Let A be a 3×3 matrix such that ∣adj(adj(adjA))∣=81. If S={n∈Z:∣adj(adjA)∣2(n−1)2=∣A∣3n2−5n−4}, then ∑n∈SAn2+n is equal to:
Get the complete, step-by-step math solution for: "Let A be a 3 × 3 matrix such that | {adj} ( {adj}( {adj} A))| = 81. If S = \{ n {Z} : | {adj}( {adj} A)|^{((n-1)²)/(2)} = |A|^{3n² - 5n - 4} \}, then ...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Determine the determinant of A
For a k×k matrix A, the determinant of the adjoint of the adjoint of the adjoint of A is given by the formula ∣adj(adj(adjA))∣=∣A∣(k−1)3. In this problem, A is a 3×3 matrix, so k=3. We are given that ∣adj(adj(adjA))∣=81. We can use this to find the value of ∣A∣.
Step 2: Calculate |A|
Substituting k=3 into the formula, we get ∣A∣(3−1)3=∣A∣8. We are given that this is equal to 81. Since 81=34, we have ∣A∣8=34. Taking the eighth root of both sides, we find that ∣A∣=±(34)1/8=±31/2=±3.
Step 3: Simplify the expression for |adj(adj A)|
For a k×k matrix A, the determinant of the adjoint of the adjoint of A is given by the formula ∣adj(adjA)∣=∣A∣(k−1)2. Since k=3, this simplifies to ∣A∣(3−1)2=∣A∣4.
Step 4: Substitute into the given equation for S
Now we substitute ∣adj(adjA)∣=∣A∣4 into the given equation for set S. This gives us (∣A∣4)2(n−1)2=∣A∣3n2−5n−4. Using the exponent rule (ab)c=abc, the left side becomes ∣A∣4⋅2(n−1)2=∣A∣2(n−1)2.
Step 5: Solve for n
Since ∣A∣=±3 and ∣A∣=0,1, we can equate the exponents: 2(n−1)2=3n2−5n−4. Expanding and simplifying the quadratic equation: 2(n2−2n+1)=3n2−5n−4⟹2n2−4n+2=3n2−5n−4. Rearranging terms gives n2−n−6=0. Factoring the quadratic equation yields (n−3)(n+2)=0. Thus, the possible values for n are n=3 and n=−2. So, S={−2,3}.
Step 6: Calculate the sum of the determinants
We need to calculate ∑n∈S∣An2+n∣. We know that ∣Am∣=∣A∣m. So, we need to calculate ∣A∣n2+n for n=3 and n=−2.
For n=3, the exponent is 32+3=9+3=12. So, ∣A∣12=(±3)12=(31/2)12=36=729.
For n=−2, the exponent is (−2)2+(−2)=4−2=2. So, ∣A∣2=(±3)2=3.
The sum is 729+3=732.