Match List-I with List-II

List-I	List-II
(A) If $\begin{vmatrix}\lambda-1&0\\ 0&\lambda-1\end{vmatrix}$ then $\lambda=$	(I) 0
(B) If $\Delta=\begin{vmatrix}1&2\\ 2&4\end{vmatrix}$ then $\Delta$ is	(II) 1
(C) If $A=\begin{bmatrix}1&0\\ 0&\frac{1}{2}\end{bmatrix}$ then $|A^{-1}|$ is	(III) -2
(D) If $\begin{bmatrix}a+1&1\\ 1&2\end{bmatrix}=\begin{bmatrix}-1&1\\ 1&2\end{bmatrix}$ then a is	(IV) 2

Choose the correct answer from the options given below:

Solution (Match the Columns):

(A) \(\begin{vmatrix}\lambda-1 & 0 \\ 0 & \lambda-1\end{vmatrix}=(\lambda-1)^2\). Singular ⇒ \((\lambda-1)^2=0 \Rightarrow \lambda=1\). ⇒ (A → II)

(B) \(\Delta=\begin{vmatrix}1 & 2 \\ 2 & 4\end{vmatrix}=1\cdot4-2\cdot2=0\). ⇒ (B → I)

(C) \(A=\begin{bmatrix}1&0\\0&\tfrac12\end{bmatrix}\), so \(|A|=\tfrac12\). ⇒ \(|A^{-1}|=1/|A|=2\). ⇒ (C → IV)

(D) \(\begin{bmatrix}a+1&1\\1&2\end{bmatrix}=\begin{bmatrix}-1&1\\1&2\end{bmatrix}\). ⇒ \(a+1=-1 \Rightarrow a=-2\). ⇒ (D → III)

✅ Correct option: (1) — (A-II), (B-I), (C-IV), (D-III)