Matrix Reloaded

Algebra Level 4

Let $A = \left( \begin{matrix} 4 & 2 \\ 2 & 2 \end{matrix} \right)$ Evaluate $\displaystyle \sum_{n = 1}^{\infty} \det (4(A^{-1})^{n})$ The above expression has a closed form. Give your answer to 3 decimal places.

Inspiration

The answer is 5.33333333.

1 solution

Guillermo Templado
Sep 30, 2016

$\det(A) = 4 \Rightarrow \det(A^{-1}) = \frac{1}{4}$ . Due to the property $\det(A\cdot B) = \det(A)\cdot \det(B)$ it can be proved by induction that $\det((A^{-1})^n) = \left(\frac{1}{4}\right)^n, \space \forall n \ge 1$ and $\det(4(A^{-1})^n) = 16\cdot \left(\frac{1}{4}\right)^n, \space \forall n \ge 1$ . Therefore, $\displaystyle \sum_{n = 1}^{\infty} \det(4(A^{-1})^n) = 16\cdot \sum_{n = 1}^{\infty} \left(\frac{1}{4}\right)^n = \frac{16}{3} \approx 5.33333333...$

Matrix Reloaded

The answer is 5.33333333.

1 solution

0 pending reports