Match List-I with List-II. $$\begin{array}{|c|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline (A)\ If\ \left|\begin{matrix} \lambda-1 & 0 \\ 0 & \lambda-1 \end{matrix}\right|=0,\ then\ \lambda\ is & (I)\ 0 \\ \hline (B)\ If\ \Delta=\left|\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}\right|,\ then\ \Delta\ is & (II)\ 1 \\ \hline (C)\ If\ A=\begin{bmatrix} 1 & 0 \\ 0 & \tfrac{1}{2} \end{bmatrix},\ then\ |A^{-1}|\ is & (III)\ -2 \\ \hline (D)\ If\ \begin{bmatrix} a+1 & 1 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix},\ then\ a\ is & (IV)\ 2 \\ \hline \end{array}$$
Step-by-step Solution:
(A) Evaluate \[ \left|\begin{matrix}\lambda-1 & 0 \\[4pt] 0 & \lambda-1 \end{matrix}\right| = 0. \] The determinant is \((\lambda-1)(\lambda-1)=(\lambda-1)^2\). So \((\lambda-1)^2=0 \Rightarrow \lambda=1\). ⇒ (A) → (II).
(B) Compute \[ \Delta=\left|\begin{matrix}1 & 2 \\[4pt] 2 & 4 \end{matrix}\right| = 1\cdot 4 - 2\cdot 2 = 4 - 4 = 0. \] ⇒ (B) → (I).
(C) Given \[ A=\begin{bmatrix}1 & 0 \\[4pt] 0 & \tfrac{1}{2}\end{bmatrix}. \] Then \(A^{-1}=\begin{bmatrix}1 & 0 \\[4pt] 0 & 2\end{bmatrix}\), so \[ |A^{-1}| = 1 \cdot 2 = 2. \] ⇒ (C) → (IV).
(D) From matrix equality \[ \begin{bmatrix}a+1 & 1 \\[4pt] 1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\[4pt] 1 & 2 \end{bmatrix}, \] equate corresponding entries: \(a+1=-1 \Rightarrow a=-2\). ⇒ (D) → (III).
Final matching:
(A) – (II), (B) – (I), (C) – (IV), (D) – (III)
Correct Option: (A)