Step 1

Given that the product D=ABC of three square matrices is invertible.

Also given that A must be invertible and so are B and C.

To find \(A^{-1}\) that involves only the matrices \(A, B B^{-1} , C, C^{-1} , D \text{ and/or } D^{-1}\)

Since A, B, C and D are invertible so \(A^{-1}, B^{-1}, C^{-1} \text {and } D^{-1}\) exists.

Given,

D=ABC

Post multiply this equation with \(D^{-1}\) on both sides.

\(D(D^{-1})=(ABC)(D^{-1})\)

\(DD^{-1}=ABCD^{-1} \ \ \ \ \ \ (DD^{-1}=I)\)

\(I=ABCD^{-1} \ \ \ \text{ where I is the identity matrix.}\)

Step 2

Now the equation is,

\(I=ABCD^{-1}\)

Pre multiply this equation with \(A^{-1}\) on both sides.

\((A^{-1})I=(A^{-1})(ABCD^{-1})\)

\(A^{-1}I=(A^{-1}A)(BCD^{-1})\)

\((\because \text{ Matrices are associative, } (AB)C=A(BC) )\)

\(A^{-1}=(I)BCD^{-1} \ \ \ \ \ (A^{-1}A=I)\)

\(A^{-1}=BCD^{-1}\)

Hence, the formula of \(A^{-1}\) involving the matrices \(A, B, B^{-1}, C, C^{-1}, D \text{ and/or } D^{-1}\) is,

\(A^{-1}=BCD^{-1}\)

Answer: \(A^{-1}=BCD^{-1}\)

Given that the product D=ABC of three square matrices is invertible.

Also given that A must be invertible and so are B and C.

To find \(A^{-1}\) that involves only the matrices \(A, B B^{-1} , C, C^{-1} , D \text{ and/or } D^{-1}\)

Since A, B, C and D are invertible so \(A^{-1}, B^{-1}, C^{-1} \text {and } D^{-1}\) exists.

Given,

D=ABC

Post multiply this equation with \(D^{-1}\) on both sides.

\(D(D^{-1})=(ABC)(D^{-1})\)

\(DD^{-1}=ABCD^{-1} \ \ \ \ \ \ (DD^{-1}=I)\)

\(I=ABCD^{-1} \ \ \ \text{ where I is the identity matrix.}\)

Step 2

Now the equation is,

\(I=ABCD^{-1}\)

Pre multiply this equation with \(A^{-1}\) on both sides.

\((A^{-1})I=(A^{-1})(ABCD^{-1})\)

\(A^{-1}I=(A^{-1}A)(BCD^{-1})\)

\((\because \text{ Matrices are associative, } (AB)C=A(BC) )\)

\(A^{-1}=(I)BCD^{-1} \ \ \ \ \ (A^{-1}A=I)\)

\(A^{-1}=BCD^{-1}\)

Hence, the formula of \(A^{-1}\) involving the matrices \(A, B, B^{-1}, C, C^{-1}, D \text{ and/or } D^{-1}\) is,

\(A^{-1}=BCD^{-1}\)

Answer: \(A^{-1}=BCD^{-1}\)