If matrix A and its inverse A^{-1} are given | vedclass

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(C) We know that $A \cdot A^{-1} = I$, where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Multiplyin

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$(\sqrt{2}, \frac{7\pi}{4})$

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(C) We know that $A \cdot A^{-1} = I$, where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Multiplying the second row of $A$ by the third column of $A^{-1}$ gives the element at $(2, 3)$ of $I$, which is $0$.$(1 	imes \frac{1}{2}) + (2 	imes y) + (3 	imes \frac{1}{2}) = 0 \implies \frac{1}{2} + 2y + \frac{3}{2} = 0 \implies 2 + 2y = 0 \implies y = -1$.Multiplying the third row of $A$ by the first column of $A^{-1}$ gives the element at $(3, 1)$ of $I$, which is $0$.$(3 	imes \frac{1}{2}) + (x 	imes -4) + (1 	imes \frac{5}{2}) = 0 \implies \frac{3}{2} - 4x + \frac{5}{2} = 0 \implies 4 - 4x = 0 \implies x = 1$.The Cartesian coordinates are $(1, -1)$.For polar coordinates $(r, 	heta)$, $r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.Since the point $(1, -1)$ is in the fourth quadrant, $	heta = \arctan(\frac{y}{x}) = \arctan(-1) = \frac{7\pi}{4}$.Thus, the polar coordinates are $(\sqrt{2}, \frac{7\pi}{4})$.

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