Question #42309

Consider the following system of equations:
x1−3x2−x3 = 3
x1+5x2+3x3 = 1
−x1+7x2+3x3 = 1
Check whether the system of equations have a solution or not
1

Expert's answer

2014-05-13T06:08:31-0400

Answer on Question #42309 – Math – Linear Algebra:

Check whether the following system of equations have a solution or not:


{x13x2x3=3x1+5x2+3x3=1x1+7x2+3x3=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.


Solution.


{x13x2x3=3x1+5x2+3x3=1x1+7x2+3x3=1{x13x2x3=3x1+5x2+3x3=1x1+7x2+3x3+(x1+5x2+3x3)=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.{x13x2x3=3x1+5x2+3x3=112x2+6x3=2{x13x2x3=3x1+5x2+3x3(x13x2x3)=1312x2+6x3=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.{x13x2x3=38x2+4x3=212x2+6x3=2{x13x2x3=32x2+x3=122x2+x3=13{x13x2x3=32x2+x3=1212=13contradiction.\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.


So, this system have no solutions.

Answer.

No

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