Multilinearity Of The Determinant In Its Rows

Linearity One Row at a Time

The determinant can be difficult to calculate when a matrix has complicated entries. A useful simplification is to express a row as a sum of simpler rows and expand the determinant accordingly. The justification comes directly from the Leibniz formula: when all rows except one are fixed, every term in that formula contains exactly one entry from the remaining row. This makes the determinant linear in that row.

Work over \(\mathbb{R}\), and let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). Its rows are written as \(1\times n\) arrays; equivalently, the \(i\)-th row is \(\mathbf{r}_i^{\top}\), where \(\mathbf{r}_i\in\mathbb{R}^n\) is a column. For a permutation \(\sigma\) of \(\{1,\ldots,n\}\), let \(\operatorname{sgn}(\sigma)\) be its sign. The Leibniz formula is

\det\mathbf{A} = \sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^{n}a_{i,\sigma(i)}.

Here \(S_n\) is the set of all permutations of \(\{1,\ldots,n\}\). In each product, the factor from row \(k\) is exactly \(a_{k,\sigma(k)}\). The other factors do not involve row \(k\). That separation is what makes it possible to distribute a sum, or factor out a scalar, from that row.

Definition: The determinant is linear in row \(k\) if, with every other row fixed, replacing row \(k\) by a sum of two rows gives the sum of the corresponding determinants, and multiplying row \(k\) by a scalar multiplies the determinant by that scalar. This property is also called linearity in each row separately, or row multilinearity.

Deriving the Row-Linearity Rules

For clarity, denote by \(\det(\mathbf{A}\text{ with row }k\text{ replaced by }\mathbf{x}^{\top})\) the determinant of the matrix obtained by making that replacement and leaving all other rows unchanged. The next result gives both parts of row linearity and derives them from the Leibniz formula.

Theorem (Row multilinearity): Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\), fix a row index \(k\), and let \(\mathbf{x},\mathbf{y}\in\mathbb{R}^n\) be column vectors. With the other rows fixed,

\(\det(\mathbf{A}\text{ with row }k\text{ replaced by }(\mathbf{x}+\mathbf{y})^{\top})=\det(\mathbf{A}\text{ with row }k\text{ replaced by }\mathbf{x}^{\top})+\det(\mathbf{A}\text{ with row }k\text{ replaced by }\mathbf{y}^{\top})\).

For every \(c\in\mathbb{R}\),

\(\det(\mathbf{A}\text{ with row }k\text{ replaced by }(c\mathbf{x})^{\top})=c\,\det(\mathbf{A}\text{ with row }k\text{ replaced by }\mathbf{x}^{\top})\).

Proof. Write the fixed entries in row \(i\ne k\) as \(a_{ij}\), and write the entries of \(\mathbf{x}\) and \(\mathbf{y}\) as \(x_j\) and \(y_j\). In the Leibniz formula for the matrix with row \(k\) equal to \((\mathbf{x}+\mathbf{y})^{\top}\), the factor from that row is \(x_{\sigma(k)}+y_{\sigma(k)}\). Distributing this sum gives

\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \left(\prod_{\substack{i=1\\i\ne k}}^{n}a_{i,\sigma(i)}\right) \bigl(x_{\sigma(k)}+y_{\sigma(k)}\bigr) = \sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \left(\prod_{\substack{i=1\\i\ne k}}^{n}a_{i,\sigma(i)}\right)x_{\sigma(k)} + \sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \left(\prod_{\substack{i=1\\i\ne k}}^{n}a_{i,\sigma(i)}\right)y_{\sigma(k)}.

The first sum on the right is the Leibniz formula for row \(k\) replaced by \(\mathbf{x}^{\top}\), and the second is the formula for row \(k\) replaced by \(\mathbf{y}^{\top}\). This proves additivity. If row \(k\) is instead \((c\mathbf{x})^{\top}\), its factor in each product is \(cx_{\sigma(k)}\). Since \(c\) is the same in every term, it factors out of the sum, proving homogeneity. When \(n=1\), the product over \(i\ne k\) has no factors and is interpreted as \(1\), so the same argument applies. \(\square\)

The phrase “in each row separately” matters. The theorem allows one row to vary while the other rows are held fixed. It does not say that the determinant is a linear function of the entire matrix when several rows vary together: expanding two decomposed rows can create terms involving every pairing of their summands.

