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Roadkilled75 · 11-30-2008, 04:52 PM

Enter the matrices like russiankid said. If [A][x]=[B], then [A]^-1 * [B] = [x]. A is your coefficients (3x3), B is the right hand side values(3x1). Find A^-1, then multiply it times B, there is your solution values for X, Y and Z.

I assume you know how to set it up in matrix format. If not, ask me or PM me, I'm sitting around sick and have nothing better to do...

Does he want you to solve it by hand using matrices or longhand? It's real easy to teach you to find the inverse of a 3x3 by hand if you want full credit.

Edit: in case you were wondering, there is a solution for your made-up example. Edit 2: missed that your 4 was negative. 1.76, 1.35, -.35

11-30-2008, 04:52 PM	#6
Roadkilled75 Tri-State Aficionado Join Date: Oct 2007 Location: (In the cornfields west of) Allentown Member #6074 My Ride: Two red ones and a green one. iTrader: (0)	Enter the matrices like russiankid said. If [A][x]=[B], then [A]^-1 * [B] = [x]. A is your coefficients (3x3), B is the right hand side values(3x1). Find A^-1, then multiply it times B, there is your solution values for X, Y and Z. I assume you know how to set it up in matrix format. If not, ask me or PM me, I'm sitting around sick and have nothing better to do... Does he want you to solve it by hand using matrices or longhand? It's real easy to teach you to find the inverse of a 3x3 by hand if you want full credit. Edit: in case you were wondering, there is a solution for your made-up example. Edit 2: missed that your 4 was negative. 1.76, 1.35, -.35 __________________ '90 Miata '75 Maverick '84 GTV6 Last edited by Roadkilled75; 11-30-2008 at 05:32 PM.