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  • Can someone explain - Matrices and Determinants - JEE Main-4

If the system of equations  2x+3y-z=0,x+ky-2z=0\:\:and\:\:2x-y+z=0 has a non-trivial solution \left ( x,y,z \right )  , then \frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k   is equal to :

  • Option 1)

    \frac{3}{4}

  • Option 2)

    \frac{1}{2}

  • Option 3)

    -\frac{1}{4}

  • Option 4)

    -4

 

Answers (1)

for non-trivial solution  A=0

\begin{vmatrix} 2 &3 &-1 \\ 1 &k & -1\\ 2& -1 & 1 \end{vmatrix}=0

\\2(k-2)-3(1+4)-1(-1-2k)=0\\\\\:2k-4-15+1+2k=0\\\\\:4k=18\\\\\:k=\frac{9}{2}

\\2x+3y-z=0\\\\\:x=\frac{z-3y}{2}\\\\\:2x+2y-2=0------(I)\\\\\:x+\frac{9}{2}y-2z=0-----(II)\\\\\:2x-y+z=0-----(III)\\\\\:from (I)\:and\:\:(II)\\\\\:I+III\\\\\:4x+2y=0\\\\\:2y=-4x\\\\\frac{x}{y}=-\frac{2}{4}=-\frac{1}{2}

\\I-III\\\\\:4y=2z\\\\\:\frac{y}{z}=\frac{2}{4}=\frac{1}{2}

 

I+3\:(III)

\\=8x+2z=0\\\\\:8x=-2z\\\\\:\frac{x}{z}=\frac{-1}{4}

so

\\\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=\\\\\\\:\frac{-1}{2}+\frac{1}{2}-4+\frac{9}{2}\\\\\\\:\frac{1}{2}

 

 

 


Option 1)

\frac{3}{4}

Option 2)

\frac{1}{2}

Option 3)

-\frac{1}{4}

Option 4)

-4

Posted by

Vakul

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