Expanding a Single Decomposed Row

The simplest application is to replace one row by a sum of rows whose entries are easier to work with. For example, if row \(k\) is \(\mathbf{u}^{\top}+\mathbf{v}^{\top}\), the theorem gives two determinants, each with all the original rows except row \(k\) unchanged. If the row is \(a\mathbf{u}^{\top}+b\mathbf{v}^{\top}\), first use homogeneity on each term and then additivity.

Worked Example: Splitting the First Row

Write the first row of \(\mathbf{A}=\begin{bmatrix}2&1&4\\1&0&1\\0&1&1\end{bmatrix}\) as \((1,0,1)+(1,1,3)\), and use row linearity to calculate its determinant. Both summands have three entries, and their sum is \((2,1,4)\).

By additivity, the determinant is the sum of the determinants formed by replacing the first row with each summand. For the \(3\times3\) matrices here, use the formula \(\det\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\), which follows by grouping the terms of the Leibniz formula:

\begin{aligned} \det\mathbf{A} &=\det\begin{bmatrix}1&0&1\\1&0&1\\0&1&1\end{bmatrix} +\det\begin{bmatrix}1&1&3\\1&0&1\\0&1&1\end{bmatrix}\\ &=\bigl(1(0-1)+1(1-0)\bigr) +\bigl(1(0-1)-1(1-0)+3(1-0)\bigr)\\ &=0+1=1. \end{aligned}

Check: applying the same \(3\times3\) formula directly to the original matrix gives \(2(0-1)-1(1-0)+4(1-0)=-2-1+4=1\), in agreement with the row decomposition.

For a scalar combination, the scalar coefficients must be carried through the expansion. They are not additional rows; they multiply the corresponding determinants.

Worked Example: A Scalar Combination in a Two-by-Two Determinant

Let \(\mathbf{u}=(1,2)^{\top}\), \(\mathbf{v}=(3,1)^{\top}\), and \(\mathbf{w}=(2,-1)^{\top}\). The first row of \(\mathbf{B}=\begin{bmatrix}-1&3\\2&-1\end{bmatrix}\) is \(2\mathbf{u}^{\top}-\mathbf{v}^{\top}\), while its second row is \(\mathbf{w}^{\top}\). Find the determinant using that decomposition.

Homogeneity and additivity give \(\det\mathbf{B}=2\det\begin{bmatrix}1&2\\2&-1\end{bmatrix}-\det\begin{bmatrix}3&1\\2&-1\end{bmatrix}\). Computing each \(2\times2\) determinant yields

\begin{aligned} \det\mathbf{B} &=2\bigl(1(-1)-2(2)\bigr)-\bigl(3(-1)-1(2)\bigr)\\ &=2(-5)-(-5)\\ &=-5. \end{aligned}

Check: the original matrix has determinant \((-1)(-1)-3(2)=1-6=-5\). The decomposition and the direct calculation agree.

Expanding More Than One Row

When several rows are sums, apply row linearity one row at a time. Every choice of one summand from each decomposed row contributes a determinant. More generally, if \(m\) distinct rows are each expressed as a sum of two rows, the expansion contains \(2^m\) terms before any terms happen to evaluate to zero. With more than two summands in a row, the corresponding number of choices is the product of the numbers of summands in the decomposed rows.

Corollary (Expansion of decomposed rows): Suppose \(m\) distinct rows of a square matrix are written as finite sums of row arrays. The determinant is the sum, over all choices of one summand from each of those \(m\) rows, of the determinant formed by making those choices and leaving every other row unchanged.

