Properties of determinants

Lecture 22

Minjae Park

Auburn University
MATH 2660 - Spring 2026

March 2, 2026

Attendance

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Recap

Determinants

For an \(n\times n\) matrix \(A\):
- \(\det(A)\) is defined by cofactor expansion along any row or any column; in practice, choose one with many zeros to simplify the computation.
- \(|\det(A)|\) equals the hypervolume of the hyperparallelogram formed by its \(n\) column vectors.
- The sign of \(\det(A)\) records the orientation of the column vectors.
- If \(\det(A)=0\), the columns are linearly dependent.
- If \(\det(A)\neq 0\), the columns are linearly independent and \(A\) is invertible (also called non-singular).
Computations can be carried out far more efficiently by computer algebra systems (e.g., MATLAB or Mathematica) than by hand.

Determinants for Special Matrices

Let \(D\) be an \(n\times n\) diagonal matrix with diagonal entries \(d_1,\dots,d_n\): \[\det(D)=d_1d_2\cdots d_n.\]
More generally, if \(T\) is a triangular matrix (upper or lower) with diagonal entries \(d_1,\dots,d_n\): \[\det(T)=d_1d_2\cdots d_n.\]
If \(E_{R_i\leftrightarrow R_j}\) is the elementary matrix corresponding to swapping two rows: \[\det\!\left(E_{R_i\leftrightarrow R_j}\right)=-1.\]
Geometrically, swapping two rows reverses orientation, which explains the negative sign.

Determinant and Invertible Matrices

Geometrically, \(A^{-1}\) undoes what \(A\) does to vectors in \(\mathbb R^n\).
If the columns of \(A\) are linearly independent, the associated hyperparallelogram has nonzero hypervolume — nothing collapses to a lower dimension.
Since no information is lost, the transformation can be reversed; hence \(A\) is invertible.
From a row-reduction perspective: if the \(n\) columns are linearly independent, then \(\text{rank}(A)=n\), so the RREF of \(A\) has \(n\) pivots and equals \(I_n\).
Therefore \(A\) can be transformed into \(I_n\) by a sequence of elementary matrices, and those same operations produce \(A^{-1}\).
Consequently, \(\det(A)\neq 0 \quad \Longleftrightarrow \quad A \text{ is invertible.}\)

Summary for Invertibility

Let \(A\) be an \(n\times n\) matrix. The following are equivalent:
- There exists a matrix \(B\) such that \(AB=BA=I_n\) (so \(A\) is invertible and \(B=A^{-1}\)).
- \(\det(A)\neq0\).
- \(\operatorname{RREF}(A)=I_n\).
- \(\mathop{\mathrm{rank}}(A)=n\) and \(\mathop{\mathrm{null}}(A)=\mathop{\mathrm{null}}(A^T)=0\).
- The columns of \(A\) are linearly independent.
- The rows of \(A\) are linearly independent.
- The homogeneous system \(A\vec{x}=\vec{0}\) has the unique solution \(\vec{x}=\vec{0}\).
- The linear equation \(A\vec{x}=\vec{b}\) has the unique solution \(\vec{x}=A^{-1} \vec{b}\).

Summary for Non-Invertibility

Let \(A\) be an \(n\times n\) matrix. The following are equivalent:
- There is no matrix \(B\) such that \(AB=BA=I_n\) (so \(A\) is not invertible).
- \(\det(A)=0\).
- \(\operatorname{RREF}(A)\neq I_n\) and has some zero rows.
- \(\mathop{\mathrm{rank}}(A)<n\) and \(\mathop{\mathrm{null}}(A)=\mathop{\mathrm{null}}(A^T)>0\).
- The columns of \(A\) are linearly dependent.
- The rows of \(A\) are linearly dependent.
- The homogeneous system \(A\vec{x}=\vec{0}\) has infinitely many solutions.

Properties of Determinants

Multiplicativity for \(2\times2\) Matrices

Let \(A,B:\mathbb{R}^2\to\mathbb{R}^2\) be linear transformations.
Matrix multiplication \(AB\) represents function composition \(A\circ B\) (apply \(B\) first, then \(A\)).
When we apply \(B\), the unit square is transformed into a parallelogram with signed area \(\det(B)\).
Applying \(A\) afterward scales the signed area of any region by a factor of \(\det(A)\), since linear transformations scale area uniformly.
Therefore, after applying \(B\) and then \(A\), the unit square becomes a parallelogram with signed area \(\det(A)\det(B)\).
Since \(AB\) represents the combined transformation, and determinants measure area scaling, \[\det(AB)=\det(A)\det(B).\]

Multiplicativity in General

The same geometric reasoning extends to \(n\times n\) matrices, where determinants represent signed hypervolume scaling in \(\mathbb{R}^n\).
Let \(A,B\) be two \(n\times n\) matrices, viewed as linear transformations from \(\mathbb{R}^n\) to \(\mathbb{R}^n\).
Applying \(B\) first scales hypervolume by \(\det(B)\), and then applying \(A\) scales it by \(\det(A)\).
Therefore, \[\det(AB)=\det(A)\det(B).\]
In other words, the determinant is multiplicative.

Determinant and Commutativity

Let \(A\) and \(B\) be \(n\times n\) matrices. By multiplicativity, \[ \det(AB)=\det(A)\det(B)=\det(B)\det(A)=\det(BA). \]
Caution: In general, \(AB\neq BA\) as matrices.
Thus matrix multiplication is not commutative, but the determinant behaves as if it were: \[ \det(AB)=\det(BA). \] if \(A\) and \(B\) are square matrices of the same size.
Caution: If \(A\) and \(B\) are not both \(n\times n\), then \(\det(A)\) or \(\det(B)\) may not even be defined, and multiplicativity does not apply.

