The Row Addition Rule
A row operation can change the entries of a matrix without changing its determinant. The key case here is adding a multiple of one row to a different row. The reason is more precise than a numerical pattern: multilinearity separates the new row into two parts, and the extra determinant has two identical rows. The repeated-row property then makes that extra term zero.
Throughout, \(\mathbf{A}\) is a square matrix in \(\mathbb{R}^{n\times n}\), and its rows are denoted by \(\mathbf{r}_1,\dots,\mathbf{r}_n\). For a list of rows, write \(\det(\mathbf{r}_1,\dots,\mathbf{r}_n)\) for the determinant of the matrix having those rows in that order. The row-addition operation replaces row \(i\) by row \(i\) plus \(c\) times row \(j\), where \(c\in\mathbb{R}\) and \(i\ne j\).
Proof. Keep every row except row \(i\) fixed. The \(i\)th row of \(\mathbf{B}\) is \(\mathbf{r}_i+c\mathbf{r}_j\). By linearity in row \(i\),
where in the second determinant the row in position \(i\) is \(\mathbf{r}_j\), while the row in position \(j\) is still \(\mathbf{r}_j\). Thus that second determinant has two identical rows, so it is \(0\). The first determinant is \(\det\mathbf{A}\). Therefore \(\det\mathbf{B}=\det\mathbf{A}+c\cdot0=\det\mathbf{A}\). This includes \(c=0\), when the operation changes nothing. The condition \(i\ne j\) ensures the second determinant really has two distinct positions containing the same row. \(\square\)
The proof also explains why the rule holds for every real \(c\), whether positive, negative, or zero. It does not depend on the entries of the matrix, or on whether its determinant is zero. The only ingredients are the two stated determinant properties and the fact that the source and target rows are distinct.
Checking the Rule in Small Matrices
Worked Example: A Two-by-Two Row Addition
Apply \(R_2\to R_2+2R_1\) to \(\mathbf{A}=\begin{bmatrix}2&1\\3&4\end{bmatrix}\), then check the determinant directly before and after.
The new second row is \((3,4)+2(2,1)=(3+4,4+2)=(7,6)\). Thus
For a \(2\times2\) matrix, \(\det\begin{bmatrix}a&b\\c&d\end{bmatrix}=ad-bc\). The original determinant is \(2\cdot4-1\cdot3=8-3=5\). The transformed determinant is \(2\cdot6-1\cdot7=12-7=5\).
Check: The row calculation gives \((7,6)\), and the direct determinant calculations give \(5\) on both sides. The operation preserves the determinant in this example.
Worked Example: Adding a Multiple of a Different Row
Apply \(R_1\to R_1+2R_2\) to \(\mathbf{A}=\begin{bmatrix}1&2&0\\0&1&3\\2&0&1\end{bmatrix}\). Compute both determinants by expansion along the first row.
The new first row is \((1,2,0)+2(0,1,3)=(1+0,2+2,0+6)=(1,4,6)\). Therefore
Expanding the original determinant along row 1, with signs \(+,-,+\), gives
Expanding the transformed determinant in the same way gives
Check: The row changed by adding twice row 2, and direct expansion gives \(\det\mathbf{A}=\det\mathbf{B}=13\).
Worked Example: A Row Addition with a Parameter
For any real \(t\), apply \(R_1\to R_1+tR_2\) to \(\mathbf{A}=\begin{bmatrix}1&2&1\\2&0&1\\0&1&3\end{bmatrix}\). Verify directly that the determinant does not depend on \(t\).
The first row becomes \((1,2,1)+t(2,0,1)=(1+2t,2,1+t)\), so
Expanding the original determinant along row 1 gives
For the transformed matrix, expansion along the first row gives
Check: The terms involving \(t\) are \(-2t\) and \(+2t\), which sum to zero. Thus the transformed determinant equals the original value \(-11\) for every real \(t\), as the theorem predicts.
Why the Extra Term Vanishes
In the proof, the determinant splits into an unchanged part and an extra part. The unchanged part uses the original target row. The extra part uses the source row in the target row’s position as well as in its original position. It is the repeated row—not a cancellation between numerical entries—that forces this part to vanish.
For instance, if the operation is \(R_1\to R_1+cR_2\), multilinearity gives the determinant of the original matrix plus \(c\) times the determinant of the matrix whose first and second rows are both \(\mathbf{r}_2\). The second determinant is zero even when \(c\) is nonzero, because its two rows are identical. No division by \(c\) is involved, and no assumption that the original determinant is nonzero is needed.
Proof. We use induction on the number \(m\) of row additions. If \(m=0\), the matrix has not changed, so its determinant is unchanged. Suppose a sequence of \(m\) such operations preserves the determinant. A sequence of \(m+1\) operations consists of those first \(m\) operations followed by one row addition. By the induction hypothesis, the determinant after the first \(m\) operations equals the original determinant. By the row addition theorem, the final operation preserves the determinant from that intermediate matrix. Hence the determinant after all \(m+1\) operations also equals the original determinant. This proves the claim for every finite \(m\). \(\square\)
The corollary allows the source row to have been changed by an earlier operation: the theorem applies to the rows present at the moment each operation is performed. What matters at each step is that the target and source are different row positions.
Common Mistakes
Mistake: treating the extra determinant as the original determinant. Consider \(\mathbf{A}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\), whose determinant is \(1\cdot4-2\cdot3=-2\). After \(R_2\to R_2+2R_1\), row 2 becomes \((3,4)+2(1,2)=(5,8)\), so the new determinant is \(1\cdot8-2\cdot5=8-10=-2\). A wrong use of linearity might write \(\det\mathbf{B}=\det\mathbf{A}+2\det\mathbf{A}=-6\). But the second term is not a second copy of the original matrix: it has row 1 in both row positions. Its value is \(2\det\begin{bmatrix}1&2\\1&2\end{bmatrix}=2(1\cdot2-2\cdot1)=0\). The correct calculation is \(-2+0=-2\).
Mistake: adding a multiple of a row to itself. The condition \(i\ne j\) is essential. If \(R_1\to R_1+2R_1\) is applied to the same matrix, its first row becomes \((3,6)\), and the result is \(\begin{bmatrix}3&6\\3&4\end{bmatrix}\). Its determinant is \(3\cdot4-6\cdot3=12-18=-6\), not the original \(-2\). Here the new row is three times the old row, so the repeated-row argument does not apply.
When using the row addition theorem, identify the target row and the different source row before splitting the determinant. That makes the repeated-row term visible and prevents confusing a row addition with an operation that scales the target row.
Check Your Understanding
For each question, use the row addition theorem or verify the determinant directly where requested.
- Apply \(R_2\to R_2-3R_1\) to \(\begin{bmatrix}1&2\\4&7\end{bmatrix}\). What is the new matrix, and what are the determinants before and after?
- In the proof of the theorem, which two row positions are identical in the extra determinant when the operation is \(R_3\to R_3+cR_1\)?
- For \(\mathbf{A}=\begin{bmatrix}1&0&2\\2&1&0\\0&3&1\end{bmatrix}\), apply \(R_1\to R_1+R_2\) and compute both determinants by expansion.
- Does the theorem allow \(c=0\)? Explain using the operation and the proof.
- For \(\begin{bmatrix}2&1\\1&3\end{bmatrix}\), apply \(R_1\to R_1+R_1\). Compute the new determinant and explain why the row addition theorem does not apply.
