If `A` is an invertible matrix, tehn `(a d jdotA)^(-1)` is equal to `a d jdot(A^(-1))` b. `A/(d e tdotA)` c. `A` d. `(detA)A`

Question

If `A` is an invertible matrix, tehn `(a d jdotA)^(-1)` is equal to `a d jdot(A^(-1))` b. `A/(d e tdotA)` c. `A` d. `(detA)A`

If `A` is an invertible matrix, tehn `(a d jdotA)^(-1)` is equal to `a d jdot(A^(-1))` b. `A/(d e tdotA)` c. `A` d. `(detA)A`
A. adj. `(A^(-1))`
B. `A/("det. A")`
C. `A`
D. (det. A) A

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`A^(-1)=("adj (A)")/(|A|)`
`:. ("adj A")^(-1)=("adj (adj A)")/(|"adj A"|)`
`=(|A|^(n-2) A)/(|A|^(n-1))`
`=A/(|A|)`
Also A (adj A) `=|A|I`
or `A^(-1) ("adj "A^(-1))=|A^(-1|I`
or `A^(-1) ("adj "A^(-1))=I/(|A|)`
or `A A^(-1) ("adj "A^(-1))=(A.I)/(|A|)`
or `I ("adj "A^(-1))=A/(|A|)`
or `("adj "A^(-1))=A/(|A|)=("adj "A)^(-1)`