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Q.If A=[2x0xx]A = \begin{bmatrix}2x & 0\\ x & x\end{bmatrix} and A−1=[10−12]A^{-1} = \begin{bmatrix}1 & 0\\ -1 & 2\end{bmatrix}, then the value of xx is : (A) 11 (B) 12\dfrac{1}{2} (C) −12-\dfrac{1}{2} (D) 22

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A A−1=IA\,A^{-1}=I forces 2x=12x=1, so x=12x=\tfrac12.

By definition of inverse, A A−1=I=[1001]A\,A^{-1}=I=\begin{bmatrix}1&0\\0&1\end{bmatrix}; multiply and match entries.

  1. Multiply: [2x0xx][10−12]=[2x+00+0x−x0+2x]=[2x002x]\begin{bmatrix}2x&0\\x&x\end{bmatrix}\begin{bmatrix}1&0\\-1&2\end{bmatrix} = \begin{bmatrix}2x+0 & 0+0\\ x-x & 0+2x\end{bmatrix} = \begin{bmatrix}2x&0\\0&2x\end{bmatrix}. …

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