CUET UG 2025 — Mathematics Algebra

A matrix is singular when its determinant equals zero.

For the given matrix:

$$\text{det}(A) = \begin{vmatrix} 2 & 1 & -1 \\ 0 & 1 & 2 \\ 2 & -1 & \lambda \end{vmatrix}$$det(A)=​202​11−1​−12λ​​

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Expanding along the first row:

$$\text{det}(A) = 2 \begin{vmatrix} 1 & 2 \\ -1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 0 & 2 \\ 2 & \lambda \end{vmatrix} + (-1) \begin{vmatrix} 0 & 1 \\ 2 & -1 \end{vmatrix}$$det(A)=2​1−1​2λ​​−1​02​2λ​​+(−1)​02​1−1​​

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For the first minor:

$$2 \begin{vmatrix} 1 & 2 \\ -1 & \lambda \end{vmatrix} = 2[(1)(\lambda) - (2)(-1)]$$2​1−1​2λ​​=2[(1)(λ)−(2)(−1)]

$$= 2[\lambda + 2]$$=2[λ+2]

$$= 2\lambda + 4$$=2λ+4

For the second minor:

$$-1 \begin{vmatrix} 0 & 2 \\ 2 & \lambda \end{vmatrix} = -1[(0)(\lambda) - (2)(2)]$$−1​02​2λ​​=−1[(0)(λ)−(2)(2)]

$$= -1[-4]$$=−1[−4]

$$= 4$$=4

For the third minor:

$$-1 \begin{vmatrix} 0 & 1 \\ 2 & -1 \end{vmatrix} = -1[(0)(-1) - (1)(2)]$$−1​02​1−1​​=−1[(0)(−1)−(1)(2)]

$$= -1[-2]$$=−1[−2]

$$= 2$$=2

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Combining all terms:

$$\text{det}(A) = 2\lambda + 4 + 4 + 2$$det(A)=2λ+4+4+2

$$\text{det}(A) = 2\lambda + 10$$det(A)=2λ+10

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Since the matrix is singular:

$$2\lambda + 10 = 0$$2λ+10=0

$$2\lambda = -10$$2λ=−10

$$\lambda = -5$$λ=−5

Therefore, the value of $\lambda$λ is $-5$−5.