Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Matrix | Inverse | ||
| (A) | \(\begin{bmatrix}2&1\\1&1\end{bmatrix}\) | (I) | \(\begin{bmatrix}3&-2\\-1&1\end{bmatrix}\) |
| (B) | \(\begin{bmatrix}1&1\\1&2\end{bmatrix}\) | (II) | \(\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\) |
| (C) | \(\begin{bmatrix}3&2\\1&1\end{bmatrix}\) | (III) | \(\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\) |
| (D) | \(\begin{bmatrix}1&2\\1&3\end{bmatrix}\) | (IV) | \(\begin{bmatrix}2&-1\\-1&1\end{bmatrix}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
For \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\) the value of \(ad - bc\) is 1 for every matrix in List-I, so each inverse is obtained by swapping a and d and changing the signs of b and c:
\( \displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1} = \begin{bmatrix}d&-b\\-c&a\end{bmatrix},\quad \text{when } ad - bc = 1 \)
(A) \(\begin{bmatrix}2&1\\1&1\end{bmatrix}\)
This matches (II). Here \(ad - bc = 2 - 1 = 1\).
\( \displaystyle \begin{bmatrix}1&-1\\-1&2\end{bmatrix} \)
(B) \(\begin{bmatrix}1&1\\1&2\end{bmatrix}\)
This matches (IV). Here \(ad - bc = 2 - 1 = 1\).
\( \displaystyle \begin{bmatrix}2&-1\\-1&1\end{bmatrix} \)
(C) \(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)
This matches (III). Here \(ad - bc = 3 - 2 = 1\).
\( \displaystyle \begin{bmatrix}1&-2\\-1&3\end{bmatrix} \)
(D) \(\begin{bmatrix}1&2\\1&3\end{bmatrix}\)
This matches (I). Here \(ad - bc = 3 - 2 = 1\).
\( \displaystyle \begin{bmatrix}3&-2\\-1&1\end{bmatrix} \)