The Useful Zero Pattern
A lower-triangular matrix can contain many nonzero entries, but its first row is especially simple: every entry to the right of the diagonal is zero. This makes cofactor expansion along that row a direct route to a smaller determinant. Repeating the same step eventually reaches a one-by-one matrix, revealing why the determinant is the product of the diagonal entries.
Throughout, matrices are square matrices over \(\mathbb{R}\). For an \(n\times n\) matrix \(\mathbf{L}=(\ell_{ij})\), the entry \(\ell_{ij}\) is in row \(i\), column \(j\). The distinction between entries above and below the diagonal is important: a lower-triangular matrix has zeros above the diagonal, not below it.
For an \(n\times n\) matrix, cofactor expansion along the first row is the formula \(\det\mathbf{L}=\sum_{j=1}^{n}(-1)^{1+j}\ell_{1j}\det(\mathbf{L}_{1j})\), where \(\mathbf{L}_{1j}\) is formed by deleting row 1 and column \(j\). This expansion rule was established earlier. The sign for the first entry is positive because \((-1)^{1+1}=1\).
First-Row Expansion Reduces the Matrix
For a lower-triangular matrix, every term in the first-row expansion except the first is zero: \(\ell_{1j}=0\) for \(j>1\). What remains is the entry \(\ell_{11}\) multiplied by the determinant of the minor obtained by deleting row 1 and column 1. That minor is itself lower triangular, so the same reasoning applies again.
Proof. In the first-row cofactor expansion, the term for column \(j\) contains the factor \(\ell_{1j}\). Since \(\mathbf{L}\) is lower triangular, \(\ell_{1j}=0\) for every \(j>1\). The term for \(j=1\) has positive sign, so the expansion gives \(\det\mathbf{L}=\ell_{11}\det(\mathbf{L}_{11})\). The matrix \(\mathbf{L}_{11}\) is \(\mathbf{B}\). To check its zero pattern, its entry in row \(r\), column \(c\) is \(\ell_{r+1,c+1}\). If \(r<c\), then \(r+1<c+1\), and lower triangularity gives \(\ell_{r+1,c+1}=0\). Therefore \(\mathbf{B}\) is lower triangular, as claimed. \(\square\)
This recursion explains why expansion down the first row is efficient: there is just one potentially nonzero term, and its minor retains the structure needed for the next step. The nonzero entries below \(\ell_{11}\) do not appear as terms in this first-row expansion; they remain inside the smaller determinant.
Worked Example: A Three-by-Three Matrix
Compute the determinant of the lower-triangular matrix \(\mathbf{L}=\begin{bmatrix}2&0&0\\3&-1&0\\4&5&3\end{bmatrix}\).
Expansion along the first row has only one nonzero term. Its minor is \(\begin{bmatrix}-1&0\\5&3\end{bmatrix}\), so
Check: the diagonal entries are \(2,-1,3\), and their product is \(2(-1)(3)=-6\), agreeing with the cofactor calculation.
Why the Diagonal Product Follows
The first-row recursion can be applied repeatedly. After deleting the first row and column, the next first entry is \(\ell_{22}\); after another deletion, it is \(\ell_{33}\). At the final stage, the determinant of the one-by-one matrix \(\begin{bmatrix}\ell_{nn}\end{bmatrix}\) is \(\ell_{nn}\). Induction packages this repeated calculation into a formula for every size.
Proof. We use induction on \(n\). If \(n=1\), then \(\mathbf{L}=\begin{bmatrix}\ell_{11}\end{bmatrix}\), and the determinant of a one-by-one matrix is its only entry. The formula holds. Now let \(n\ge2\), and assume the formula holds for lower-triangular matrices of size \(n-1\). By the first-row recursion, \(\det\mathbf{L}=\ell_{11}\det\mathbf{B}\), where \(\mathbf{B}\) is lower triangular of size \(n-1\). Its diagonal entries are \(\ell_{22},\ell_{33},\ldots,\ell_{nn}\). Applying the induction hypothesis to \(\mathbf{B}\) gives \(\det\mathbf{B}=\ell_{22}\ell_{33}\cdots\ell_{nn}\). Hence \(\det\mathbf{L}=\ell_{11}\ell_{22}\cdots\ell_{nn}\). This proves the formula for every \(n\ge1\). \(\square\)
The formula does not require the entries below the diagonal to be zero. It requires only the stated zero condition above the diagonal. Consequently, a lower-triangular matrix can have substantial nonzero entries below its diagonal and still have a determinant found from its diagonal alone.
