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

(a) |AB^(-1) |=(|A|)/(|B|)                            (b) |(AB)^(-1) |=1/(|A||B|)

(c) (AB)^(-1)=B^(-1) A^(-1)                       (d) (A+B)^(-1)=B^(-1)+A^(-1)

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Let’s dicuss each option one by one Option (a) - |〖𝑨𝑩〗^(βˆ’πŸ) |=(|𝑨|)/(|𝑩|) Solving LHS |〖𝐴𝐡〗^(βˆ’1) |=|𝐴||𝐡^(βˆ’1) | = |𝐴| 1/(|𝐡|) = (|𝐴|)/(|𝐡|) So, option (a) is correct Option (b) - |(𝑨𝑩)^(βˆ’πŸ) |=𝟏/(|𝑨||𝑩|) Solving L.H.S |γ€–(𝐴𝐡)γ€—^(βˆ’1) | = | 𝐡^(βˆ’1) 𝐴^(βˆ’1)| = | 𝑩^(βˆ’πŸ) γ€–| |𝑨〗^(βˆ’πŸ)| = 1/(|𝐡|) 1/(|𝐴|) = 𝟏/(|𝑨||𝑩|) = R.H.S So, option (b) is correct Option (c) - (𝐴𝐡)^(βˆ’1)=𝐡^(βˆ’1) 𝐴^(βˆ’1) This is a correct property. So, option (c) is correct Option (d) - (𝑨+𝑩)^(βˆ’πŸ)=𝑩^(βˆ’πŸ)+𝑨^(βˆ’πŸ) Let’s check this with the help of an example Let’s consider A = [β– (1&3@0&2)] and B = [β– (2&1@1&2)] Since L.H.L β‰  R.H.L Thus, (𝐴+𝐡)^(βˆ’1)≠𝐡^(βˆ’1)+𝐴^(βˆ’1) ∴ So, option (d) is incorrect So, the correct answer is (d)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo