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If we have three equations say in three variables say $a_1x+b_1y+c_1z+d_1=0$ , $a_2x+b_2y+c_2z+d_2=0$,$a_3x+b_3y+c_3z+d_3=0$, can we use method of "cross multiplication " ? If so, then how can we use it ?

Arthur
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  • do you know how we use cross multiplication for solving a pair of equations in 2 variables? Can you extend this into 3 variables? Hint: Try eliminating 1 variable each from the first and second equations by the third equation, and then eliminate 1 variable from the second by the first, and you are done. – D S Jan 02 '23 at 10:19

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$$a_1x+b_1y+c_1z+d_1=0\tag{1}\label{1}$$ $$a_2x+b_2y+c_2z+d_2=0\tag{2}\label{2}$$ $$a_3x+b_3y+c_3z+d_3=0\tag{3}\label{3}$$ Multiply $\eqref{1}$ by $a_2$ and $\eqref{2}$ by $a_1$. $$a_1a_2x+a_2b_1y + a_2c_1z + a_2d_1 = 0\tag{4}\label{4}$$ $$a_1a_2x+a_1b_2y+a_1c_2z+a_1d_2=0\tag{5}\label{5}$$ Subtract $\eqref{5}$ from $\eqref{4}$: $$(a_2b_1-a_1b_2)y + (a_2c_1-a_1c_2)z+(a_2d_1-a_1d_2)=0\tag{6}\label{6}$$ Similar process with $\eqref{1}$ and $\eqref{3}$ yields: $$(a_3b_1-a_1b_3)y + (a_3c_1-a_1c_3)z+(a_3d_1-a_1d_3)=0\tag{7}\label{7}$$ Multiplying $\eqref{6}$ by $(a_3b_1-a_1b_3)$ and $\eqref{7}$ by $(a_2b_1-a_1b_2)$: $$(a_3b_1-a_1b_3)(a_2b_1-a_1b_2)y + (a_3b_1-a_1b_3)(a_2c_1-a_1c_2)z+(a_3b_1-a_1b_3)(a_2d_1-a_1d_2)=0\tag{8}\label{8}$$ $$(a_2b_1-a_1b_2)(a_3b_1-a_1b_3)y + (a_2b_1-a_1b_2)(a_3c_1-a_1c_3)z+(a_2b_1-a_1b_2)(a_3d_1-a_1d_3)=0\tag{9}\label{9}$$Subtracting $\eqref{9}$ from $\eqref{8}$: $$((a_3b_1-a_1b_3)(a_2c_1-a_1c_2) - (a_2b_1-a_1b_2)(a_3c_1-a_1c_3))z + (a_3b_1-a_1b_3)(a_2d_1-a_1d_2) - (a_2b_1-a_1b_2)(a_3d_1-a_1d_3) = 0$$So, we get : $$z = \frac{(a_2b_1-a_1b_2)(a_3d_1-a_1d_3)-(a_3b_1-a_1b_3)(a_2d_1-a_1d_2)}{(a_3b_1-a_1b_3)(a_2c_1-a_1c_2) - (a_2b_1-a_1b_2)(a_3c_1-a_1c_3)}\tag{10}\label{10}$$Plug $\eqref{10}$ in $\eqref{6}$ to get $y$, and plug both in $\eqref{1}$ to get $x$. The formulae will be messy and inconvenient and pretty hard to memorise, probably that's why you've never seen them before.

D S
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