Answer on Question #42309 – Math – Linear Algebra:
Check whether the following system of equations have a solution or not:
{ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 − x 1 + 7 x 2 + 3 x 3 = 1 \left\{ \begin{array}{l} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ x _ {1} + 5 x _ {2} + 3 x _ {3} = 1 \\ - x _ {1} + 7 x _ {2} + 3 x _ {3} = 1 \end{array} \right. ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 − x 1 + 7 x 2 + 3 x 3 = 1
Solution.
{ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 − x 1 + 7 x 2 + 3 x 3 = 1 ⇒ { x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 − x 1 + 7 x 2 + 3 x 3 + ( x 1 + 5 x 2 + 3 x 3 ) = 1 + 1 ⇒ \left\{ \begin{array}{l} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ x _ {1} + 5 x _ {2} + 3 x _ {3} = 1 \\ - x _ {1} + 7 x _ {2} + 3 x _ {3} = 1 \end{array} \Rightarrow \left\{ \begin{array}{c} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ x _ {1} + 5 x _ {2} + 3 x _ {3} = 1 \\ - x _ {1} + 7 x _ {2} + 3 x _ {3} + (x _ {1} + 5 x _ {2} + 3 x _ {3}) = 1 + 1 \end{array} \right. \Rightarrow \right. ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 − x 1 + 7 x 2 + 3 x 3 = 1 ⇒ ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 − x 1 + 7 x 2 + 3 x 3 + ( x 1 + 5 x 2 + 3 x 3 ) = 1 + 1 ⇒ ⇒ { x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 12 x 2 + 6 x 3 = 2 ⇒ { x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 − ( x 1 − 3 x 2 − x 3 ) = 1 − 3 ⇒ 12 x 2 + 6 x 3 = 2 \Rightarrow \left\{ \begin{array}{l} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ x _ {1} + 5 x _ {2} + 3 x _ {3} = 1 \\ 1 2 x _ {2} + 6 x _ {3} = 2 \end{array} \Rightarrow \left\{ \begin{array}{c} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ x _ {1} + 5 x _ {2} + 3 x _ {3} - (x _ {1} - 3 x _ {2} - x _ {3}) = 1 - 3 \Rightarrow \\ 1 2 x _ {2} + 6 x _ {3} = 2 \end{array} \right. \right. ⇒ ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 = 1 12 x 2 + 6 x 3 = 2 ⇒ ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 x 1 + 5 x 2 + 3 x 3 − ( x 1 − 3 x 2 − x 3 ) = 1 − 3 ⇒ 12 x 2 + 6 x 3 = 2 ⇒ { x 1 − 3 x 2 − x 3 = 3 8 x 2 + 4 x 3 = − 2 12 x 2 + 6 x 3 = 2 ⇒ { x 1 − 3 x 2 − x 3 = 3 2 x 2 + x 3 = − 1 2 2 x 2 + x 3 = 1 3 ⇒ { x 1 − 3 x 2 − x 3 = 3 2 x 2 + x 3 = − 1 2 − 1 2 = 1 3 − contradiction . \Rightarrow \left\{ \begin{array}{l} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ 8 x _ {2} + 4 x _ {3} = - 2 \\ 1 2 x _ {2} + 6 x _ {3} = 2 \end{array} \Rightarrow \left\{ \begin{array}{c} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ 2 x _ {2} + x _ {3} = - \frac {1}{2} \\ 2 x _ {2} + x _ {3} = \frac {1}{3} \end{array} \Rightarrow \left\{ \begin{array}{c} x _ {1} - 3 x _ {2} - x _ {3} = 3 \\ 2 x _ {2} + x _ {3} = - \frac {1}{2} \\ - \frac {1}{2} = \frac {1}{3} \end{array} \right. - \text{contradiction}. \right. \right. ⇒ ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 8 x 2 + 4 x 3 = − 2 12 x 2 + 6 x 3 = 2 ⇒ ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 2 x 2 + x 3 = − 2 1 2 x 2 + x 3 = 3 1 ⇒ ⎩ ⎨ ⎧ x 1 − 3 x 2 − x 3 = 3 2 x 2 + x 3 = − 2 1 − 2 1 = 3 1 − contradiction .
So, this system have no solutions.
Answer.
No
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