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Invertible Element Binary Operation

For the follo...

Question

For the following pairs of matrices verity that ${(A B)}^{- 1} = B^{- 1} A^{- 1} :$

(i) $A = [\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] and B [\begin{matrix} 4 & 6 \\ 3 & 2 \end{matrix}]$

(ii) $A = [\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}] and B [\begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix}]$

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Solution

$(i) We have, A = [\begin{matrix} 3 & 2 \\ 7 & 5 \end{matrix}] and B = [\begin{matrix} 4 & 6 \\ 3 & 2 \end{matrix}] ∴ A B = [\begin{matrix} 18 & 22 \\ 43 & 52 \end{matrix}] |A B| = - 10 Since, |A B| \neq 0 Hence, A B is invertible . Let C_{i j} be the cofactor of a_{i n} in A B = [a_{i j}] C_{11} = 52, C_{12} = - 43, C_{21} = - 22 and C_{22} = 18 adj (A B) = {[\begin{matrix} 52 & - 43 \\ - 22 & 18 \end{matrix}]}^{T} = [\begin{matrix} 52 & - 22 \\ - 43 & 18 \end{matrix}] ∴ {(A B)}^{- 1} = - \frac{1}{10} [\begin{matrix} 52 & - 22 \\ - 43 & 18 \end{matrix}] . . . (1) Now, B = [\begin{matrix} 4 & 6 \\ 3 & 2 \end{matrix}] |B| = - 10 Since, |B| \neq 0 Hence, B is invertible . Let C_{i j} be the cofactor of a_{i n} in B = [a_{i j}] C_{11} = 2, C_{12} = - 3, C_{21} = - 6 and C_{22} = 4 adj B = {[\begin{matrix} 2 & - 3 \\ - 6 & 4 \end{matrix}]}^{T} = [\begin{matrix} 2 & - 6 \\ - 3 & 4 \end{matrix}] ∴ B^{- 1} = - \frac{1}{10} [\begin{matrix} 2 & - 6 \\ - 3 & 4 \end{matrix}] |A| = 1 Since, |A| \neq 0 Hence, A is invertible . Let C_{i j} be the cofactor of a_{i n} in A = [a_{i j}] C_{11} = 5, C_{12} = - 7, C_{21} = - 2 and C_{22} = 3 adj A = {[\begin{matrix} 5 & - 7 \\ - 2 & 3 \end{matrix}]}^{T} = [\begin{matrix} 5 & - 2 \\ - 7 & 3 \end{matrix}] ∴ A^{- 1} = [\begin{matrix} 5 & - 2 \\ - 7 & 3 \end{matrix}] Now, B^{- 1} A^{- 1} = - \frac{1}{10} [\begin{matrix} 52 & - 22 \\ - 43 & 18 \end{matrix}] . . . (2) From eq . (1) and (2), we have {(A B)}^{- 1} = B^{- 1} A^{- 1} Hence verified . (ii) We have, A = [\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}] and B = [\begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix}] ∴ A B = [\begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix}] Now, |A B| = 1 Since, |A B| \neq 0 Hence, A B is invertible . Let C_{i j} be the cofactor of a_{i n} in A B = [a_{i j}] C_{11} = 37, C_{12} = - 29, C_{21} = - 14 and C_{22} = 11 adj (A B) = [\begin{matrix} 37 & - 14 \\ - 29 & 11 \end{matrix}] ∴ {(A B)}^{- 1} = [\begin{matrix} 37 & - 14 \\ - 29 & 11 \end{matrix}] . . . (1) |B| = 1 Since, |B| \neq 0 Hence, B is invertible . Let C_{i j} be the cofactor of a_{i n} in B = [a_{i j}] C_{11} = 4, C_{12} = - 3, C_{21} = - 5 and C_{22} = 4 adj B = [\begin{matrix} 4 & - 5 \\ - 3 & 4 \end{matrix}] ∴ B^{- 1} = [\begin{matrix} 4 & - 5 \\ - 3 & 4 \end{matrix}] |A| = 1 Since, |A| \neq 0 Hence, A is invertible . Let C_{i j} be the cofactor of a_{i n} in A = [a_{i j}] C_{11} = 3, C_{12} = - 5, C_{21} = - 1 and C_{22} = 2 adj A = [\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}] ∴ A^{- 1} = [\begin{matrix} 3 & - 1 \\ - 5 & 2 \end{matrix}] Now, B^{- 1} A^{- 1} = [\begin{matrix} 37 & - 14 \\ - 29 & 11 \end{matrix}] . . . (2) From eq . (1) and (2), we have {(A B)}^{- 1} = B^{- 1} A^{- 1} Hence verified .$
(AB)−1=B−1 A−1 (AB)−1=B−1 A−1(AB)−1=B−1 A−1

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A

B

are invertible matrices of the same order, then prove that

(A B)^{- 1} = B^{- 1} A^{- 1}

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adj A = |A| A^{- 1}

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

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{(A B)}^{- 1} = B^{- 1} A^{- 1}

If A and B are invertible matrices, then which of the following is not correct?

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|A|

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