\( \left\{ \begin{array} { l } { F ( 1 ; 0 ; 0 ) - F ( 0 ; 1 ; 0 ) = ( 0 ; 0 ; 0 ) } \\ { F ( 1 ; 0 ; 0 ) + F ( 0 ; 0 ; 1 ) = ( 0 ; 0 ; 0 ) } \\ { F ( 1 ; 0 ; 0 ) + F ( 0 ; 1 ; 0 ) - F ( 0 ; 0 ; 1 ) = ( - 3 ; - 3 ; 3 ) } \end{array} \right. \)
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To solve this system of equations, we can denote the vectors \( F(1; 0; 0) \), \( F(0; 1; 0) \), and \( F(0; 0; 1) \) as \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), respectively. The equations can then be rewritten as follows: 1. \( \mathbf{a} - \mathbf{b} = \mathbf{0} \) (which means \( \mathbf{a} = \mathbf{b} \)) 2. \( \mathbf{a} + \mathbf{c} = \mathbf{0} \) (which means \( \mathbf{c} = -\mathbf{a} \)) 3. \( \mathbf{a} + \mathbf{b} - \mathbf{c} = (-3; -3; 3) \) Using what we know from the first equation \( \mathbf{a} = \mathbf{b} \) in the third equation gives us \( \mathbf{a} + \mathbf{a} - \mathbf{c} = (-3; -3; 3) \). Substituting \( \mathbf{c} \) from the second equation into the modified third equation results in: \( 2\mathbf{a} - (-\mathbf{a}) = (-3; -3; 3) \) \( 3\mathbf{a} = (-3; -3; 3) \) Dividing both sides by 3, we find: \( \mathbf{a} = (-1; -1; 1) \). Now, using \( \mathbf{a} \): 1. From \( \mathbf{b} = \mathbf{a} \), we have \( \mathbf{b} = (-1; -1; 1) \). 2. From \( \mathbf{c} = -\mathbf{a} \), we find \( \mathbf{c} = (1; 1; -1) \). In conclusion, the vectors are: \( F(1; 0; 0) = (-1; -1; 1) \), \( F(0; 1; 0) = (-1; -1; 1) \), and \( F(0; 0; 1) = (1; 1; -1) \).
