How Column Operations Change Determinants

Why Column Operations Have Their Own Rules

When computing a determinant, it is useful to change a matrix in ways whose effects on the determinant are known. Earlier material established the effects of swapping two rows, scaling a row, and adding a multiple of one row to a different row. The corresponding column rules can be derived without a new determinant formula: transpose the matrix, apply the row rule, and transpose back. This works because transposition preserves the determinant.

Throughout, let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\). A column operation changes one or more of these column vectors. The row-operation rules we use are: swapping two rows multiplies the determinant by \(-1\); multiplying a row by \(c\in\mathbb{R}\) multiplies the determinant by \(c\); and adding \(c\) times one row to a different row leaves the determinant unchanged. These rules apply to square matrices. We will also use the established result \(\det(\mathbf{A}^{\top})=\det(\mathbf{A})\), which holds for every square matrix.

The Three Column Rules

For a column swap, exchanging columns \(i\) and \(j\) in \(\mathbf{A}\) exchanges rows \(i\) and \(j\) in \(\mathbf{A}^{\top}\). The row swap changes the determinant’s sign, so the column swap does too. For scaling, multiplying column \(j\) by \(c\) multiplies row \(j\) of the transpose by \(c\), so the determinant is multiplied by \(c\). This includes \(c=0\): the resulting matrix has a zero column and its determinant is zero.

For a column replacement, choose distinct indices \(j\) and \(k\), and replace column \(j\) by \(\mathbf{a}_j+c\mathbf{a}_k\). In the transpose, this is exactly the row replacement \(R_j\to R_j+cR_k\), which leaves the determinant unchanged. The indices must be distinct: replacing a column by itself plus a multiple of itself would instead scale that column by \(1+c\).

Proof. Write \(\mathbf{A}'\) for the matrix after the indicated operation. In the first case, \((\mathbf{A}')^{\top}\) is obtained from \(\mathbf{A}^{\top}\) by swapping rows \(i\) and \(j\). The row-swap rule gives \(\det((\mathbf{A}')^{\top})=-\det(\mathbf{A}^{\top})\). Since transposition preserves determinants, \(\det(\mathbf{A}')=-\det(\mathbf{A})\). In the second case, \((\mathbf{A}')^{\top}\) is obtained by scaling row \(j\) of \(\mathbf{A}^{\top}\) by \(c\), so \(\det((\mathbf{A}')^{\top})=c\det(\mathbf{A}^{\top})\), and hence \(\det(\mathbf{A}')=c\det(\mathbf{A})\). In the third case, \((\mathbf{A}')^{\top}\) is obtained by replacing row \(j\) by row \(j\) plus \(c\) times row \(k\). Because \(j\ne k\), the row-replacement rule applies and gives \(\det((\mathbf{A}')^{\top})=\det(\mathbf{A}^{\top})\). Thus \(\det(\mathbf{A}')=\det(\mathbf{A})\). Each conclusion uses a row rule for a square matrix and the determinant-preserving transpose identity. \(\square\)

The theorem describes the effect of one operation at a time. When several operations are performed in sequence, apply the appropriate rule at each step: a swap contributes a factor of \(-1\), a scaling contributes its scale factor, and a replacement contributes a factor of \(1\). A replacement is not a scaling just because one column changes; its determinant effect depends on the specific form of the replacement.

Worked Column Operations

A Consequence for Columns That Are Combinations

The replacement rule can also reveal when a determinant must be zero. A list of columns is linearly dependent if one of its columns is a linear combination of the others. For the consequence below, the combination may involve any number of the other columns, including none; the empty combination is the zero vector.

Proof. Suppose column \(j\) is \(\sum_{k\ne j}c_k\mathbf{a}_k\). Replace column \(j\), one term at a time, by subtracting \(c_k\) times column \(k\), for each \(k\ne j\). Each replacement uses a different column and leaves the determinant unchanged by the theorem. After all these replacements, column \(j\) is \(\mathbf{a}_j-\sum_{k\ne j}c_k\mathbf{a}_k=\mathbf{0}\). A matrix with a zero column has determinant zero: its transpose has a zero row, whose determinant is zero by linearity of the determinant in that row. Since transposition preserves the determinant, the original matrix with its zero column also has determinant zero. The determinant was unchanged through the replacements, so the starting matrix has determinant zero as well. \(\square\)

This consequence gives a useful check on a calculation, but it does not say that every matrix with determinant zero must have a column that is visibly a combination of its other columns. The theorem here is only the stated implication; deciding the converse requires additional results about determinants and invertibility.

Common Mistake: Treating Replacement as Scaling

A column replacement does not multiply the determinant by the coefficient used in the replacement. For example, take \(\mathbf{D}=\begin{bmatrix}1&2\\3&5\end{bmatrix}\), whose determinant is \(1\cdot5-2\cdot3=-1\). Replace column 2 by column 2 plus twice column 1. A mistaken calculation might say that the determinant is multiplied by \(2\), giving \(-2\). But the operation does not scale column 2 by \(2\); it adds a multiple of a different column.

The actual new column 2 is \(\begin{bmatrix}2\\5\end{bmatrix}+2\begin{bmatrix}1\\3\end{bmatrix}=\begin{bmatrix}4\\11\end{bmatrix}\), so the new matrix is \(\mathbf{D}'=\begin{bmatrix}1&4\\3&11\end{bmatrix}\). Its determinant is \(1\cdot11-4\cdot3=11-12=-1\), unchanged. By contrast, if column 2 were actually multiplied by \(2\), the result would be \(\begin{bmatrix}1&4\\3&10\end{bmatrix}\), with determinant \(10-12=-2\). The operation’s form determines which rule applies.