AcademyDeterminants
Academy
Determinant Calculations
Level 1 - Linear Algebra topic page in Determinants.
Principle
Determinants can be calculated efficiently by using row operations to create zeros. The best method is usually to reduce the matrix toward triangular form while tracking how each row operation changes the determinant.
For a triangular matrix, the determinant is the product of its diagonal entries.
Notation
\(A\)
a square matrix
\(U\)
an upper triangular matrix
\(R_i\)
row i
\(u_{ii}\)
diagonal entry i of U
\(\det A\)
determinant of A
The Core Method
A triangular matrix has determinant equal to the product of diagonal entries:
Triangular determinant
\[\det U=u_{11}u_{22}\cdots u_{nn}\]
Use row replacement operations to create zeros, because they do not change the determinant:
Safe row replacement
\[R_i\leftarrow R_i+\lambda R_j\quad\text{does not change }\det\]
If you swap rows or scale rows, track the change carefully:
- Row swap: determinant changes sign.
- Row scaling by \(\lambda\): determinant is multiplied by \(\lambda\).
Worked Cases
Question
Compute \(\det\begin{pmatrix}1&2&1\\3&8&5\\0&4&2\end{pmatrix}\).
Answer
Use row replacement operations. Start with \(A\). Apply \(R_2\leftarrow R_2-3R_1\), which does not change the determinant: row \(2\) becomes \((0,2,2)\). The matrix is now \(\begin{pmatrix}1&2&1\\0&2&2\\0&4&2\end{pmatrix}\). Apply \(R_3\leftarrow R_3-2R_2\), giving row \(3\) as \((0,0,-2)\). The triangular matrix is \(\begin{pmatrix}1&2&1\\0&2&2\\0&0&-2\end{pmatrix}\). Its determinant is \(1\cdot2\cdot(-2)=-4\). Therefore \(\det A=-4\).
Examples
Question
Which row is best for cofactor expansion in \(\begin{pmatrix}4&0&0\\1&2&3\\5&0&6\end{pmatrix}\)?
Answer
The first row has two zeros, so expanding along the first row leaves only one non-zero term. This minimises calculation. The determinant is \(4\det\begin{pmatrix}2&3\\0&6\end{pmatrix}=4(12)=48\).