Exercise 902Linear algebra › DeterminantsHigh schoolAdded 3 Nov 2026

Expansion along a column

#cofactor expansion#minors#determinant

Statement

Compute the determinant of
$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{pmatrix}$
by expanding along the first column.

Hint

Use the cofactor expansion along column $1$ : $\det A = \sum_i (-1)^{i+1}a_{i1}\det M_{i1}$ .

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Solution

Expansion along column 1.
$\det A = 1\cdot\begin{vmatrix}4&5\\0&6\end{vmatrix} - 0\cdot\begin{vmatrix}2&3\\0&6\end{vmatrix} + 1\cdot\begin{vmatrix}2&3\\4&5\end{vmatrix}.$

Minor computations.
$\begin{vmatrix}4&5\\0&6\end{vmatrix} = 24, \qquad \begin{vmatrix}2&3\\4&5\end{vmatrix} = 10-12=-2.$

Conclusion. $\det A = 1\cdot24 + 1\cdot(-2) = 22$ .

Check by Sarrus. $1\cdot4\cdot6 + 2\cdot5\cdot1 + 3\cdot0\cdot0 - 3\cdot4\cdot1 - 2\cdot0\cdot6 - 1\cdot5\cdot0 = 24+10+0-12-0-0 = 22$ . Consistent.

Expansion along a column

Statement

Solution

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