Carilah solusi persamaan linear berikut menggunakan cramer's rule X-3y+z=4 2x-y=-2 4x-z=0

Cramer,
[tex] \left[\begin{array}{ccc}1&-3&1\\2&-1&0\\4&0&-1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}4\\-2\\0\end{array}\right] \\ \\ x=\frac{\left|\begin{array}{ccc}4&-3&1\\-2&-1&0\\0&0&-1\end{array}\right|}{\left|\begin{array}{ccc}1&-3&1\\2&-1&0\\4&0&-1\end{array}\right|}=\frac{4+0+0-(0+0-6)}{1+0+0-(-4+0+6)}=\frac{4+6}{1-2} = -10 \\[/tex]

[tex]y=\frac{\left|\begin{array}{ccc}1&4&1\\2&-2&0\\4&0&-1\end{array}\right|}{\left|\begin{array}{ccc}1&-3&1\\2&-1&0\\4&0&-1\end{array}\right|}=\frac{2+0+0-(-8-8))}{1+0+0-(-4+0+6)}=\frac{18}{1-2} = -18[/tex]

[tex]z=\frac{\left|\begin{array}{ccc}1&-3&4\\2&-1&-2\\4&0&0\end{array}\right|}{\left|\begin{array}{ccc}1&-3&1\\2&-1&0\\4&0&-1\end{array}\right|}=\frac{0+24+0-(-16+0+0)}{1+0+0-(-4+0+6)}=\frac{24+16}{1-2} = -40[/tex]