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  4. The system x+4y-2z=3, 3x+y+5z=7, 2x+3y+z=5 has:

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Q. The system $ x+4y-2z=3,\, $ $ 3x+y+5z=7, $ $ 2x+3y+z=5 $ has:

KEAMKEAM 2000

Solution:

Given system of equations are $ x+4y-2z=3 $ $ 3x+y+5z=7 $ $ 2x+3y+z=5 $ $ \therefore $ $ \Delta =\left| \begin{matrix} 1 & 4 & -2 \\ 3 & 1 & 5 \\ 2 & 3 & 1 \\ \end{matrix} \right| $ $ =1(1-15)-4(3-10)-2(9-2) $ $ =-14+28-14=0 $ Since, $ \Delta =0 $ and $ {{\Delta }_{2}}\ne 0 $ $ \therefore $ No solution will exist.