silemus.

23×33\times 3 determinant expansion

Find the determinant of A=[123045106]\displaystyle A = \begin{bmatrix}1 & 2 & 3\\0 & 4 & 5\\1 & 0 & 6\end{bmatrix}.

Expanding along the top row, det A = 1(24) − 2(−5) + 3(−4) = 24 + 10 − 12 = 22. The sign pattern alternates +, −, + across the row.

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