JEE Main 2025 8 April Evening Shift
Exam JEE Main
Year 2025
Date 8 April
Shift Evening Shift
Subject Mathematics
Chapter Matrices JEE Main 2025 · 8 April · Evening Shift · Mathematics · Matrices
Let A=[ array ccc 2 & 2+p & 2+p+q 4 & 6+2 p & 8+3 p+2 q 6 & 12+3 p...
Let
A=\left[ 2 2 + p 2 + p + q 4 6 + 2 p 8 + 3 p + 2 q 6 12 + 3 p 20 + 6 p + 3 q \begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array} 2 4 6 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q \right] .
If
det ( adj ( adj ( 3 A ) ) ) = 2 m ⋅ 3 n , m , n ∈ N \operatorname{det}(\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A})))=2^{\mathrm{m}} \cdot 3^{\mathrm{n}}, \mathrm{m}, \mathrm{n} \in \mathrm{N} det ( adj ( adj ( 3 A ))) = 2 m ⋅ 3 n , m , n ∈ N , then
m + n \mathrm{m}+\mathrm{n} m + n is equal to
Step-by-step solution
|A|=\left| 2 2 + p 2 + p + q 4 6 + 2 p 8 + 3 p + 2 q 6 12 + 3 p 20 + 6 p + 3 q \begin{array}{ccc}2 & 2+p & 2+p+q \\ 4 & 6+2 p & 8+3 p+2 q \\ 6 & 12+3 p & 20+6 p+3 q\end{array} 2 4 6 2 + p 6 + 2 p 12 + 3 p 2 + p + q 8 + 3 p + 2 q 20 + 6 p + 3 q \right| C 3 → C 3 − C 2 − C 1 × q 2 \mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_2-\mathrm{C}_1 \times \frac{\mathrm{q}}{2} C 3 → C 3 − C 2 − C 1 × 2 q Then
C 3 → C 2 − C 1 X ( 1 + p 2 ) \mathrm{C}_3 \rightarrow \mathrm{C}_2-\mathrm{C}_1 \mathrm{X}\left(1+\frac{\mathrm{p}}{2}\right) C 3 → C 2 − C 1 X ( 1 + 2 p ) < b r / > ⇒ ∣ A ∣ = ∣ < b r / > 2 0 0 4 2 2 + p 6 6 8 + 3 p < b r / > ∣ ⇒ ∣ A ∣ = 2 ( 16 + 6 p − 12 − 6 p ) = 8 = 2 3 ∣ adj ( adj ( 3 A ) ) ∣ = ∣ 3 A ∣ ( 3 − 1 ) 2 = ∣ 3 A ∣ 4 = ( 3 3 ∣ A ∣ ) 4 = ( 3 3 × 2 3 ) 4 = 2 12 × 3 12 ⇒ m + n = 24 < b r / > \begin{aligned}<br/>& \Rightarrow|\mathrm{A}|=\left|\begin{array}{ccc}<br/>2 & 0 & 0 \\
4 & 2 & 2+\mathrm{p} \\
6 & 6 & 8+3 \mathrm{p}<br/>\end{array}\right| \\
& \Rightarrow|\mathrm{A}|=2(16+6 \mathrm{p}-12-6 \mathrm{p})=8=2^3 \\
& |\operatorname{adj}(\operatorname{adj}(3 \mathrm{~A}))|=|3 \mathrm{~A}|^{(3-1)^2}=|3 \mathrm{~A}|^4 \\
& =\left(3^3|\mathrm{~A}|\right)^4=\left(3^3 \times 2^3\right)^4=2^{12} \times 3^{12} \\
& \Rightarrow \mathrm{~m}+\mathrm{n}=24<br/>\end{aligned} < b r / > ⇒ ∣ A ∣ = < b r / > 2 4 6 0 2 6 0 2 + p 8 + 3 p < b r / > ⇒ ∣ A ∣ = 2 ( 16 + 6 p − 12 − 6 p ) = 8 = 2 3 ∣ adj ( adj ( 3 A )) ∣ = ∣3 A ∣ ( 3 − 1 ) 2 = ∣3 A ∣ 4 = ( 3 3 ∣ A ∣ ) 4 = ( 3 3 × 2 3 ) 4 = 2 12 × 3 12 ⇒ m + n = 24 < b r / >
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