CUET UG 2025 — Mathematics Calculus

For a system of equations to have a unique solution, the determinant of the coefficient matrix must be non-zero.

The coefficient matrix from the three equations is:

$$A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & -2 & 1 \\ k & 1 & -1 \end{pmatrix}$$A=​21k​−1−21​31−1​​

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Using first row expansion to calculate the determinant:

$$\det(A) = 2\begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} - (-1)\begin{vmatrix} 1 & 1 \\ k & -1 \end{vmatrix} + 3\begin{vmatrix} 1 & -2 \\ k & 1 \end{vmatrix}$$det(A)=2​−21​1−1​​−(−1)​1k​1−1​​+3​1k​−21​​

For the first term:

$$2[(-2)(-1) - (1)(1)]$$2[(−2)(−1)−(1)(1)]

$$= 2[2 - 1]$$=2[2−1]

$$= 2$$=2

For the second term:

$$+1[(1)(-1) - (1)(k)]$$+1[(1)(−1)−(1)(k)]

$$= -1 - k$$=−1−k

For the third term:

$$3[(1)(1) - (-2)(k)]$$3[(1)(1)−(−2)(k)]

$$= 3[1 + 2k]$$=3[1+2k]

$$= 3 + 6k$$=3+6k

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Combining all terms:

$$\det(A) = 2 + (-1 - k) + (3 + 6k)$$det(A)=2+(−1−k)+(3+6k)

$$= 2 - 1 - k + 3 + 6k$$=2−1−k+3+6k

$$= 4 + 5k$$=4+5k

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For a unique solution:

$$\det(A) \neq 0$$det(A)=0

$$4 + 5k \neq 0$$4+5k=0

$$5k \neq -4$$5k=−4

$$k \neq -\dfrac{4}{5}$$k=−54​

Therefore, the system has a unique solution when $k \neq -\dfrac{4}{5}$k=−54​.