CUET UG 2025 — Mathematics Algebra

The determinant is given as:

$$\begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86$$​120​−2a4​5−12a​​=86

Expanding along Row 1:

$$\text{Det} = 1 \cdot M_{11} - (-2) \cdot M_{12} + 5 \cdot M_{13}$$Det=1⋅M11​−(−2)⋅M12​+5⋅M13​

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For $M_{11}$M11​, remove row 1 and column 1:

$$M_{11} = \begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix}$$M11​=​a4​−12a​​

$$= a(2a) - (-1)(4)$$=a(2a)−(−1)(4)

$$= 2a^2 + 4$$=2a2+4

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For $M_{12}$M12​, remove row 1 and column 2:

$$M_{12} = \begin{vmatrix} 2 & -1 \\ 0 & 2a \end{vmatrix}$$M12​=​20​−12a​​

$$= 2(2a) - (-1)(0)$$=2(2a)−(−1)(0)

$$= 4a$$=4a

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For $M_{13}$M13​, remove row 1 and column 3:

$$M_{13} = \begin{vmatrix} 2 & a \\ 0 & 4 \end{vmatrix}$$M13​=​20​a4​​

$$= 2(4) - a(0)$$=2(4)−a(0)

$$= 8$$=8

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Substituting into the determinant formula:

$$\text{Det} = 1(2a^2 + 4) + 2(4a) + 5(8)$$Det=1(2a2+4)+2(4a)+5(8)

$$= 2a^2 + 4 + 8a + 40$$=2a2+4+8a+40

$$= 2a^2 + 8a + 44$$=2a2+8a+44

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Setting the determinant equal to 86:

$$2a^2 + 8a + 44 = 86$$2a2+8a+44=86

$$2a^2 + 8a - 42 = 0$$2a2+8a−42=0

$$a^2 + 4a - 21 = 0$$a2+4a−21=0

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Factoring the quadratic:

$$(a + 7)(a - 3) = 0$$(a+7)(a−3)=0

Therefore:

$a = -7$a=−7 or $a = 3$a=3

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The product of all values of $a$a:

$$(-7) \times 3 = -21$$(−7)×3=−21