If A=left[begin{array}{ccc}1 & 2 & 3 1 & 1 & | vedclass

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(A) We are given the matrices $A, B, C$. We need to evaluate the expression $X = \left(\left(\left((A B C)^{-1}\right)^T\right)^{-1}\right)^T$.Using t

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$\left[\begin{array}{ccc}64 & 39 & 28 \ 29 & 16 & 11 \ 11 & 2 & 5\end{array}ight]$

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(A) We are given the matrices $A, B, C$. We need to evaluate the expression $X = \left(\left(\left((A B C)^{-1}ight)^Tight)^{-1}ight)^T$.Using the properties of matrix transpose and inverse: $(P Q)^{-1} = Q^{-1} P^{-1}$,$(P^T)^{-1} = (P^{-1})^T$,and $(P^T)^T = P$.First,$(A B C)^{-1} = C^{-1} B^{-1} A^{-1}$.Then,$((A B C)^{-1})^T = (C^{-1} B^{-1} A^{-1})^T = (A^{-1})^T (B^{-1})^T (C^{-1})^T = (A^T)^{-1} (B^T)^{-1} (C^T)^{-1}$.Next,$(((A B C)^{-1})^T)^{-1} = ((A^T)^{-1} (B^T)^{-1} (C^T)^{-1})^{-1} = ((C^T)^{-1})^{-1} ((B^T)^{-1})^{-1} ((A^T)^{-1})^{-1} = C^T B^T A^T$.Finally,$X = (C^T B^T A^T)^T = (A^T)^T (B^T)^T (C^T)^T = A B C$.Now,calculate $AB$:$AB = \left[\begin{array}{ccc}1 & 2 & 3 \ 1 & 1 & 1 \ 1 & -1 & 1\end{array}ight] \left[\begin{array}{lll}1 & 1 & 0 \ 0 & 1 & 3 \ 3 & 0 & 4\end{array}ight] = \left[\begin{array}{ccc}1+0+9 & 1+2+0 & 0+6+12 \ 1+0+3 & 1+1+0 & 0+3+4 \ 1+0+3 & 1-1+0 & 0-3+4\end{array}ight] = \left[\begin{array}{ccc}10 & 3 & 18 \ 4 & 2 & 7 \ 4 & 0 & 1\end{array}ight]$.Now,calculate $(AB)C$:$(AB)C = \left[\begin{array}{ccc}10 & 3 & 18 \ 4 & 2 & 7 \ 4 & 0 & 1\end{array}ight] \left[\begin{array}{lll}2 & 0 & 1 \ 0 & 1 & 0 \ 3 & 2 & 1\end{array}ight] = \left[\begin{array}{ccc}20+0+54 & 0+3+36 & 10+0+18 \ 8+0+21 & 0+2+14 & 4+0+7 \ 8+0+3 & 0+0+2 & 4+0+1\end{array}ight] = \left[\begin{array}{ccc}64 & 39 & 28 \ 29 & 16 & 11 \ 11 & 2 & 5\end{array}ight]$.

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