AcademyDeterminants
Academy
Small Determinants
Level 1 - Linear Algebra topic page in Determinants.
Principle
A determinant is a scalar attached to a square matrix. For small matrices, it is a direct calculation. For \(2\) by \(2\) matrices, it measures signed area scaling. For \(3\) by \(3\) matrices, it measures signed volume scaling.
Determinants are defined only for square matrices.
Notation
\(\det A\)
the determinant of a square matrix A
\(|A|\)
another notation for \det A, not an absolute value unless context says so
\(a,b,c,d\)
entries of a 2 by 2 matrix
\(\mathbf a,\mathbf b,\mathbf c\)
column vectors of a 3 by 3 matrix
\(\mathbf a\cdot(\mathbf b\times\mathbf c)\)
scalar triple product
The Core Method
For a \(1\) by \(1\) matrix, the determinant is its only entry:
One by one determinant
\[\det\begin{pmatrix}a\end{pmatrix}=a\]
For a \(2\) by \(2\) matrix,
Two by two determinant
\[\det\begin{pmatrix}a&b\\c&d\end{pmatrix}=ad-bc\]
For a \(3\) by \(3\) matrix with columns \(\mathbf a,\mathbf b,\mathbf c\),
Three by three determinant
\[\det(\mathbf a\;\mathbf b\;\mathbf c)=\mathbf a\cdot(\mathbf b\times\mathbf c)\]
The sign depends on column order. Swapping two columns changes the sign.
Worked Cases
Question
Compute \(\det\begin{pmatrix}3&-2\\5&4\end{pmatrix}\).
Answer
Use \(ad-bc\). Here \(a=3\), \(b=-2\), \(c=5\), and \(d=4\). Then \(ad=3\cdot4=12\), and \(bc=(-2)\cdot5=-10\). Therefore \(ad-bc=12-(-10)=22\). The determinant is \(22\).
Examples
Question
Compute \(\det\begin{pmatrix}2&6\\1&3\end{pmatrix}\) and interpret the result.
Answer
The determinant is \(2\cdot3-6\cdot1=6-6=0\). The columns \((2,1)\) and \((6,3)\) are parallel because \((6,3)=3(2,1)\). They span zero area, which matches the zero determinant.