Let A be a 3×3 real matrix such that A2(A−2I)−4(A−I)=O, where I and O are the identity and null matrices, respectively. If A5=αA2+βA+γI, where α, β and γ are real constants, then α+β+γ is equal to:
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Step-by-Step Solution
Step 1: Simplify the given matrix equation
First, we expand the given matrix equation A2(A−2I)−4(A−I)=O. We distribute A2 and 4 into the parentheses to simplify the expression. This gives us a polynomial equation in terms of matrix A.
Step 2: Express A3 in terms of lower powers of A
From the simplified equation, we can express A3 in terms of A2, A, and I. This relationship will be crucial for reducing higher powers of A to a polynomial of degree at most 2.
Step 3: Calculate A4
Now we calculate A4 by multiplying A with the expression for A3. After substituting the expression for A3 again, we simplify to get A4 as a polynomial of degree 2 in A.
Step 4: Calculate A5
Similarly, we calculate A5 by multiplying A with the expression for A4. We substitute the expression for A3 one last time and simplify to obtain A5 in the desired form A5=αA2+βA+γI.
Step 5: Determine α,β,γ and their sum
By comparing A5=20A2+24A−32I with the given form A5=αA2+βA+γI, we identify the values of α, β, and γ. Finally, we sum these values to get the required result.