Let A, B and C are n×n matrices such that A=2, B=3 and C=5. If 2A^23B5C^(-1)=1728/125, then the value of n is equal to

Correct answer: (A) 2. 2A^23B5C^(-1)=(2A^23B)/5C=(2^2n3^n)/(5^n)(A^2B)/C ⇒125^(n+1)=125^3⇒n=2

Q27 / 124 · NTA Abhyas (JEE Main) 2020 · Mathematics · Determinants

Let $A, B$ and $C$ are $n \times n$ matrices such that $|A| = 2, |B| = 3$ and $|C| = 5 .$ If $|{(2 A)}^{2} (3 B) {(5 C)}^{- 1}| = \frac{1728}{125}$ , then the value of $n$ is equal to

|{(2 A)}^{2} (3 B) {(5 C)}^{- 1}| = \frac{{|2 A|}^{2} |3 B|}{|5 C|} = \frac{2^{2 n} 3^{n}}{5^{n}} \frac{{|A|}^{2} |B|}{|C|}

\Rightarrow {(\frac{12}{5})}^{n + 1} = {(\frac{12}{5})}^{3} \Rightarrow n = 2

Hence, the correct answer is option: 'A'.