Have you tried to solve the system? The first thing I notice is that if you multiply the second equation by "a" and add that to the first equation, you eliminate "x": (ax- 2y+ 3z)+ (-ax+ ay- abz)= (2- a)y+ (3- ab)z= 5- 3a. And that if you multiply the second equation by "2" and add that to the third equation, you also eliminate x: (2x- cy- 2z)+ (-2x+ 2y- 2bz)= (2- c)y- (2+ 2b)z= d- 6. Can you solve those two equations for y and z? If not what would stop you?Ankit said:Find the surface in terms of a,b,c on which the following system of equations has no
solution
ax-2y+3z=5
-x+y-bz=-3
2x+cy-2z=d
Could there be any values of a,b,c,d for which the system has infinite solution? (Justify).
When graphing the equations, if the lines do not intersect at any point, then the system has no solution. This means that there is no combination of values for the variables that can satisfy both equations simultaneously.
If the equations are graphed and the lines overlap, then the system has infinite solutions. This means that any value for the variables that satisfies one equation will also satisfy the other equation.
No, a system of equations can only have either no solution or infinite solutions. It is not possible for it to have both at the same time.
If, when solving the system of equations algebraically, you end up with a statement that is always false, such as 0=3, then the system has no solution. If you end up with a statement that is always true, such as 0=0, then the system has infinite solutions.
Knowing whether a system has no solution or infinite solutions can help in finding the best approach to solving the system. It can also provide important information about the relationship between the equations and the variables involved.