Proof. For \(m=1\), the statement is additivity in one row, applied repeatedly if that row has more than two summands. Suppose the statement holds for \(m-1\) decomposed rows. Apply additivity to the remaining decomposed row, producing one determinant for each of its summands. In each resulting determinant, the other \(m-1\) rows are still decomposed as before. By the induction hypothesis, each such determinant expands into all choices from those \(m-1\) rows. Collecting the terms gives exactly all choices from the \(m\) rows, each appearing once. This proves the statement for every finite \(m\). \(\square\)

Worked Example: Decomposing Two Rows

In \(\mathbf{C}=\begin{bmatrix}1&1&0\\0&1&1\\1&2&3\end{bmatrix}\), write row 1 as \((1,0,0)+(0,1,0)\) and row 2 as \((0,1,0)+(0,0,1)\). Expand the determinant into all choices and evaluate each term.

There are two choices for each of the first two rows, so there are four determinant terms:

\begin{aligned} \det\mathbf{C} ={}&\det\begin{bmatrix}1&0&0\\0&1&0\\1&2&3\end{bmatrix} +\det\begin{bmatrix}1&0&0\\0&0&1\\1&2&3\end{bmatrix}\\ &+\det\begin{bmatrix}0&1&0\\0&1&0\\1&2&3\end{bmatrix} +\det\begin{bmatrix}0&1&0\\0&0&1\\1&2&3\end{bmatrix}\\ ={}&3+(-2)+0+1=2. \end{aligned}

For the four values, the \(3\times3\) formula gives respectively \(3\), \(-2\), \(0\), and \(1\). For instance, the second determinant is \(1(0\cdot3-1\cdot2)=-2\); the third is \(-1(0\cdot3-0\cdot1)=0\); and the fourth is \(-1(0\cdot3-1\cdot1)=1\).

Check: the original matrix has determinant \(1(1\cdot3-1\cdot2)-1(0\cdot3-1\cdot1)+0=1-(-1)=2\). This matches the four-term expansion.

Common Pitfall: Omitting Cross-Terms

A common error is to decompose two rows but pair only the first summand in one row with the first summand in the other, then pair the second summands. That is not the expansion rule. Each summand in the first row must be paired with each summand in the second. In the preceding example, keeping only the terms with matching positions gives \(3+1=4\), which is not the determinant. The two omitted terms are \(-2\) and \(0\); the full sum is \(3-2+0+1=2\).

There is a second useful check: every determinant in an expansion must have the same size as the original, and every replacement row must have the same number of entries as the row it replaces. In the first worked example, both pieces of the first-row decomposition have three entries. A two-entry row cannot be added to a three-entry row, and such an expression cannot represent a valid row decomposition.

Why Row Multilinearity Is Useful

Row multilinearity is a calculation tool, but it is also a way to expose the structure of a determinant. A complicated row can be separated into simpler pieces, and a scalar coefficient can be taken outside the corresponding determinant. When multiple rows are decomposed, the expansion records every possible combination of the pieces. This is useful when many terms in that expansion are simple to calculate, even though no individual term should be omitted merely because it does not match another row’s choice.

The scope of the result should remain precise: the determinant is linear when one row varies and the remaining rows are fixed. For several varying rows, the full expansion is obtained by applying that one-row rule repeatedly. The next calculation should therefore begin by listing the summands in each decomposed row and counting the combinations they produce.

Check Your Understanding

For each question, keep the other rows fixed whenever you apply row linearity, and include every choice when more than one row is decomposed.

State the additivity rule for replacing row \(k\) by \((\mathbf{x}+\mathbf{y})^{\top}\).
If one row is \(3\mathbf{u}^{\top}-2\mathbf{v}^{\top}\), what two determinant terms result?
How many determinant terms arise if three rows are each written as a sum of two rows?
Expand a determinant whose first row is \((1,0)+(0,1)\), whose second row is \((1,0)+(0,1)\), and whose other rows are fixed. What choices must be included?
Explain why linearity in each row separately does not permit omitting cross-terms when two rows are decomposed.

Multilinearity Of The Determinant In Its Rows

What You'll Learn

Linearity One Row at a Time

Deriving the Row-Linearity Rules