If \( \mathrm{A}=\begin{bmatrix}1 & a\\b & 2\end{bmatrix} \) and \( \mathrm{B}=\begin{bmatrix}a & 1\\2 & -3b\end{bmatrix} \) such that \( |AB|=|2B| \) and \( ab\ne-\displaystyle\frac{2}{3} \) then:
A. \( |A|=2 \)
B. \( |A|=4 \)
C. \( |B|=2 \)
D. \( |B|=4 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B and D only
\( \displaystyle |A|=2-ab \)
\( \displaystyle |B|=-3ab-2 \)
The condition \( ab\ne-\frac{2}{3} \) means \( |B|\ne 0 \). For 2 × 2 matrices \( |2B|=2^2|B|=4|B| \), and \( |AB|=|A|\,|B| \), so
\( \displaystyle |A|\,|B|=4|B|\ \Rightarrow\ |A|=4 \)
\( \displaystyle 2-ab=4\ \Rightarrow\ ab=-2 \)
\( \displaystyle |B|=-3(-2)-2=4 \)
So |A| = 4 and |B| = 4.
Checking the statements
(A) \( |A|=2 \) — Incorrect. |A| = 4.
(B) \( |A|=4 \) — Correct. |A| |B| = 4|B| with |B| ≠ 0 gives |A| = 4.
(C) \( |B|=2 \) — Incorrect. |B| = 4.
(D) \( |B|=4 \) — Correct. With ab = −2, |B| = 6 − 2 = 4.
Tip: For a square matrix of order n, \( |kB|=k^n|B| \); taking \( |2B|=2|B| \) for a 2 × 2 matrix would wrongly give |A| = 2.