
(x + (a^2 - 8)y) - (x + y) = a - 3
(a^2 - 9)y = a - 3
y(a^2 - 9) = a - 3
y = (a - 3) / (a^2 - 9)
Now we can substitute this value of y back into equation 1 to find x:
x = 3 - y
x = 3 - (a - 3) / (a^2 - 9)
So, the system of equations has a unique solution for any value of a except those that make the denominator of the expression for x equal to zero. In other words, we need to find the values of a for which (a^2 - 9) is not equal to zero:
(a^2 - 9) ≠ 0
To solve this equation, we can find the values of a that make the numerator and denominator equal to zero:
Factorizing (a^2 - 9), we get:
(a - 3)(a + 3) = 0
Setting each factor equal to zero:
a - 3 = 0 -> a = 3
a + 3 = 0 -> a = -3
So, the values of a that make the denominator zero are a = 3 and a = -3. Thus, the system of equations has a unique solution for any value of a except a = 3 and a = -3.