A set of homogeneous simultaneous equations is given by

x + ky = 0
kx + 3y = 0
Calculate the two values of k that lead to non-trivial solutions to these
equations and express y in terms of x for the two values.

I thought that for the solutions to be no trivial the determinant of the coefficients had to equal zero. From this I got k=+/- sqrt 3 but I am not sure of this is right.

Rewrite as

y = (-1/k)x and
y = (-k/3)x
There are an infinite number of solutions if the lines coincide, which happens when

1/k = k/3
k = +/-sqrt3
In this case,
y = x/sqrt3 or -x/sqrt3

Otherwise, there are no solutions at all, other than x=y=0. That may be what they call the "trivial" solution. Nevertheless, it is still a solution.

To find the non-trivial solutions of a set of homogeneous simultaneous equations, you are correct that the determinant of the coefficients must be equal to zero. Let's go through the steps to solve this problem.

Given equations:
1) x + ky = 0
2) kx + 3y = 0

Step 1: Write the coefficient matrix and calculate its determinant
The coefficient matrix is formed by taking the coefficients of the variables x and y in the given equations:

| 1 k |
| k 3 |

The determinant of this matrix must be equal to zero for non-trivial solutions.

Step 2: Calculate the determinant
To find the determinant of this matrix, we can use the rule for a 2x2 matrix:
Det = (a*d) - (b*c)

In our case:
Det = (1 * 3) - (k * k) = 3 - k^2

Step 3: Set determinant equal to zero and solve for k
To find the values of k that lead to non-trivial solutions, we set the determinant equal to zero and solve the equation:

3 - k^2 = 0

Step 4: Solve the equation for k
Rearranging the equation:
k^2 = 3

Taking the square root of both sides (remember to consider both positive and negative square roots):
k = ± sqrt(3)

So, you are correct that the values of k that lead to non-trivial solutions are k = ± sqrt(3).

Step 5: Express y in terms of x for the two values of k
For k = sqrt(3):
From equation 1 (x + ky = 0):
x + sqrt(3)y = 0
x = -sqrt(3)y

For k = -sqrt(3):
From equation 1 (x + ky = 0):
x - sqrt(3)y = 0
x = sqrt(3)y

In both cases, we have expressed y in terms of x: y = -(1/sqrt(3))x and y = (1/sqrt(3))x.

Therefore, the two values of k that lead to non-trivial solutions are k = ± sqrt(3), and y can be expressed in terms of x as y = -(1/sqrt(3))x and y = (1/sqrt(3))x for the two respective values of k.