If x − y = 2 and y − z = 3 then value of, \(\begin{vmatrix}1 & x & x^2\\1 & y & y^2\\1 & z & z^2\end{vmatrix}=\)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −30
Applying \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\), and using \(y^2 - x^2 = (y - x)(y + x)\):
\( \displaystyle \Delta = \begin{vmatrix}1 & x & x^2\\0 & y-x & (y-x)(y+x)\\0 & z-x & (z-x)(z+x)\end{vmatrix} \)
Taking out \(y - x\) from \(R_2\) and \(z - x\) from \(R_3\):
\( \displaystyle \Delta = (y-x)(z-x)\begin{vmatrix}1 & x & x^2\\0 & 1 & y+x\\0 & 1 & z+x\end{vmatrix} \)
Expanding along \(C_1\):
\( \Delta = (y-x)(z-x)\,[(z+x) - (y+x)] \)
\( = (y-x)(z-x)(z-y) \)
\( = (x-y)(y-z)(z-x) \)
Adding the two given equations, \(x - z = 5\), so \(z - x = -5\):
\( \Delta = (2)(3)(-5) = -30 \)