Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 112 Question 112 of 472 Medium NUMERICAL
Let A be a 3 × 3 3\times 3 3 × 3 matrix such that A + A T = O A+{A}^{T}=O A + A T = O . If A [ [ ∣ 1 − 1 0 ] ] ∣ = [ [ ∣ 3 3 2 ] ] ∣ , A 2 [ [ ∣ 1 − 1 0 ] ] ∣ = [ [ ∣ − 3 19 − 24 ] ] ∣ A[[|\begin{gathered} 1 \\ -1 \\ 0 \end{gathered}]]|=[[|\begin{gathered} 3 \\ 3 \\ 2 \end{gathered}]]|,{A}^{2}[[|\begin{gathered} 1 \\ -1 \\ 0 \end{gathered}]]|=[[|\begin{gathered} -3 \\ 19 \\ -24 \end{gathered}]]| A [[ ∣ 1 − 1 0 ]] ∣ = [[ ∣ 3 3 2 ]] ∣ , A 2 [[ ∣ 1 − 1 0 ]] ∣ = [[ ∣ − 3 19 − 24 ]] ∣ and det ( a d j ( 2 a d j ( A + I ) ) ) = ( 2 ) α ⋅ ( 3 ) β ⋅ ( 11 ) γ , α , β , γ \det (adj(2adj(A+I)))=(2{)}^{\alpha }\cdot (3{)}^{\beta }\cdot (11{)}^{\gamma },\alpha ,\beta ,\gamma det ( a d j ( 2 a d j ( A + I ))) = ( 2 ) α ⋅ ( 3 ) β ⋅ ( 11 ) γ , α , β , γ are non-negative integers, then α + β + γ \alpha +\beta +\gamma α + β + γ is equal to ____
Answer: 18
Solutions · 1 Post your solution Official solution
A = − A T A = [ [ ∣ 0 α β − α 0 γ − β − γ 0 ] ] ∣ [ [ ∣ 0 α β − α 0 γ − β − γ 0 ] ] ∣ [ [ ∣ 1 − 1 0 ] ] ∣ = [ [ ∣ 3 3 2 ] ] ∣ α = − 3 , − β + γ = 2 [ [ ∣ 0 α β − α 0 γ − β − γ 0 ] ] ∣ [ [ ∣ 3 3 2 ] ] ∣ = [ [ ∣ − 3 19 β = 3 ] ] ∣ \begin{gathered} A=-{A}^{T} \\ A=[[|\begin{array}{|c|}0 \quad \alpha \quad \beta \\ -\alpha \quad 0 \quad \gamma \\ -\beta \quad -\gamma \quad 0\end{array}]]| \\ [[|\begin{array}{|c|}0 \quad \alpha \quad \beta \\ -\alpha \quad 0 \quad \gamma \\ -\beta \quad -\gamma \quad 0\end{array}]]|[[|\begin{array}{|c|}1 \\ -1 \\ 0\end{array}]]|=[[|\begin{array}{|c|}3 \\ 3 \\ 2\end{array}]]| \\ \alpha =-3,-\beta +\gamma =2 \\ [[|\begin{array}{|c|}0 \quad \alpha \quad \beta \\ -\alpha \quad 0 \quad \gamma \\ -\beta \quad -\gamma \quad 0\end{array}]]|[[|\begin{array}{|c|}3 \\ 3 \\ 2\end{array}]]|=[[|\begin{array}{|c|}-3 \\ 19 \\ \beta =3\end{array}]]| \end{gathered} A = − A T A = [[ ∣ 0 α β − α 0 γ − β − γ 0 ]] ∣ [[ ∣ 0 α β − α 0 γ − β − γ 0 ]] ∣ [[ ∣ 1 − 1 0 ]] ∣ = [[ ∣ 3 3 2 ]] ∣ α = − 3 , − β + γ = 2 [[ ∣ 0 α β − α 0 γ − β − γ 0 ]] ∣ [[ ∣ 3 3 2 ]] ∣ = [[ ∣ − 3 19 β = 3 ]] ∣
∣ A + I ∣ = ∣ 1 − 3 3 3 1 5 − 3 − 5 1 ∣ = 44 det ( a d j ( 2 a d j ( A + I ) ) ) = 2 6 det ( a d j ( a d j ( A + I ) ) ) = 2 6 ( 44 ) 4 = 2 14 × 11 4 \begin{gathered} |A+I|=|\begin{array}{|c|}1 \quad -3 \quad 3 \\ 3 \quad 1 \quad 5 \\ -3 \quad -5 \quad 1\end{array}|=44 \\ \det (adj(2adj(A+I)))={2}^{6}\det (adj(adj(A+I)))={2}^{6}(44{)}^{4}={2}^{14}\times {11}^{4} \end{gathered} ∣ A + I ∣ = ∣ 1 − 3 3 3 1 5 − 3 − 5 1 ∣ = 44 det ( a d j ( 2 a d j ( A + I ))) = 2 6 det ( a d j ( a d j ( A + I ))) = 2 6 ( 44 ) 4 = 2 14 × 11 4 Answer the question to unlock