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If ∣3xx1∣=∣3241∣\begin{vmatrix}3 & x\\ x & 1\end{vmatrix}=\begin{vmatrix}3 & 2\\ 4 & 1\end{vmatrix}, then value of xx is/are

(A) x=2x=2

(B) x=−2x=-2

(C) x=22x=2\sqrt2

(D) x=−22x=-2\sqrt2

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

The left determinant is 3−x23-x^2 and the right one is 3−8=−53-8=-5. So 3−x2=−53-x^2=-5, giving x2=8x^2=8 and x=±22x=\pm2\sqrt2.

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