# Infinite series of matrix product

If $$t$$ is real and positive and matrix $$A$$ and $$B$$ are such that $$A = \begin{pmatrix} \dfrac{t^2+1}{t} & 0 & 0 \\ 0 & t & 0 \\ 0 & 0 & 25 \end{pmatrix}$$ and $$B=\begin{pmatrix} \dfrac{2t}{t^2+1} & 0 & 0 \\ 0 & \dfrac{3}{t} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}$$ and $$X=(AB)^{-1} + (AB)^{-2} + (AB)^{-3} + \cdots\infty, Y=X^{-1}$$ then which option is correct?

$$1.$$ $$\text{det}(Y)=10$$

$$2.$$ $$X\cdot \text{adj(adj}(Y)=8I$$

$$3.$$ $$\text{det}(Y)=20$$

$$4.$$ $$X\cdot \text{adj(adj}(Y)=5I$$

My attempt:

I managed to calculate some of the important matrices, but am stuck in a part where I have to find the determinant of a matrix that has some elements tending to infinity.

As $$t>0$$ $$AB= \begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix}$$

The inverse of a diagonal matrix is another diagonal matrix that's diagonal elements are reciprocals of the diagonal elements of the original matrix, so

$$(AB)^{-1}= \begin{pmatrix} \dfrac{1}{2} & 0 & 0 \\ 0 & \dfrac{1}{3} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}, (AB)^{-2}=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5 \end{pmatrix}, (AB)^{-3}=\begin{pmatrix} \dfrac{1}{2} & 0 & 0 \\ 0 & \dfrac{1}{3} & 0 \\ 0 & 0 & \dfrac{1}{5} \end{pmatrix}, \cdots$$

My problem is, adding these matrices results in a matrix that's diagonal elements tend to infinity, as $$X=\begin{pmatrix} 2+1/2+2+1/2+\cdots \infty & 0 & 0 \\ 0 & 3+1/3+3+1/3+\cdots \infty \\ 0 & 0 & 5+1/5+5+1/5+\cdots \infty\end{pmatrix} = \begin{pmatrix} S_2 & 0 & 0 \\ 0 & S_3 & 0\\ 0 & 0 & S_5 \end{pmatrix}$$

And $$\text{adj}(X) = \begin{pmatrix} S_3S_5 & 0 & 0 \\ 0 & S_2S_5 & 0\\ 0 & 0 & S_2S_3 \end{pmatrix}$$

I know that $$Y=\dfrac{\text{adj}(X)}{|X|}$$ but I can't figure out how to calculate $$|X|$$. The determinant of a diagonal matrix is simply the product of the diagonal elements, but how to calculate this value when the diagonal elements tend to infinity?

• $$(AB)^{-2}\neq {(AB)^{-1}}^{-1}$$ Jul 21, 2021 at 10:19

$$AB = diag(2,3,5)$$
$$(AB)^{-1}=diag\left(\frac12, \frac13, \frac15\right)$$
$$(AB)^{-n}=diag\left(\frac1{2^n}, \frac1{3^n}, \frac1{5^n}\right)$$
$$X=diag\left(\frac{1/2}{1-1/2}, \frac{1/3}{1-1/3},\frac{1/5}{1-1/5} \right)=diag(1, 1/2, 1/4)$$
Hence now, you can compute quantity about $$X$$ and $$Y$$.