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  • Need clarity, kindly explain! For all real numbers x,y and z the determinantis equal to:

For all real numbers x,y and z the determinant

is equal to:

  • Option 1)

    \left ( y-xz \right ) (z-x)

  • Option 2)

    zero

  • Option 3)

    (x-y)\: (y-z)\: (z-x)

  • Option 4)

    (x-yz)\: (y-z)

 

Answers (2)

As we learnt in 

Value of determinants of order 3 -

-

 

 \begin{vmatrix} 2x & xy-xz & y\\ 2x+z+1 & xy-xz+yz-z^{2} & 1+y\\ 3x+1 & 2xy-2xz & 1+y \end{vmatrix}

\Rightarrow \begin{vmatrix} 2x & x\left ( y-z \right ) & y\\ 2x+z+1 & \left ( x+z \right )\left ( y-z \right ) & 1+y\\ 3x+1 & 2x\left ( y-z \right ) & 1+y \end{vmatrix}

\Rightarrow \left ( y-z \right )\begin{vmatrix} 2x & x & y\\ 2x+z+1 & x+z & 1+y\\ 3x+1 & 2x & 1+y \end{vmatrix}

R_{3}\rightarrow R_{3}-R_{2}

\Rightarrow \left ( y-z \right )\begin{vmatrix} 2x & x & y\\ 2x+z+1 & x+z & 1+y\\ x-z & x-z & 0 \end{vmatrix}

\Rightarrow \left ( y-z \right )\left ( x-z \right )\begin{vmatrix} 2x & x & y\\ 2x+z+1 & x+z & 1+y\\ 1 & 1 & 0 \end{vmatrix}

\Rightarrow \left ( y-z \right )\left ( x-z \right )\begin{vmatrix} x & x & y\\ x+1 & x+z & 1+y\\ 0 & 1 & 0 \end{vmatrix}

\Rightarrow \left ( y-z \right )\left ( x-z \right )\left ( x-y \right )


Option 1)

\left ( y-xz \right ) (z-x)

This option is incorrect.

Option 2)

zero

This option is incorrect.

Option 3)

(x-y)\: (y-z)\: (z-x)

This option is correct.

Option 4)

(x-yz)\: (y-z)

This option is incorrect.

Posted by

Sabhrant Ambastha

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