An algebra problem by أحمد الحلاق

Algebra Level 5

Let A = [ 3 2 2 2 ] A = \begin{bmatrix}{3} & {2} \\ {2} & {2}\end{bmatrix} . Evaluate det ⁡ ( 4 ( A − 1 ) ) + det ⁡ ( 4 ( A − 1 ) 2 ) + det ⁡ ( 4 ( A − 1 ) 3 ) + ⋯ + det ⁡ ( 4 ( A − 1 ) 6 ) \det (4(A^{-1})) + \det (4(A^{-1})^2) + \det (4(A^{-1})^3) + \cdots + \det (4(A^{-1})^6)


The answer is 15.75.

أحمد الحلاق by أحمد الحلاق , 28, Palestine

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1 solution

det ⁡ ( A ) = 2 ⇒ det ⁡ ( A − 1 ) = 1 2 ⇒ det ⁡ ( 4 ( A − 1 ) ) = 16 ⋅ 1 2 \det (A) = 2 \Rightarrow \det (A^{-1}) = \frac{1}{2} \Rightarrow \det (4(A^{-1})) = 16\cdot\frac{1}{2} Now due to the property det ⁡ ( A ⋅ B ) = det ⁡ ( A ) ⋅ det ⁡ ( B ) \det (A \cdot B) = \det(A)\cdot \det(B) , we get det ⁡ ( 4 ( A − 1 ) ) + det ⁡ ( 4 ( A − 1 ) 2 ) + det ⁡ ( 4 ( A − 1 ) 3 ) + ⋯ + det ⁡ ( 4 ( A − 1 ) 6 ) = \det (4(A^{-1})) + \det (4(A^{-1})^2) + \det (4(A^{-1})^3) + \cdots + \det (4(A^{-1})^6) = = 16 ⋅ ( 1 2 + 1 4 + 1 8 + 1 16 + 1 32 + 1 64 ) = 8 + 4 + 2 + 1 + 1 / 2 + 1 / 4 = 15.75 = 16\cdot(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}) = 8 + 4 + 2 +1 + 1/2 + 1/4 = \boxed{15.75}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

highly over rated question.

rajdeep brahma - 3 years ago

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