Example

Let \[ A= \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, \qquad B= \begin{bmatrix} 2 & 0 \\ 1 & 3 \end{bmatrix}. \]
Compute \(BA\).
Then compute \(\det(BA)\) and submit your answer on iClicker.
Next, we will compute \(AB\) and \(\det(AB)\) and compare the results together.

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Determinant of the Inverse

If \(A\) is invertible, then by multiplicity \[ \det(A)\det(A^{-1}) = \det(AA^{-1}) = \det(I_n) = 1. \]
Therefore, \[ \det(A^{-1})=\frac{1}{\det(A)}. \]
In particular, \(A\) is invertible if and only if \(\det(A)\neq0\).

RREF and Determinants

If \(\operatorname{RREF}(A)=R\), then \[ R=E_kE_{k-1}\cdots E_1 A, \] where each \(E_i\) is an elementary row matrix.
Taking determinants, \[ \det(A) = \det(R)/\left(\det(E_k)\cdots\det(E_1)\right). \]
Each elementary row operation has a simple determinant effect:
- Row swap \(\Rightarrow\) multiply by \(-1\),
- Row scaling by \(c\) \(\Rightarrow\) multiply by \(c\),
- Row replacement (adding a multiple of another row) \(\Rightarrow\) no change.
Thus determinants can be computed efficiently via row reduction.

Example

Let \[ A= \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 0 \\ 1 & 1 & 1 \end{bmatrix}. \]
Row reduce to \(\operatorname{RREF}(A)=I_3\) while tracking determinant changes.
Step 1: \(R_2\leftarrow R_2-2R_1\), \(R_3\leftarrow R_3-R_1\)
(row replacement → determinant unchanged) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -2 \\ 0 & -1 & 0 \end{bmatrix}. \]
Step 2: \(R_3\leftarrow R_3-R_2\)
(row replacement → determinant unchanged) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 2 \end{bmatrix}. \]
Step 3: \(R_2\leftarrow -R_2\)
(row scaling by \(-1\) → determinant multiplies by \(-1\)) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{bmatrix}. \]
Step 4: \(R_3\leftarrow \frac12 R_3\)
(row scaling by \(\frac12\) → determinant multiplies by \(\frac12\)) \[ \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}. \]
Step 5: \(R_2\leftarrow R_2-2R_3\), \(R_1\leftarrow R_1-R_3\)
(row replacement → determinant unchanged) \[ \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}. \]
Step 6: \(R_1\leftarrow R_1-2R_2\)
(row replacement → determinant unchanged) \[ \operatorname{RREF}(A)= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=I_3. \]
Let \(E\) be the product of elementary matrices for these row operations. Then \(EA=I_3\), so \(E=A^{-1}\).
Taking determinants: \[ \det(E)\det(A)=\det(I_3)=1 \quad\Rightarrow\quad \det(A)=\frac{1}{\det(E)}. \]
Only the scaling steps change \(\det(E)\): \[ \det(E)=(-1)\left(\frac12\right)=-\frac12. \]
Therefore, \[ \boxed{\det(A)=\frac{1}{-\frac12}=-2}. \]

Transpose of Multiplication

Let \(A=[\vec{u}_1\ \dots\ \vec{u}_n]\) be a \(k\times n\) matrix and \(B=[\vec{v}_1\ \dots\ \vec{v}_m]\) a \(k\times m\) matrix.
Then \(A^TB\) is an \(n\times m\) matrix with entries \[ (A^TB)_{ij}=\vec{u}_i\cdot \vec{v}_j. \]
Similarly, \(B^TA\) is an \(m\times n\) matrix with \[ (B^TA)_{ij}=\vec{v}_i\cdot \vec{u}_j. \]
Since the dot product is commutative, \(\vec{u}_i\cdot\vec{v}_j=\vec{v}_j\cdot\vec{u}_i\), we obtain \[ (B^TA)^T=A^TB. \]

Transpose of a Product

Let \(C=B^T\). Then \(C^T=B\), and the previous identity \[ (B^TA)^T=A^TB \] becomes \[ (CA)^T=A^TC^T. \]
Applying this rule repeatedly, for three matrices \(A,B,C\) (whose product is defined), \[ (ABC)^T = (AB\,C)^T = C^T(AB)^T = C^TB^TA^T. \]
More generally, for matrices whose product is defined, \[ (A_1A_2\cdots A_\ell)^T = A_\ell^T\cdots A_2^T A_1^T. \]

Determinant and Transpose

If \(\operatorname{RREF}(A)=R\), then \(A=E_kE_{k-1}\cdots E_1 R\) with some elementary row matrices \(E_i\).
Taking transpose, \(A^T=R^T E_1^T\cdots E_k^T\).
Taking determinants, by multiplicativity, \[ \det(A^T)=\det(R^T)\det(E_1^T)\cdots\det(E_k^T). \]
Each elementary matrix and its transpose have the same determinant.
Also, \(\det(R)=\det(R^T)\) (either \(1\) if \(R=I_n\), or \(0\) otherwise).
Therefore \[\det(A^T)=\det(A).\]
Consequently, cofactor expansion along rows gives the same determinant as expansion along columns.

Summary

For \(n\times n\) matrices \(A\) and \(B\),
\[\det(AB)=\det(A)\det(B),\]
and therefore \(\det(AB)=\det(BA)\).
\(A\) is invertible if and only if \(\det(A)\neq0\); in that case, \[\det(A^{-1})=\frac{1}{\det(A)}.\]
Determinants of elementary matrices are easy to compute, so \(\det(A)\) can be found efficiently using Gaussian or Gauss–Jordan elimination.
The determinant is invariant under transpose: \(\det(A^T)=\det(A)\).