Ada yang tau dan bisa jabarkan penyelesaiannya gimana dapat nilai a b dan c kalo bisa cara crammer ya 1) 4a-3b+2c=-5 3a+6b+4c=10 -4a-3b+5c=-6 Tolong dijawab y p

penyelesaian dengan Cramer's rule:
1.)
[tex]a= \frac{\left|\begin{array}{ccc}-5&-3&2\\10&6&4\\-6&-3&5\end{array}\right|}{\left|\begin{array}{ccc}4&-3&2\\3&6&4\\-4&-3&5\end{array}\right|} = \frac{24}{291} [/tex]
[tex]b= \frac{\left|\begin{array}{ccc}4&-5&2\\3&10&4\\-4&-6&5\end{array}\right|}{\left|\begin{array}{ccc}4&-3&2\\3&6&4\\-4&-3&5\end{array}\right|} = \frac{495}{291} [/tex]
[tex]c = \frac{\left|\begin{array}{ccc}4&-3&-5\\3&6&10\\-4&-3&-6\end{array}\right|}{\left|\begin{array}{ccc}4&-3&2\\3&6&4\\-4&-3&5\end{array}\right|} = \frac{-33}{291}[/tex]

2.)
[tex]a= \frac{\left|\begin{array}{ccc}3&5&-2\\7&-3&5\\-10&6&3\end{array}\right|}{\left|\begin{array}{ccc}4&5&-2\\6&-3&5\\5&6&3\end{array}\right|} = \frac{-496}{-223} [/tex]
[tex]b= \frac{\left|\begin{array}{ccc}4&3&-2\\6&7&5\\5&-10&3\end{array}\right|}{\left|\begin{array}{ccc}4&5&-2\\6&-3&5\\5&6&3\end{array}\right|} = \frac{495}{-223} [/tex]
[tex]c = \frac{\left|\begin{array}{ccc}4&5&3\\6&-3&7\\5&6&-10\end{array}\right|}\left|\begin{array}{ccc}4&5&-2\\6&-3&5\\5&6&3\end{array}\right|} = \frac{580}{-223}[/tex]

3.)
[tex]a= \frac{\left|\begin{array}{ccc}10&-2&5\\5&4&-1\\-10&-3&5\end{array}\right|}{\left|\begin{array}{ccc}3&-2&5\\5&4&-1\\6&-3&5\end{array}\right|}= \frac{325}{-82} [/tex]
[tex]b= \frac{\left|\begin{array}{ccc}3&10&5\\5&5&-1\\6&-10&5\end{array}\right|} {\left|\begin{array}{ccc}3&-2&5\\5&4&-1\\6&-3&5\end{array}\right|} = \frac{-665}{-82} [/tex]
[tex]c = \frac{\left|\begin{array}{ccc}3&-2&10\\5&4&5\\6&-3&-10\end{array}\right|}{\left|\begin{array}{ccc}3&-2&5\\5&4&-1\\6&-3&5\end{array}\right|} = \frac{-625}{-82}[/tex]

tadi baru mau dijawab, udah dilaporin aja