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Example · Example 9

Q.Find the inverse of A=[2111]A=\begin{bmatrix}2&1\\1&1\end{bmatrix} using the adjoint-determinant method, and verify that AA−1=IAA^{-1}=I.

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For A=[21\11]A=\begin{bmatrix}2&1\1&1\end{bmatrix}, det⁡A=2(1)−1(1)=1\det A=2(1)-1(1)=1. Since det⁡Aeq0\det A eq0, AA is invertible, and A−1=1det⁡A[1−1\-12]=[1−1\-12]A^{-1}=\dfrac{1}{\det A}\begin{bmatrix}1&-1\-1&2\end{bmatrix}=\begin{bmatrix}1&-1\-1&2\end{bmatrix} (diagonal entries swapped, off-diagonal entries negated, divided by 11). Verifying AA−1AA^{-1}: position (1,1)(1,1): 2(1)+1(−1)=12(1)+1(-1)=1; position (1,2)(1,2): 2(−1)+1(2)=02(-1)+1(2)=0; position (2,1)(2,1): 1(1)+1(−1)=01(1)+1(-1)=0; po …

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