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I have a task which teacher gave me and I can't figure out this determinant with Cramers rule. It's 2x2 , but there is Z. I move Z and 6 to other end of equals, but what do I do with Z?

\begin{cases} 2x-2y+6=4 \\ x-3y-z=1\end{cases}

I change it to

\begin{cases} 2x-2y=4-6 \\ x-3y=1+z\end{cases}

What to do when I need determinant of X and Y ? For determinant it would be: 2*(-3)-(-2*1)... But what about X? Is it: ((4-6)(-3))-(-2(1+z)).. ?

Emilio Novati
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  • I guess that you are not using cramer's rule correctly. The determinant of x includes only coefficients of y and the constants. Not the coefficients of x and likewise with y. – Adesh Tamrakar Oct 04 '15 at 12:44
  • You are correct: now find $x=\dfrac{\Delta_x}{\Delta}$. – Emilio Novati Oct 04 '15 at 12:50
  • Hint: use the all the things except having x and y coefficient as constant. I think now you shouldn't have any trouble in solving until you don't know cramer's rule correctly. But if you ask I can solve it for you. – Adesh Tamrakar Oct 04 '15 at 12:58

1 Answers1

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Let's rewrite this system of equations $$\begin{cases} 2x-2y=-2 \\ x-3y=1+z\end{cases}$$ as a matrix equation: $$\pmatrix{2 & -2 \\ 1 & -3}\pmatrix{x \\ y} = \pmatrix{-2 \\ 1+z}$$

For Cramer's rule, we'll need $\Delta$, $\Delta_x$, and $\Delta_y$. So let's find those:

$$\begin{align}\Delta &= \left|\begin{array}{cc} 2 & -2 \\ 1 & -3 \end{array}\right| = -6-(-2) = -4 \\ \Delta_x &= \left|\begin{array}{cc} -2 & -2 \\ 1+z & -3 \end{array}\right| = 6-(-2-2z) = 8+2z \\ \Delta_y &= \left|\begin{array}{cc} 2 & -2 \\ 1 & 1+z \end{array}\right| = (2+2z)-(-2) = 4+2z\end{align}$$

From this we can immediately find $x$ and $y$ in terms of $z$:

$$\color{red}{x} = \frac{\Delta_x}{\Delta} = \frac{8+2z}{-4} \color{red}{= -2-\frac12z} \\ \color{red}{y} = \frac{\Delta_y}{\Delta} = \frac{4+2z}{-4} \color{red}{= -1-\frac12z}$$