I obtained only one value v = -1.5.

v = 2 gets infinitely many solutions.

A non matrix method is easily available for this question.

3x + vy = 4
(4v - 2)x + 4y = 8

For the equation to NOT have exactly one solution, the gradients of the lines must be the same.

Or in other words,
The ratio of the coefficients of x in both equations must be equal to the ratio of the coefficients of y in both equations.

(4v - 2)/3 = 4/v

If the two lines have infinitely many solutions, the ratio above also equals the ratio of the two numbers on the other side.

(4v - 2)/3 = 4/v = 8/4

Otherwise, the lines do not meet.

For the lines to not meet,

(4v - 2)/3 = 4/v is NOT equal to 8/4 = 2.

So v = 2 is rejected if it comes up.

4v2 - 2v = 12
4v2 - 2v - 12 = 0
2v2 - v - 6 = 0
(2v + 3) (v - 2) = 0
v = -3/2 is the only answer

This question should not appear in Sec 4 A Maths.

This question can be seen as a secondary 2 or secondary 3 E Math question if the ratio method is used. It’s based on the understanding that two parallel lines do not meet while two identical lines meet everywhere.

Look at the following.

x + y = 5
2x + 2y = 10

These two are the same when simplified, so obviously they intersect everywhere along the line.

Look at the following.

x + y = 5
2x + 2y = 50

The second equation simplifies to x + y = 25. Upon inspection, both equations are parallel to each other and do not meet.

This is the basic idea behind the ratio method which I have used a moment ago.