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\(A=\begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\)

\(R_2\to R_2-R_1,\ R_3\to R_3-R_1\): \(\begin{bmatrix}2&3&4\\1&1&1\\2&2&2\end{bmatrix}\); \(R_3\to R_3-2R_2\): \(\begin{bmatrix}2&3&4\\1&1&1\\0&0&0\end{bmatrix}\)

\(R_1\leftrightarrow R_2,\ R_2\to R_2-2R_1\): \(\begin{bmatrix}1&1&1\\0&1&2\\0&0&0\end{bmatrix}\) (echelon form, 2 non-zero rows).

\(|A|=0\), but the 2×2 minor \(\begin{vmatrix}2&3\\3&4\end{vmatrix}=-1\ne0\). Rank of A = 2.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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