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Q.The inverse of matrix A=[4−121]A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix} is (A) 16[−42−1−1]\dfrac{1}{6}\begin{bmatrix} -4 & 2 \\ -1 & -1 \end{bmatrix} (B) [131623−16]\begin{bmatrix} \frac{1}{3} & \frac{1}{6} \\ \frac{2}{3} & -\frac{1}{6} \end{bmatrix} (C) [1616−1323]\begin{bmatrix} \frac{1}{6} & \frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} \end{bmatrix} (D) [−2316−13−16]\begin{bmatrix} -\frac{2}{3} & \frac{1}{6} \\ -\frac{1}{3} & -\frac{1}{6} \end{bmatrix}

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det⁡A=6\det A = 6 and adj⁡A=[11−24]\operatorname{adj}A=\begin{bmatrix}1&1\\-2&4\end{bmatrix}, so A−1=[1/61/6−1/32/3]A^{-1}=\begin{bmatrix}1/6&1/6\\-1/3&2/3\end{bmatrix}.

For A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}: A−1=1ad−bc[d−b−ca]A^{-1}=\dfrac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}.

  1. Compute the determinant: det⁡A=(4)(1)−(−1)(2)=4+2=6\det A = (4)(1)-(-1)(2) = 4+2 = 6 (non-zero, so the inverse exists).
  2. Form the adjoint by swapping diagonal entries and negating off-diagonal: adj⁡A=[11−24]\operatorname{adj}A = \begin{bmatrix}1 & 1\\ -2 & 4\end{bmatrix}. …

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