CUET UG 2025 — Mathematics Algebra

For $A^{-1}$A−1 to exist, the determinant of $A$A must be non-zero.

$$\det(A) \neq 0$$det(A)=0

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Given: $A = \begin{bmatrix} 1 & \lambda & -1 \\ -1 & 1 & 0 \\ \lambda & 1 & 1 \end{bmatrix}$A=​1−1λ​λ11​−101​​

Expanding along Row 1:

$$\det(A) = 1 \cdot \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} - \lambda \cdot \begin{vmatrix} -1 & 0 \\ \lambda & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} -1 & 1 \\ \lambda & 1 \end{vmatrix}$$det(A)=1⋅​11​01​​−λ⋅​−1λ​01​​+(−1)⋅​−1λ​11​​

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Computing each 2×2 determinant:

$$\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = (1)(1) - (0)(1) = 1$$​11​01​​=(1)(1)−(0)(1)=1

$$\begin{vmatrix} -1 & 0 \\ \lambda & 1 \end{vmatrix} = (-1)(1) - (0)(\lambda) = -1$$​−1λ​01​​=(−1)(1)−(0)(λ)=−1

$$\begin{vmatrix} -1 & 1 \\ \lambda & 1 \end{vmatrix} = (-1)(1) - (1)(\lambda) = -1 - \lambda$$​−1λ​11​​=(−1)(1)−(1)(λ)=−1−λ

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Substituting back:

$$\det(A) = 1(1) - \lambda(-1) + (-1)(-1 - \lambda)$$det(A)=1(1)−λ(−1)+(−1)(−1−λ)

$$= 1 + \lambda + 1 + \lambda$$=1+λ+1+λ

$$= 2 + 2\lambda$$=2+2λ

$$= 2(1 + \lambda)$$=2(1+λ)

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For $A^{-1}$A−1 to exist:

$$\det(A) \neq 0$$det(A)=0

$$2(1 + \lambda) \neq 0$$2(1+λ)=0

$$1 + \lambda \neq 0$$1+λ=0

$$\lambda \neq -1$$λ=−1

Therefore, $A^{-1}$A−1 exists when $\lambda \neq -1$λ=−1.