# Wrong solution using Cramer's rule?

I have tried to solve the following systems of equations: $$\left\{ \begin{array}{rrrcl} x & + y & + z &=& 1 \\ 2x & + 5y & + 2z &=& 4 \\ 4x & + 2y & + 3z &=& 5 \end{array} \right.$$

The determinant of the matrix of the coefficients is $$3$$ and the other determinants are $$-8$$, $$-2$$, $$7$$. So the solutions are $$x = - \frac{8}{3} \,, \quad y = - \frac{2}{3} \,, \quad z = \frac{7}{3} \,.$$

However, these values doesn't satisfy the system. Furthermore, Wolfram Alpha says that the solutions are $$x = 8/3, y = 2/3, z = -7/3$$. The roots are the same in numerical value, but with opposite sign. What is the error here?

• The determinant of $\left[\begin{smallmatrix}1&1&1\\2&5&2\\4&2&3\end{smallmatrix}\right]$ is $-3$, not $3$. – JMoravitz Jul 26 '17 at 4:39

At least$$\Delta=1\cdot5\cdot3+1\cdot2\cdot2+4\cdot1\cdot2-4\cdot5\cdot1-1\cdot2\cdot2-3\cdot2\cdot1=-3$$ $$\Delta_x=1\cdot5\cdot3+1\cdot2\cdot5+4\cdot1\cdot2-5\cdot5\cdot1-1\cdot2\cdot2-3\cdot4\cdot1=-8$$ $$\Delta_y=1\cdot4\cdot3+1\cdot4\cdot2+5\cdot1\cdot2-4\cdot4\cdot1-1\cdot2\cdot5-3\cdot2\cdot1=-2$$ and $$\Delta_z=1\cdot5\cdot5+1\cdot2\cdot2+4\cdot1\cdot4-4\cdot5\cdot1-1\cdot2\cdot4-5\cdot2\cdot1=7.$$