Question

Find the value of x if \(\begin{vmatrix}3&x\\2&x^2 \end{vmatrix}\)=\(\begin{vmatrix}5&3\\3&2\end{vmatrix}\).

(a) x=1, –\(\frac{1}{3}\)

(b) x=-1, –\(\frac{1}{3}\)

(c) x=1, \(\frac{1}{3}\)

(d) x=-1, \(\frac{1}{3}\)

I got this question in class test.

This question is from Determinant in section Determinants of Mathematics – Class 12

Answer 1

The correct choice is (a) x=1, –\(\frac{1}{3}\)

For explanation I would say: Given that \(\begin{vmatrix}3&x\\2&x^2 \end{vmatrix}\)=\(\begin{vmatrix}5&3\\3&2\end{vmatrix}\)

⇒3x^2-2x=5(2)-3(3)

⇒3x^2-2x=1

Solving for x, we get

x=1, –\(\frac{1}{3}\).