pickslides, If it were true that deleting y from the equations resulted in the two equations 4x+ z= 0, 4x+ z= 0, then the matrix would NOT be invertible. That is the problem mmounds is compaining about. If that were the case then 4x+ z= 0 would be satisfied for any x, z satisfying z= -4x. Pick any value for x, and calculate z from that. Put that into either of the other two equations and solve for y: there exist an infinite number of solutions.
However while one equation certainly is 4x+ z= 0, the other two are x-y+3z=-4 and x+2y+z=8.
Multiply the first equation by 2 to get 2x- 2y+ 6z= -8. Add that to the second equation and y is eliminated: 2x+ x-2y+ 2y+ 6z+ z= -8+ 8 or 3x+ 7z= 0, NOT 4x+ z= 0. It is easy to see that the only x and z satisfying 4x+ z= 0 and 3x+ 7z= 0 is x= z= 0. Putting that into x-y+3z=-4 we get -y= -4 so y= 4. Putting it into x+2y+z=8, 2y= 8 which again is satisfied by y= 4. All three equations are satisfied by x= 0, y= 4, z= 0.
In fact, finding the inverse matrix is seldom the best way to solve matrix equations. Row reduction of the matrix is typically much simpler.
Thank you for the response, i have a similar question
i know using the substitution method plays out like so...
I have to express the answer in interval notation....how would I do that, and since the answer is 0=0 is their a value for x and y at all
can by any real number. Try plugging in any real number for , and the equations will always hold true.
I think in interval notation, this would be .
If you were to choose any value of x and then calculate y=2x+4, you would get x and y that satisfy the system of equations. For example, taking x= 1, y= 6 which satisfy both y= 2x+ 4 and 10x-5y=-20. If you take x= 5, y= 14 also satisfy both equations.
In general, if, solving a system of equations, you arrive at "0= 0" there are an infinite number of solutions. If you arrive at "a= 0" where a is not 0, there are no solutions.