Related Formula
adj(kA) = kⁿ⁻¹adj(A)$$\text{adj}(kA) = k^{n-1}\text{adj}(A)$$
A · adj(A) = |A|Iₙ$$A \cdot \text{adj}(A) = |A|I_n$$
|adj(A)| = |A|ⁿ⁻¹$$|\text{adj}(A)| = |A|^{n-1}$$
Core Logic
Analyze the innermost term:
adj(2A) = 2³⁻¹ adj(A) = 4adj(A)$\text{adj}(2A) = 2^{3-1} \text{adj}(A) = 4\text{adj}(A)$
So, A² · adj(2A) = 4A² · adj(A) = 4A(A · adj(A))$A^2 \cdot \text{adj}(2A) = 4A^2 \cdot \text{adj}(A) = 4A(A \cdot \text{adj}(A))$
Using A · adj(A) = |A|I₃$A \cdot \text{adj}(A) = |A|I_3$:
= 4A(|A|I₃) = 4|A|A = 4(6)A = 24A$$= 4A(|A|I_3) = 4|A|A = 4(6)A = 24A$$
Step 1: Expand the Adjoint Layers
Substitute this back into the next layer:
3adj(24A) = 3 · (24)³⁻¹adj(A) = 3(24)²adj(A)$3\text{adj}(24A) = 3 \cdot (24)^{3-1}\text{adj}(A) = 3(24)^2\text{adj}(A)$
Step 2: Evaluate the Determinant
Now we need |adj(3(24)²adj(A))|$|\text{adj}(3(24)^2\text{adj}(A))|$. Let k = 3(24)²$k = 3(24)^2$.
The expression is |adj(k adj(A))|$|\text{adj}(k \text{adj}(A))|$.
Since adj(kX) = k²adj(X)$\text{adj}(kX) = k^2\text{adj}(X)$ for a 3 × 3$3 \times 3$ matrix, the matrix inside the determinant is k² adj(adj(A))$k^2 \text{adj}(\text{adj}(A))$.
Wait, taking the determinant directly is simpler: |adj(X)| = |X|²$|\text{adj}(X)| = |X|^2$.
Let X = k adj(A)$X = k \text{adj}(A)$. Then |X| = |k adj(A)| = k³ |adj(A)| = k³ |A|²$|X| = |k \text{adj}(A)| = k^3 |\text{adj}(A)| = k^3 |A|^2$.
Thus, |adj(X)| = |X|² = (k³ |A|²)² = k⁶ |A|⁴$|\text{adj}(X)| = |X|^2 = (k^3 |A|^2)^2 = k^6 |A|^4$.
Step 3: Calculate Powers of 2 and 3
Substitute k = 3 · 24² = 3 · (2³ · 3)² = 3 · 2⁶ · 3² = 2⁶ · 3³$k = 3 \cdot 24^2 = 3 \cdot (2^3 \cdot 3)^2 = 3 \cdot 2^6 \cdot 3^2 = 2^6 \cdot 3^3$.
k⁶ = (2⁶ · 3³)⁶ = 2³⁶ · 3¹⁸$$k^6 = (2^6 \cdot 3^3)^6 = 2^{36} \cdot 3^{18}$$
We know |A| = 6 = 2 · 3$|A| = 6 = 2 \cdot 3$. So |A|⁴ = 2⁴ · 3⁴$|A|^4 = 2^4 \cdot 3^4$.
Total determinant = (2³⁶ · 3¹⁸) × (2⁴ · 3⁴) = 2⁴⁰ · 3²²$= (2^{36} \cdot 3^{18}) \times (2^4 \cdot 3^4) = 2^{40} \cdot 3^{22}$.
Equating to 2^m · 3ⁿ$2^m \cdot 3^n$, we get m = 40$m = 40$ and n = 22$n = 22$.
Step 4: Final Sum
m + n = 40 + 22 = 62$$m + n = 40 + 22 = 62$$
Pattern Recognition
Telescoping determinant operations demand rigorous sequential application of |kA| = kⁿ|A|$|kA| = k^n|A|$ and adj(kA) = kⁿ⁻¹adj(A)$\text{adj}(kA) = k^{n-1}\text{adj}(A)$. Collapsing A · adj(A)$A \cdot \text{adj}(A)$ into |A|I$|A|I$ early prevents chaotic polynomial matrix expressions.
Chapter Mix
Class 12 Maths: Matrices
Class 12 Maths: Determinants