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If AA is a square matrix of order 3 such that A(adj A)=[−3000−3000−3]A(adj\,A)=\begin{bmatrix}-3 & 0 & 0\\0 & -3 & 0\\0 & 0 & -3\end{bmatrix}, then which of the following is NOT correct?

Solution

✅ Correct Option: 4

Since A(adj A)=∣A∣IA(adj\,A)=|A|I, here ∣A∣=−3≠0|A|=-3\neq0, so AA is invertible and A−1A^{-1} is non-singular, and ∣adj A∣=∣A∣2=9|adj\,A|=|A|^{2}=9. Hence the claim that AA is singular is the incorrect one.

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