The Effect Of Scaling A Row

One Row Contributes One Scalar Factor

A determinant depends on all the rows of a square matrix, but it responds to a change in one row in a particularly simple way. If one row is multiplied by \(c\), the determinant is multiplied by \(c\)—not by a power of \(c\) depending on the matrix size. The matrix size matters only when the same scaling is applied to every row: then the factor is counted once for each row.

Throughout this tutorial, matrices are in \(\mathbb{R}^{n\times n}\), where \(n\geq 1\). For a matrix \(\mathbf{A}=(a_{ij})\), let \(\mathbf{A}'\) denote the matrix obtained by multiplying row \(i\) by a scalar \(c\in\mathbb{R}\), leaving every other row unchanged. Thus the entries in row \(i\) of \(\mathbf{A}'\) are \(c a_{ij}\). The determinant is defined for square matrices; the row-scaling rule below holds whether \(c\) is positive, negative, or zero.

The distinction between one row and every row is essential. A row scaling by \(c\) contributes one factor of \(c\) to the determinant. Scaling all \(n\) rows by the same \(c\) contributes \(n\) such factors, so the total factor is \(c^n\). We will prove both statements and compute examples directly.

Proof from the Determinant Formula

The Leibniz formula expresses the determinant as a sum indexed by permutations. Here, \(\mathfrak{S}_n\) is the set of permutations of \(\{1,\dots,n\}\), and \(\operatorname{sgn}(\sigma)\) is the sign of the permutation \(\sigma\). Each product in the sum selects exactly one entry from each row.

Proof. Write the entries of \(\mathbf{A}'\) as \(a'_{rj}\). By construction, \(a'_{ij}=c a_{ij}\) for every column \(j\), and \(a'_{rj}=a_{rj}\) when \(r\ne i\). Apply the Leibniz formula to \(\mathbf{A}'\). In each permutation term, the product contains exactly one entry from row \(i\), namely \(a'_{i,\sigma(i)}=c a_{i,\sigma(i)}\). Every other row entry in that term is unchanged. Therefore each term is \(c\) times the corresponding term for \(\mathbf{A}\), and factoring \(c\) out of the sum gives \(\det\mathbf{A}'=c\det\mathbf{A}\). This argument also applies when \(c=0\), since no division by \(c\) was used. \(\square\)

The proof identifies why the factor occurs only once: every product in the determinant formula uses one entry from each row, including the row that was scaled. It also shows that the rule does not require the original determinant to be nonzero. If \(\det\mathbf{A}=0\), then \(c\det\mathbf{A}=0\) for every scalar \(c\).

Zero and Negative Determinants

The sign description needs care. A negative factor reverses the sign of a nonzero determinant and scales its magnitude by the absolute value of the factor. If the determinant is zero, multiplying by a negative factor leaves it zero; zero has no positive or negative sign to reverse. A zero scaling factor also gives a determinant of zero, because the scaled row becomes a zero row and the theorem gives \(0\cdot\det\mathbf{A}=0\).

In elementary row reduction, a row scaling is usually taken to use a nonzero scalar so the operation can be reversed. The determinant identity itself has no such restriction: it remains valid for \(c=0\). These are different questions—whether the determinant formula applies, and whether the row operation is reversible.

Scaling Several Rows

Proof. For \(k=0\), no rows have yet been scaled, so the determinant is \(\det\mathbf{A}\), which is the product of the first zero factors (the empty product \(1\)) times \(\det\mathbf{A}\). Suppose that after scaling rows \(1,\dots,k\), the determinant is \((c_1\cdots c_k)\det\mathbf{A}\). Scaling row \(k+1\) by \(c_{k+1}\) multiplies the current determinant by \(c_{k+1}\), by the one-row theorem. The new determinant is therefore \[ c_{k+1}(c_1\cdots c_k)\det\mathbf{A} =(c_1\cdots c_k c_{k+1})\det\mathbf{A}. \] Applying this step through row \(n\) gives the stated formula. No factor needs to be nonzero; if one factor is zero, the product and the resulting determinant are both zero. \(\square\)

When every row is scaled by the same \(c\), there are \(n\) factors of \(c\), one from each row. Thus the determinant is multiplied by \(c^n\). This is why the single-row rule cannot be applied as though changing all the rows were just one row scaling.

Common Pitfall: Counting the Scaling Once

A common error is to say, “Every entry was multiplied by \(2\), so the determinant is multiplied by \(2\).” For an \(n\times n\) matrix, multiplying every entry by \(2\) means multiplying all \(n\) rows by \(2\), not just one row. In the \(3\times3\) example above, the incorrect calculation would be \(\det\mathbf{C}'=2\det\mathbf{C}=12\). The correction is to apply the one-row rule three times: \(2\cdot2\cdot2\cdot6=48\), which also matches the direct determinant calculation.

A second error is to say that a negative factor always changes the determinant’s sign. If the starting determinant is zero, the result remains zero. For example, the rule gives \((-3)\cdot0=0\), not a determinant with a reversed sign. Always distinguish the sign of a nonzero determinant from the case where the determinant equals zero.

The practical method is to identify exactly which rows changed, record one factor for each of those rows, and multiply those factors together. An unchanged row contributes a factor of \(1\). A zero factor makes the resulting determinant zero, and a negative factor reverses the sign only when the determinant before scaling is nonzero. These checks help keep row operations and determinant factors aligned.