CUET UG 2025 — Mathematics Algebra

For any $2 \times 2$2×2 matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$(ac​bd​), the determinant is $ad - bc$ad−bc.

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Determinant of the left-hand side:

$$\det \begin{pmatrix} 2x & 5 \\ 8 & x \end{pmatrix} = (2x)(x) - (5)(8) = 2x^2 - 40$$det(2x8​5x​)=(2x)(x)−(5)(8)=2x2−40

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Determinant of the right-hand side:

$$\det \begin{pmatrix} 6 & -2 \\ 1 & 1 \end{pmatrix} = (6)(1) - (-2)(1) = 6 + 2 = 8$$det(61​−21​)=(6)(1)−(−2)(1)=6+2=8

Note: subtracting a negative means we add, so $6 - (-2) = 6 + 2$6−(−2)=6+2.

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Setting the two determinants equal:

$$2x^2 - 40 = 8$$2x2−40=8

$$2x^2 = 48$$2x2=48

$$x^2 = 24$$x2=24

$$x = \pm\sqrt{24}$$x=±24​

Simplifying: $\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$24​=4×6​=26​

$$\boxed{x = \pm\, 2\sqrt{6}}$$x=±26​​