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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?

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

1 Answer 1

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$$AB = diag(2,3,5)$$

$$(AB)^{-1}=diag\left(\frac12, \frac13, \frac15\right)$$

$$(AB)^{-n}=diag\left(\frac1{2^n}, \frac1{3^n}, \frac1{5^n}\right)$$

$$X=diag\left(\frac{1/2}{1-1/2}, \frac{1/3}{1-1/3},\frac{1/5}{1-1/5} \right)=diag(1, 1/2, 1/4)$$

Hence now, you can compute quantity about $X$ and $Y$.

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