If \( A=\begin{bmatrix}a & 2\\2 & 3\end{bmatrix},B=\begin{bmatrix}b & c\\c & b\end{bmatrix} \) such that \( A^2=4A+B,c\gt a\gt b \), then
Match List-I with List-II
| List-I | List-II | ||
| (A) | \( a+b \) | (I) | 1 |
| (B) | \( b+c \) | (II) | 2 |
| (C) | \( c-a \) | (III) | 4 |
| (D) | \( a-b \) | (IV) | 5 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
The values of a, b and c come from comparing \( A^2 \) with \( 4A+B \).
\( \displaystyle A^2=\begin{bmatrix}a & 2\\2 & 3\end{bmatrix}\begin{bmatrix}a & 2\\2 & 3\end{bmatrix} \)
\( \displaystyle =\begin{bmatrix}a^2+4 & 2a+6\\2a+6 & 13\end{bmatrix} \)
\( \displaystyle 4A+B=\begin{bmatrix}4a+b & 8+c\\8+c & 12+b\end{bmatrix} \)
Equating the entries:
\( \displaystyle (2,2):\ 13=12+b\ \Rightarrow\ b=1 \)
\( \displaystyle (1,1):\ a^2+4=4a+1 \)
\( \displaystyle a^2-4a+3=0\ \Rightarrow\ a=1\ \text{or}\ a=3 \)
Since a > b = 1, a = 3.
\( \displaystyle (1,2):\ 2a+6=8+c\ \Rightarrow\ c=4 \)
Check c > a > b: 4 > 3 > 1.
(A) a + b
This matches (III) 4: 3 + 1 = 4.
(B) b + c
This matches (IV) 5: 1 + 4 = 5.
(C) c − a
This matches (I) 1: 4 − 3 = 1.
(D) a − b
This matches (II) 2: 3 − 1 = 2.
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).