x-y+z=1..........................(1)
2ax+3y+az=4a.................(2)
x+(a-1)y=a......................(3).
To find the conditions for which there is no solutions to the system of equations.
The no solution to the above case is there only when the determinant formed of the coefficients of x,y and z of the equations is nonzero. I.e.,:
|[(1,-1,1),(2a,3,a),(1,a-1,0)]| is not equal to zero.
Solving, we get:
|[(1,-1,1),(2a,3,a),(1,a-1,0)]| = (3+2a)(-1)+a^2 =a^2-2a-3 is non zero. Or (a-3)(a+1) is nonzero.
So if a is not equal to 3 and a is not equal to -1, then the system of equations has unique solution.
Other than unique solution cases:
When a=3 or a=-1:
When a=3, the given equations become:
x-y+z=1...........................(4)
6x+3y+3z=12 or
2x+y+z = 4.......................(5) and
x+2y=3.............................(6)
Eliminating z between (4) and (5) by subtraction, we get:
x+2y=3 and this is exacly the equation (6). So, the two equations have reduced to one equation, resulting in x and y having many solutions like: For any given x, y = (x-3)/2 or,
For any given x like: 0, 1, 2, 3, 4, 5
y value = (x-3)/2 : -3/2, -1, -1/2, 0, 1/2, 1
Thus many solutions, the rank of the 3X4 matrix,
[(1,-1,1,1),(2a,3,a,3a),(x,a-1,0,a],should be less than 3. In this case, it is so as the 3 three equations reduces to only 2 equations since the third equation is dirivable from the first 2 equations. Hence the infinite solution case exist when a=3.
No solution case:
When a =-1, the equatios a become:
x-y+z =1..................(i)
-2x+3y-z = -3 .........(ii)
x-2y = -1................(iii)
Adding (i) and (ii): -x+2y=-2 or
x-2y = 2 .............(iv), but from (iii) x-2y = -1 . Try solve this .You cannot eliminate one of x or y. Further (iii) - (iv) gives an absurd result like:
0 = 3 an all time false statement. So a = -1 is a no solution case.
Hope this helps.
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