Worked Example: A Four-by-Four Recursion
Compute \(\det\mathbf{L}\) for \(\mathbf{L}=\begin{bmatrix}1&0&0&0\\2&3&0&0\\-1&4&2&0\\5&0&7&-2\end{bmatrix}\) by tracing the successive first-row expansions.
The first expansion contributes the factor \(1\) and leaves the lower-triangular minor \(\begin{bmatrix}3&0&0\\4&2&0\\0&7&-2\end{bmatrix}\). Expanding that minor along its first row contributes \(3\) and leaves \(\begin{bmatrix}2&0\\7&-2\end{bmatrix}\). Expanding once more contributes \(2\), leaving the one-by-one determinant \(\det\begin{bmatrix}-2\end{bmatrix}=-2\). Thus
Check: the diagonal entries of the original matrix are \(1,3,2,-2\), whose product is \(1\cdot3\cdot2\cdot(-2)=-12\). The successive minors also show explicitly that each recursion step retains the lower-triangular pattern.
Zero Diagonal Entries and Parameters
The diagonal-product formula includes zero entries without needing a separate rule. If even one diagonal entry is zero, the product is zero. In the recursion, that zero eventually appears as the first entry of a remaining lower-triangular matrix, making that stage’s determinant zero. The earlier factors then multiply a zero result.
Worked Example: A Zero on the Diagonal
Compute the determinant of \(\mathbf{L}=\begin{bmatrix}0&0&0\\2&5&0\\7&-1&3\end{bmatrix}\).
Every entry in the first row is zero, so every term in the first-row cofactor expansion is zero. Equivalently, the first-row recursion gives
Check: the diagonal product is \(0\cdot5\cdot3=0\), as obtained from expansion. The nonzero entries below the diagonal do not change this result.
The same method works when diagonal entries contain a parameter, provided the matrix remains lower triangular for every parameter value under consideration. In that setting, the formula gives an exact expression in the parameter; particular values can then be substituted to determine when the determinant vanishes.
Worked Example: A Determinant Depending on a Parameter
For real \(t\), find the determinant of \(\mathbf{L}(t)=\begin{bmatrix}t&0&0\\2&4&0\\-1&3&t-1\end{bmatrix}\), and evaluate it at \(t=2\).
Expansion along the first row gives the factor \(t\), followed by the minor \(\begin{bmatrix}4&0\\3&t-1\end{bmatrix}\). Therefore
At \(t=2\), this expression gives \(4(2)(2-1)=8\). Directly substituting into the matrix gives \(\mathbf{L}(2)=\begin{bmatrix}2&0&0\\2&4&0\\-1&3&1\end{bmatrix}\), and first-row expansion gives \(2\det\begin{bmatrix}4&0\\3&1\end{bmatrix}=2(4\cdot1-0\cdot3)=8\). The two calculations agree.
Common Mistakes and a Condition Check
Do not confuse lower triangular with upper triangular. In a lower-triangular matrix, entries above the diagonal are zero, while entries below it may be nonzero. Also, the diagonal-product formula should not be applied to a matrix that fails the lower-triangular condition.
For example, consider \(\mathbf{A}=\begin{bmatrix}2&1&0\\3&4&0\\0&5&6\end{bmatrix}\). A tempting but wrong calculation is to multiply the diagonal entries and report \(2\cdot4\cdot6=48\). This matrix is not lower triangular because its entry in row 1, column 2 is \(1\ne0\). Expanding along the first row instead gives
The correction is to check the zero pattern before using the formula. Here the nonzero entry in the first row, second column contributes a second cofactor term, so the diagonal product alone does not give the determinant.
When the matrix is lower triangular, a reliable calculation has two parts: identify the first-row nonzero entry, then check that the remaining minor is lower triangular before repeating. This explains the product formula rather than treating it as a pattern to memorize.
Check Your Understanding
For each question, use the zero pattern and the first-row recursion to support your answer.
- Write the condition on the entries \(\ell_{ij}\) that defines an \(n\times n\) lower-triangular matrix.
- For a lower-triangular \(3\times3\) matrix, which entry can contribute to the first-row cofactor expansion, and what minor remains?
- Compute the determinant of \(\begin{bmatrix}3&0&0\\-2&1&0\\4&5&-2\end{bmatrix}\).
- For which real values of \(t\) does the lower-triangular matrix \(\begin{bmatrix}t&0\\7& t+1\end{bmatrix}\) have determinant zero?
- Explain why the diagonal-product formula cannot be applied directly to \(\begin{bmatrix}1&2\\0&3\end{bmatrix}\) as a lower-triangular matrix.
