CUET UG 2025 — Mathematics Algebra

A matrix is not invertible when its determinant equals 0.

For the matrix $$\begin{bmatrix}2 & -1 & 3\\ \lambda & 0 & 7\\-1 & 1 & 4\end{bmatrix}$$​2λ−1​−101​374​​, expanding along the first row:

$$\text{det}(A) = 2 \begin{vmatrix}0 & 7\\1 & 4\end{vmatrix} - (-1) \begin{vmatrix}\lambda & 7\\-1 & 4\end{vmatrix} + 3 \begin{vmatrix}\lambda & 0\\-1 & 1\end{vmatrix}$$det(A)=2​01​74​​−(−1)​λ−1​74​​+3​λ−1​01​​

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For the 2×2 determinants:

$$\begin{vmatrix}0 & 7\\1 & 4\end{vmatrix} = (0)(4) - (7)(1)$$​01​74​​=(0)(4)−(7)(1)

$$= -7$$=−7

$$\begin{vmatrix}\lambda & 7\\-1 & 4\end{vmatrix} = (\lambda)(4) - (7)(-1)$$​λ−1​74​​=(λ)(4)−(7)(−1)

$$= 4\lambda + 7$$=4λ+7

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

$$= \lambda$$=λ

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Substituting back:

$$\text{det}(A) = 2(-7) + 1(4\lambda + 7) + 3(\lambda)$$det(A)=2(−7)+1(4λ+7)+3(λ)

$$= -14 + 4\lambda + 7 + 3\lambda$$=−14+4λ+7+3λ

$$= -7 + 7\lambda$$=−7+7λ

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Since the matrix is not invertible:

$$-7 + 7\lambda = 0$$−7+7λ=0

$$7\lambda = 7$$7λ=7

$$\lambda = 1$$λ=1

Therefore, the value of $\lambda$λ is $1$1.