For each matrix A in List-I, k is the real number such that \(A^2 = kA\).
Match List-I with List-II
| List-I (Matrix A) | List-II (Value of k) | ||
|---|---|---|---|
| (A) | \(\begin{bmatrix}2&1\\2&1\end{bmatrix}\) | (I) | 5 |
| (B) | \(\begin{bmatrix}1&3\\1&3\end{bmatrix}\) | (II) | 3 |
| (C) | \(\begin{bmatrix}3&6\\1&2\end{bmatrix}\) | (III) | 6 |
| (D) | \(\begin{bmatrix}4&2\\4&2\end{bmatrix}\) | (IV) | 4 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Each \(A^2\) is found by row-by-column multiplication; k is then read off by comparing \(A^2\) with A.
(A) \(\begin{bmatrix}2&1\\2&1\end{bmatrix}\)
This matches (II) 3.
\( \displaystyle A^2 = \begin{bmatrix}4 + 2&2 + 1\\4 + 2&2 + 1\end{bmatrix} = \begin{bmatrix}6&3\\6&3\end{bmatrix} = 3A \)
(B) \(\begin{bmatrix}1&3\\1&3\end{bmatrix}\)
This matches (IV) 4.
\( \displaystyle A^2 = \begin{bmatrix}1 + 3&3 + 9\\1 + 3&3 + 9\end{bmatrix} = \begin{bmatrix}4&12\\4&12\end{bmatrix} = 4A \)
(C) \(\begin{bmatrix}3&6\\1&2\end{bmatrix}\)
This matches (I) 5.
\( \displaystyle A^2 = \begin{bmatrix}9 + 6&18 + 12\\3 + 2&6 + 4\end{bmatrix} = \begin{bmatrix}15&30\\5&10\end{bmatrix} = 5A \)
(D) \(\begin{bmatrix}4&2\\4&2\end{bmatrix}\)
This matches (III) 6.
\( \displaystyle A^2 = \begin{bmatrix}16 + 8&8 + 4\\16 + 8&8 + 4\end{bmatrix} = \begin{bmatrix}24&12\\24&12\end{bmatrix} = 6A \)