Question

Match List-I with List-II

List-I (Matrix)	List-II (Inverse of the Matrix)
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\)	(I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\)
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\)	(II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\)
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\)	(III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\)
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\)	(IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\)

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (I), (B) - (III), (C) - (II), (D) - (IV)
• (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Hint:

For finding the inverse of a 2x2 matrix, remember the simple rule: swap the elements on the main diagonal, change the sign of the off-diagonal elements, and divide the resulting matrix by the determinant. This is much faster than using the full cofactor method.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept

The task is to find the inverse of each matrix in List-I and match it with the correct inverse from List-II.

Step 2: Key Formula or Approach

For a \(2\times2\) matrix \( M=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the inverse is: \[ M^{-1}=\frac{1}{\det(M)}\,\mathrm{adj}(M) =\frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \]

Step 3: Detailed Explanation

(A) For \(A=\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\):
\(\det(A)=(1)(-2)-(7)(4)=-2-28=-30\).
\[ A^{-1}=\frac{1}{-30}\begin{bmatrix}-2 & -7 \\ -4 & 1\end{bmatrix} =\begin{bmatrix}\tfrac{1}{15} & \tfrac{7}{30} \\[4pt] \tfrac{2}{15} & -\tfrac{1}{30}\end{bmatrix}. \] This matches (III).

(B) For \(B=\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\):
\(\det(B)=(6)(4)-(-3)(2)=24+6=30\).
\[ B^{-1}=\frac{1}{30}\begin{bmatrix}4 & 3 \\ -2 & 6\end{bmatrix} =\begin{bmatrix}\tfrac{2}{15} & \tfrac{1}{10} \\[4pt] -\tfrac{1}{15} & \tfrac{1}{5}\end{bmatrix}. \] This matches (I).

(C) For \(C=\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\):
\(\det(C)=(5)(4)-(2)(-5)=20+10=30\).
\[ C^{-1}=\frac{1}{30}\begin{bmatrix}4 & -2 \\ 5 & 5\end{bmatrix} =\begin{bmatrix}\tfrac{2}{15} & -\tfrac{1}{15} \\[4pt] \tfrac{1}{6} & \tfrac{1}{6}\end{bmatrix}. \] This matches (IV).

(D) For \(D=\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\):
\(\det(D)=(7)(6)-(4)(3)=42-12=30\).
\[ D^{-1}=\frac{1}{30}\begin{bmatrix}6 & -4 \\ -3 & 7\end{bmatrix} =\begin{bmatrix}\tfrac{1}{5} & -\tfrac{2}{15} \\[4pt] -\tfrac{1}{10} & \tfrac{7}{30}\end{bmatrix}. \] This matches (II).

Step 4: Final Answer

\((A \to \text{III}),\ (B \to \text{I}),\ (C \to \text{IV}),\ (D \to \text{II})\)  →  Option (C).