Let A, B, C, D and E be n times n matrices,ea | vedclass

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(B) Given the equation $ABCDE = I$. Multiply both sides by $A^{-1}$ from the left: $A^{-1}(ABCDE) = A^{-1}I \Rightarrow BCDE = A^{-1}$. Multiply both

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$DEAB$

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(B) Given the equation $ABCDE = I$. Multiply both sides by $A^{-1}$ from the left: $A^{-1}(ABCDE) = A^{-1}I \Rightarrow BCDE = A^{-1}$. Multiply both sides by $B^{-1}$ from the left: $B^{-1}(BCDE) = B^{-1}A^{-1} \Rightarrow CDE = B^{-1}A^{-1}$. Multiply both sides by $E^{-1}$ from the right: $(CDE)E^{-1} = B^{-1}A^{-1}E^{-1} \Rightarrow CD = B^{-1}A^{-1}E^{-1}$. Multiply both sides by $D^{-1}$ from the right: $(CD)D^{-1} = B^{-1}A^{-1}E^{-1}D^{-1} \Rightarrow C = B^{-1}A^{-1}E^{-1}D^{-1}$. Taking the inverse of both sides: $C^{-1} = (B^{-1}A^{-1}E^{-1}D^{-1})^{-1}$. Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we get $C^{-1} = (D^{-1})^{-1}(E^{-1})^{-1}(A^{-1})^{-1}(B^{-1})^{-1} = DEAB$.

Let $A, B, C, D$ and $E$ be $n \times n$ matrices,each with non-zero determinant. If $ABCDE=I$,then $C^{-1}=$

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