# Find the values of h that makes the following system inconsistent: x + 2y + hz = hx + 3y + 4z = h2x + 4y + 5z = 5

Tibor Pejić | Certified Educator

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There are number of different ways to solve this: using Cramer's ruleRouché–Capelli theorem, some forms of matrix decomposition... But probably the best way for you would be to simply thy to solve the system and see what you get.

`x+2y+hz=h`

`x+3y+4z=h`

`2x+4y+5z=5`

Now 1) subtract first row from the second row and

2) subtract first row multiplied by 2 from the third row

`x+2y+hz=h`

`0+y+(4-h)z=0`

`0+0+(5-2h)z=5-2h`

Now, only way we would have some problem with solving this system of equations is if `5-2h=0=>h=5/2`  (then the last row would be only zeros).

If we put that value of `h` back into our system we would get infinitely many solutions, to be exact solutions would be defined with

`y=15/2-3x`

`z=-5+2x,` `x in RR`

In all the other cases the system has unique solution. And since a system of linear equations is inconsistent when it has no solution there exist no real number h for which this system is inconsistent.

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