Let A be a matrix of order 3×3 with ∣A∣=5. If 2adj(3Aadj(2A))=2α3β5γ, where α,β,γ∈N, then α+β+γ is equal to:
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Step-by-Step Solution
Step 1: Recall Properties of Determinants and Adjoint
We begin by recalling some fundamental properties of determinants and adjoints for an n×n matrix X. These properties are crucial for simplifying the given expression. Specifically, we'll use how scalars interact with determinants and adjoints, and the determinant of an adjoint matrix.
Step 2: Simplify the innermost adjoint term
The matrix A is of order 3×3, so n=3. We first simplify the determinant of the innermost term, adj(2A). Using the property ∣adj(X)∣=∣X∣n−1 and ∣kX∣=kn∣X∣, we find that ∣adj(2A)∣=∣2A∣2=(23∣A∣)2. Given ∣A∣=5, this simplifies to (8⋅5)2=402=1600.
Step 3: Simplify the expression inside the outermost adjoint
Next, we evaluate the determinant of the expression 3Aadj(2A). Using the property ∣XY∣=∣X∣∣Y∣, we can write this as ∣3A∣∣adj(2A)∣. We know ∣3A∣=33∣A∣=27⋅5=135, and we calculated ∣adj(2A)∣=1600. Multiplying these values gives 135⋅1600=216000.
Step 4: Evaluate the full expression
Finally, we evaluate the determinant of the entire expression. Using the property ∣kX∣=kn∣X∣ and ∣adj(X)∣=∣X∣n−1, we get 23∣adj(3Aadj(2A))∣=8⋅∣3Aadj(2A)∣2. Substituting the value 216000 for ∣3Aadj(2A)∣, we have 8⋅(216000)2. We then express 216000 in its prime factorization: 216000=216⋅1000=63⋅103=(2⋅3)3⋅(2⋅5)3=23⋅33⋅23⋅53=26⋅33⋅53. Squaring this gives (26⋅33⋅53)2=212⋅36⋅56. Multiplying by 23 gives 23⋅212⋅36⋅56=215⋅36⋅56.
Step 5: Determine α,β,γ and their sum
By comparing the final expression 215⋅36⋅56 with the given form 2α3β5γ, we can identify the values of α, β, and γ. We find α=15, β=6, and γ=6. The problem asks for the sum α+β+γ, which is 15+6+6=27.