Find the value of x, if \(\begin{vmatrix}1&-1\\3&-5\end{vmatrix}\)=\(\begin{vmatrix}x&x^2\\3&5\end{vmatrix}\).(a) x=2, –\(\frac{1}{3}\)(b) x=-1, –\(\frac{1}{3}\)(c) x=-2, –\(\frac{1}{3}\)(d) x=0, –\(\frac{1}{3}\)I had been asked this question at a job interview.The above asked question is from Determinant topic in division Determinants of Mathematics – Class 12

Find the value of x, if \(\begin{vmatrix}1&-1\\3&-5\end{vmatrix}\)=\(\

1.	Find the value of x, if \(\begin{vmatrix}1&-1\\3&-5\end{vmatrix}\)=\(\begin{vmatrix}x&x^2\\3&5\end{vmatrix}\).(a) x=2, –\(\frac{1}{3}\)(b) x=-1, –\(\frac{1}{3}\)(c) x=-2, –\(\frac{1}{3}\)(d) x=0, –\(\frac{1}{3}\)I had been asked this question at a job interview.The above asked question is from Determinant topic in division Determinants of Mathematics – Class 12
Answer» Right OPTION is (a) x=2, –\(\frac{1}{3}\) The explanation: Given that, \(\BEGIN{vmatrix}1&-1\\3&-5\end{vmatrix}\)=\(\begin{vmatrix}x&x^2\\3&5\end{vmatrix}\) -5—(-3)=5x-3x^2 -2=5x-3x^2 3x^2-5x-2=0 Solving for x, we get x=2, –\(\frac{1}{3}\).

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