2. For which value(s) of a does the system of equations
(a-3)x+y=0
x+(a-3)y=0
have non-trivial solutions
This is a homogeneous equation of which the right-hand side is populated with zeroes.
For a homogeneous equation to have non-trivial roots, the determinant of the left-hand side must be zero.
So expand the determinant in terms of a, equate to zero and solve for a.
Since the resulting equation is a quadratic, you should expect to have two roots. In this case, the two roots are distinct.
To find the values of "a" for which the given system of equations has non-trivial solutions, we can start by setting up the augmented matrix:
[ a-3 | 1 ]
[ 1 | a-3 ]
Next, we will perform row operations to bring the matrix to row-echelon form:
1. Multiply Row 1 by (a-3) and subtract Row 2 from it.
R1 ← (a-3)R1 - R2
This will give us:
[ a(a-3) - 1(a-3) | (a-3) - (a-3) ]
[ 1(a-3) | a-3 ]
2. Simplify the result:
[ a^2 - 3a - a + 3 | 0 ]
[ a - 3 | 0 ]
3. Factor out common terms in the first row:
[ (a^2 - 4a + 3) | 0 ]
[ a - 3 | 0 ]
Now, we need to find the "a" values that make the system of equations have non-trivial solutions. For a non-trivial solution, the system must have either infinitely many solutions or no solutions at all. This can happen when the determinant of the coefficient matrix is zero.
The determinant of the coefficient matrix is given by (a^2 - 4a + 3)(a - 3).
Setting the determinant equal to zero:
(a^2 - 4a + 3)(a - 3) = 0
Now, we can solve for "a":
(a^2 - 4a + 3) = 0 OR (a - 3) = 0
Using the quadratic formula for the first equation, we have:
a = (4 ± √(16 - 4(1)(3))) / 2
a = (4 ± √(16 - 12)) / 2
a = (4 ± √4) / 2
a = (4 ± 2) / 2
a = 3 OR a = 1
Therefore, the system of equations has non-trivial solutions for a = 1 and a = 3.
To determine the values of 'a' for which the system of equations has non-trivial solutions, we need to find when the coefficients of 'x' and 'y' become zero simultaneously.
Let's start by solving the first equation: (a-3)x + y = 0.
To make the coefficient of 'x' zero, we need a - 3 to equal zero:
a - 3 = 0
a = 3
Now, let's solve the second equation: x + (a-3)y = 0.
Substituting the value of 'a' we just found, we have:
x + (3-3)y = 0
x + 0y = 0
x = 0
So, when 'a' is equal to 3, both the coefficients of 'x' and 'y' become zero simultaneously. As a result, the system of equations has non-trivial solutions when a = 3.