If \( \begin{vmatrix}1+x^2 & 1 & 1\\1 & 1+y^2 & 1\\1 & 1 & 1+z^2\end{vmatrix}=k\left(1+\displaystyle\frac{1}{x^2}+\displaystyle\frac{1}{y^2}+\displaystyle\frac{1}{z^2}\right) \) then \( k \) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( x^2y^2z^2 \)
Taking \( x^2 \), \( y^2 \), \( z^2 \) out of rows 1, 2, 3 respectively:
\( \displaystyle \begin{vmatrix}1+x^2 & 1 & 1\\1 & 1+y^2 & 1\\1 & 1 & 1+z^2\end{vmatrix} \)
\( \displaystyle =x^2y^2z^2\begin{vmatrix}1+\frac{1}{x^2} & \frac{1}{x^2} & \frac{1}{x^2}\\\frac{1}{y^2} & 1+\frac{1}{y^2} & \frac{1}{y^2}\\\frac{1}{z^2} & \frac{1}{z^2} & 1+\frac{1}{z^2}\end{vmatrix} \)
With \( R_1\to R_1+R_2+R_3 \), every entry of the first row becomes \( s=1+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \). Taking s out, and subtracting the first column from the other two columns:
\( \displaystyle =x^2y^2z^2\,s\begin{vmatrix}1 & 0 & 0\\\frac{1}{y^2} & 1 & 0\\\frac{1}{z^2} & 0 & 1\end{vmatrix} \)
\( \displaystyle =x^2y^2z^2\,s \)
Comparing with k · s, \( k=x^2y^2z^2 \).
Shortcut: Put x = y = z = 2. The determinant becomes
\( \displaystyle \begin{vmatrix}5 & 1 & 1\\1 & 5 & 1\\1 & 1 & 5\end{vmatrix}=5(24)-1(4)+1(-4) \)
\( \displaystyle =112 \)
and the bracket becomes \( 1+\frac{3}{4}=\frac{7}{4} \), so k = 64. Of the options, only \( x^2y^2z^2 \) equals 64 here.