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If ∣23−xx1∣=0\begin{vmatrix} 2 & 3-x \\ x & 1 \end{vmatrix} = 0, then the values of xx are:

Solution

✅ Correct Option: 2

Expanding the determinant: 2(1)−x(3−x)=02(1) - x(3-x) = 0, giving 2−3x+x2=02 - 3x + x^2 = 0, i.e. x2−3x+2=0x^2 - 3x + 2 = 0. Factoring: (x−1)(x−2)=0(x-1)(x-2) = 0, so x=1x = 1 or 22.

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