Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 455 Question 455 of 472 Hard MCQ
If A A A is a square matrix of order 3 such that \[\operatorname{det}(A)=3\] and \[\operatorname{det}\left(\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 \mathrm{~A})^{-1}\right)\right)\right)\right)\right)=2^{\mathrm{m}} 3^{\mathrm{n}}\], then \[\mathrm{m}+2 \mathrm{n}\] is equal to :
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\[ ∣ A ∣ = 3 ∣ adj ( − 4 adj ( − 3 adj ( 3 adj ( 2 A ) − 1 ) ) ) ∣ = ∣ − 4 adj ( − 3 adj ( 3 adj ( ( 2 A ) − 1 ) ) ) ∣ 2 \begin{aligned}
& |A|=3 \\
& \left|\operatorname{adj}\left(-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}(2 A)^{-1}\right)\right)\right)\right| \\
& =\left|-4 \operatorname{adj}\left(-3 \operatorname{adj}\left(3 \operatorname{adj}\left((2 A)^{-1}\right)\right)\right)\right|^2
\end{aligned} ∣ A ∣ = 3 adj ( − 4 adj ( − 3 adj ( 3 adj ( 2 A ) − 1 ) ) ) = − 4 adj ( − 3 adj ( 3 adj ( ( 2 A ) − 1 ) ) ) 2 \]
\[ = 4 6 ∣ − 3 adj ( aadj ( ( 2 A ) − 1 ) ) ∣ 4 = 4 6 ⋅ 3 12 ∣ aadj ( ( 2 A ) − 1 ) ∣ 8 = 4 6 ⋅ 3 12 ⋅ 3 24 ∣ ( 2 A ) − 1 ∣ 16 = 4 6 ⋅ 3 36 2 − 48 ∣ A − 1 ∣ 16 = 2 − 36 3 36 3 16 = 2 − 36 3 20 m = − 36 n = 20 m + 2 n = 4 \begin{aligned}
& =4^6\left|-3 \operatorname{adj}\left(\operatorname{aadj}\left((2 A)^{-1}\right)\right)\right|^4 \\
& =4^6 \cdot 3^{12}\left|\operatorname{aadj}\left((2 A)^{-1}\right)\right|^8 \\
& =4^6 \cdot 3^{12} \cdot 3^{24}\left|(2 A)^{-1}\right|^{16} \\
& =4^6 \cdot 3^{36} 2^{-48}\left|A^{-1}\right|^{16} \\
& =\frac{2^{-36} 3^{36}}{3^{16}}=2^{-36} 3^{20} \\
& m=-36 \\
& n=20 \\
& m+2 n=4
\end{aligned} = 4 6 − 3 adj ( aadj ( ( 2 A ) − 1 ) ) 4 = 4 6 ⋅ 3 12 aadj ( ( 2 A ) − 1 ) 8 = 4 6 ⋅ 3 12 ⋅ 3 24 ( 2 A ) − 1 16 = 4 6 ⋅ 3 36 2 − 48 A − 1 16 = 3 16 2 − 36 3 36 = 2 − 36 3 20 m = − 36 n = 20 m + 2 n = 4 \] Answer the question to unlock