Match List-I with List-II
| List-I | List-II | ||
| (Matrix) | (Inverse of the Matrix) | ||
| (A) | \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\) | (I) | \(\begin{bmatrix} 2/15 & 1/10 \\ -1/15 & 1/5 \end{bmatrix}\) |
| (B) | \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\) | (II) | \(\begin{bmatrix} 1/5 & -2/15 \\ -1/10 & 7/30 \end{bmatrix}\) |
| (C) | \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\) | (III) | \(\begin{bmatrix} 1/15 & 7/30 \\ 2/15 & -1/30 \end{bmatrix}\) |
| (D) | \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\) | (IV) | \(\begin{bmatrix} 2/15 & -1/15 \\ 1/6 & 1/6 \end{bmatrix}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
For a 2 × 2 matrix, the inverse is found by swapping the diagonal entries, changing the signs of the other two, and dividing by the determinant:
\( \displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad - bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \)
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\)
This matches (III). The determinant is \(-2 - 28 = -30\):
\( \displaystyle \frac{1}{-30}\begin{bmatrix} -2 & -7 \\ -4 & 1 \end{bmatrix} = \begin{bmatrix} 1/15 & 7/30 \\ 2/15 & -1/30 \end{bmatrix} \)
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\)
This matches (I). The determinant is \(24 + 6 = 30\):
\( \displaystyle \frac{1}{30}\begin{bmatrix} 4 & 3 \\ -2 & 6 \end{bmatrix} = \begin{bmatrix} 2/15 & 1/10 \\ -1/15 & 1/5 \end{bmatrix} \)
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\)
This matches (IV). The determinant is \(20 + 10 = 30\):
\( \displaystyle \frac{1}{30}\begin{bmatrix} 4 & -2 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 2/15 & -1/15 \\ 1/6 & 1/6 \end{bmatrix} \)
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\)
This matches (II). The determinant is \(42 - 12 = 30\):
\( \displaystyle \frac{1}{30}\begin{bmatrix} 6 & -4 \\ -3 & 7 \end{bmatrix} = \begin{bmatrix} 1/5 & -2/15 \\ -1/10 & 7/30 \end{bmatrix} \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).