If the product D=ABC of three square matrices is invertible , then A must be invertible (so are B and C). Find a formula for A^{-1} (i.e. A^{-1}=dotsb) that involves only the matrices A, B B^{-1} , C, C^{-1} , D text{ and/or } D^{-1}

If the product D=ABC of three square matrices is invertible , then A must be invertible (so are B and C). Find a formula for A^{-1} (i.e. A^{-1}=dotsb) that involves only the matrices A, B B^{-1} , C, C^{-1} , D text{ and/or } D^{-1}

Question
Matrices
asked 2020-12-15
If the product D=ABC of three square matrices is invertible , then A must be invertible (so are B and C). Find a formula for \(A^{-1} (i.e. A^{-1}=\dotsb) that involves only the matrices \(A, B B^{-1} , C, C^{-1} , D \text{ and/or } D^{-1}\)

Answers (1)

2020-12-16
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: \(A^{-1}=BCD^{-1}\)
0

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