If A^{-1}=left[begin{array}{lll}3 & 2 & 6 1 | vedclass

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(A) To find $A$,we use the property $(A^{-1})^{-1} = A$. We calculate the inverse of the given matrix $A^{-1}$ using the adjoint method or row operati

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$\left[\begin{array}{ccc}-5 & 20 & -2 \ -1 & 3 & 0 \ 3 & -11 & 1\end{array}ight]$

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(A) To find $A$,we use the property $(A^{-1})^{-1} = A$. We calculate the inverse of the given matrix $A^{-1}$ using the adjoint method or row operations. Let $M = A^{-1} = \left[\begin{array}{lll}3 & 2 & 6 \ 1 & 1 & 2 \ 2 & 5 & 5\end{array}ight]$.First,find the determinant $|M| = 3(5-10) - 2(5-4) + 6(5-2) = 3(-5) - 2(1) + 6(3) = -15 - 2 + 18 = 1$.Next,find the matrix of cofactors $C_{ij}$:$C_{11} = +(5-10) = -5, C_{12} = -(5-4) = -1, C_{13} = +(5-2) = 3$$C_{21} = -(10-30) = 20, C_{22} = +(15-12) = 3, C_{23} = -(15-4) = -11$$C_{31} = +(4-6) = -2, C_{32} = -(6-6) = 0, C_{33} = +(3-2) = 1$The adjoint matrix $adj(M) = \left[\begin{array}{ccc}-5 & 20 & -2 \ -1 & 3 & 0 \ 3 & -11 & 1\end{array}ight]$.Since $A = M^{-1} = \frac{1}{|M|} adj(M) = \frac{1}{1} \left[\begin{array}{ccc}-5 & 20 & -2 \ -1 & 3 & 0 \ 3 & -11 & 1\end{array}ight]$.Thus,$A = \left[\begin{array}{ccc}-5 & 20 & -2 \ -1 & 3 & 0 \ 3 & -11 & 1\end{array}ight]$.

If $A^{-1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 5 & 5\end{array}\right]$,then $A=$

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