Question

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 1

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, –\(\frac{1}{3}\).