MHT CET · Maths · Matrices
If \(A\) is a square matrix of order \(n \times n\), then \(\operatorname{adj}(\operatorname{adj} A)\) is equal to
- A | \(\left.A\right|^{n} A\)
- B \(|A|^{n-1} A\)
- C \(|A|^{n-2} A\)
- D \(|A|^{n-3} A\)
Answer & Solution
Correct Answer
(C) \(|A|^{n-2} A\)
Step-by-step Solution
Detailed explanation
For any square matrix \(B\), we have \(B(\operatorname{adj} B)=|B| I_{n}\)
On taking \(B=\operatorname{adj} A\), we get \((\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|\operatorname{adj} A| I_{n}\)
\(
\Rightarrow \quad \operatorname{adj} A[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} I_{n}
\)
\(\left(\because|\operatorname{adj} A|=|A|^{n-1}\right)\)
\(\Rightarrow(A \operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A\)
\(\Rightarrow\left(|A| I_{n}\right)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A\)
\(\Rightarrow\)
\((\operatorname{adj} A)=|A|^{n-2} A\)
On taking \(B=\operatorname{adj} A\), we get \((\operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|\operatorname{adj} A| I_{n}\)
\(
\Rightarrow \quad \operatorname{adj} A[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} I_{n}
\)
\(\left(\because|\operatorname{adj} A|=|A|^{n-1}\right)\)
\(\Rightarrow(A \operatorname{adj} A)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A\)
\(\Rightarrow\left(|A| I_{n}\right)[\operatorname{adj}(\operatorname{adj} A)]=|A|^{n-1} A\)
\(\Rightarrow\)
\((\operatorname{adj} A)=|A|^{n-2} A\)
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