CUET UG 2025 — Mathematics Algebra

$$f(x) = \begin{vmatrix} 0 & x-1 & x-2 \\ x+1 & 0 & x-3 \\ x+2 & x+3 & 0 \end{vmatrix}$$f(x)=​0x+1x+2​x−10x+3​x−2x−30​​

To find $f(0)$f(0), substitute $x = 0$x=0 into the matrix:

$$f(0) = \begin{vmatrix} 0 & 0-1 & 0-2 \\ 0+1 & 0 & 0-3 \\ 0+2 & 0+3 & 0 \end{vmatrix}$$f(0)=​00+10+2​0−100+3​0−20−30​​

$$f(0) = \begin{vmatrix} 0 & -1 & -2 \\ 1 & 0 & -3 \\ 2 & 3 & 0 \end{vmatrix}$$f(0)=​012​−103​−2−30​​

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Expanding along the first row:

$$f(0) = (0) \cdot \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix}$$f(0)=(0)⋅​03​−30​​−(−1)⋅​12​−30​​+(−2)⋅​12​03​​

$$f(0) = 0 + 1 \cdot \begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix}$$f(0)=0+1⋅​12​−30​​−2⋅​12​03​​

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For the first $2 \times 2$2×2 determinant:

$$\begin{vmatrix} 1 & -3 \\ 2 & 0 \end{vmatrix} = (1)(0) - (-3)(2)$$​12​−30​​=(1)(0)−(−3)(2)

$$= 0 + 6$$=0+6

$$= 6$$=6

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For the second $2 \times 2$2×2 determinant:

$$\begin{vmatrix} 1 & 0 \\ 2 & 3 \end{vmatrix} = (1)(3) - (0)(2)$$​12​03​​=(1)(3)−(0)(2)

$$= 3 - 0$$=3−0

$$= 3$$=3

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$$f(0) = 1(6) - 2(3)$$f(0)=1(6)−2(3)

$$f(0) = 6 - 6$$f(0)=6−6

$$f(0) = 0$$f(0)=0