a question with linear equations that could be solved by determinants?

Question

i would like to find out that given the following equations
6 x + p y = -3

3 x + 3 y = q

i would like to determine values for p and q for which the equations have no solutions (for x and y).

There are many right answers to this question. ijust would like to know one correct pair of values.

thanks for any help

p, q must be exact numbers eg not decimals

Answer 1

The (simultaneous) equations have no solution where the determinant equals 0. This is because division by 0 leaves no result. Since the determinant is 18 - 3p, then set this equal to 0 and solve for p. You will get p = 6. It doesn't matter what the value of q is since for any value of q you will not get a result so long as p = 6. You can therefore choose any value for q.

Answer 2

This system can be written in the form Az = b, where z is the vector of unknowns, [ x y ], b is the right-hand side vector, [ -3 q ], and A is the matrix
[ 6 p
3 3 ]

The determinant of this matrix is 6*3-3*p = 18-3p.

If the determinant is nonzero, then this system has a unique solution for any right-hand side b, so the value of q is irrelevant in this case. The system is said to be "nonsingular", meaning that the only solution to the system Az = 0 is the "trivial" solution z = [ 0 0 ].

If the determinant is zero, then the system is said to be "singular". That means it is possible to find a nonzero vector z = [ x y ] such that Az = 0. In that case, any multiple of z would also be a solution to Az=0.

If A is singular, then a given system Az = b will either have no solution, or infinitely many solutions. The determinant cannot tell you which is the case; it depends on the right-hand side b. Here, the value of q is relevant.

To have infinitely many solutions, it must be possible to express b as a linear combination of the columns of A. That is, there must exist constants x and y such that
[ 6 3 ]x + [ 6 3 ]y = [ -3 q ].
We say that b must be an element of the "column space" of A.

The easiest way to check this is to perform Gaussian elimination on the augmented matrix
[ A b ] =
[ 6 6 -3
3 3 q ].

Subtract 1/2*first row from the second row and you get
[ 6 6 -3
0 0 q+3/2 ]
For there to be any solution, we must have q+3/2=0, for otherwise we must solve the equation 0*x + 0*y = nonzero, which is impossible. Therefore, q=-3/2.

If q=-3/2, then we have infinitely many solutions. To describe them, we can examine the remaining equation after elimination,

6x + 6y = -3.

First, we can set y=0, and obtain 6x = -3, so x=-1/2. So one solution is z = [ -1/2 0 ].

To this solution, we can add any solution to the homogeneous system Az=0, of which there are infinitely many. To describe them, we can set y=1, and solve the homogeneous system 6x+6y = 0. We get x=-1, so our general solution is

z = [ -1/2 0 ] + y*[ -1 1 ]

where y is arbitrary.

Some of this is overkill for this 2-by-2 case, but the ideas readily extend to larger systems.

Answer 3

If there's no solution, the two lines represented by the equations have to be parallel, so the slopes have to be the same. That tells you that p has to be 6.

As long as the slopes are the same, the value of q can be any number. Just pick one.

Answer 4

try p=6 and q=6

Answer 5

p=-6,q =0; anything but ... well u dont want to know