TS EAMCET 10 May 2024 Shift 2 Solved Paper

Given that $\left|\operatorname{adj} A\right|=x$ and $\left|\operatorname{adj} B\right|=y$, we are tasked with finding $\left|\left(\operatorname{adj}(AB)\right)^{-1}\right|$.First, recognize that the determinant of the adjugate of a matrix can be expressed with respect to the determinants of the matrix itself. Using this relation, we have:$\left|\operatorname{adj}(AB)\right|=\left|\operatorname{adj}(\operatorname{adj}(A))\right|\cdot\left|\operatorname{adj}(\operatorname{adj}(B))\right|$The inverse of the adjugate adjoint product can be decomposed as:$\left|\left(\operatorname{adj}(AB)\right)^{-1}\right|=\left|(\operatorname{adj} B)^{-1}(\operatorname{adj} A)^{-1})\right|$Because $\left|\operatorname{adj} A\right|=x$ and $\left|\operatorname{adj} B\right|=y$, the properties of determinants give us:$\; \left|\left(\operatorname{adj} A\right)^{-1}\right| = \frac{1}{\left|\operatorname{adj} A\right|} = \frac{1}{x}$$\; \left|\left(\operatorname{adj} B\right)^{-1}\right| = \frac{1}{\left|\operatorname{adj} B\right|} = \frac{1}{y}$Thus, the determinant of the inverse of the product is:$\left|\left(\operatorname{adj}(AB)\right)^{-1}\right| = \frac{1}{x} \times \frac{1}{y} = \frac{1}{x y}$Therefore, the result is $\; \frac{1}{x y}$.