Let α be a solution of x2+x+1=0, and for some a and b in R, [4ab]1−1−216−1−14132−8=[000]. If α44+αm+α2n=3, then m + n is equal to
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Step-by-Step Solution
Step 1: Analyze the quadratic equation
The given quadratic equation is x2+x+1=0. The roots of this equation are the complex cube roots of unity, denoted by ω and ω2. Since α is a solution, we can say α=ω. We know that ω3=1 and 1+ω+ω2=0.
Step 2: Simplify the expression involving alpha
Substitute α=ω into the given expression. Since ω3=1, we have ω4=ω3⋅ω=1⋅ω=ω. Also, ω1=ω3ω2=ω2 and ω21=ω3ω=ω. Substitute these simplified terms back into the expression.
Step 3: Solve for m and n using properties of roots of unity
From 1+ω+ω2=0, we know that ω2=−1−ω. Substitute this into the equation. Since 1,ω,ω2 are linearly independent over real numbers, for an equation of the form A+Bω=0 where A and B are real, we must have A=0 and B=0. Here, we have (−4−m−3)+(n−4−m)ω=0.
Step 4: Formulate and solve system of equations
Equating the real and imaginary parts to zero, we get two linear equations. Solving the first equation for m gives m=−7. Substituting this value into the second equation allows us to solve for n, yielding n=−3.
Step 5: Calculate m + n
Finally, we need to find the sum m+n. Substitute the values of m and n we found.