Linear Algebra – Determinants – Properties

Definition:: A matrix \(A=(a_{ij})\) with \(a_{ij}=0\) for all \(i > j\) is called an upper triangular matrix. A matrix \(A=(a_{ij})\) with \(a_{ij}=0\) for all \(j > i\) is called a lower triangular matrix

Remark: A triangular matrix needs not to be square.

Theorem: If \(A\) is a square (upper or lower) triangular matrix, then \(\det(A)\) is the product of the entries on the main diagonal of \(A\).

We may use row operations to compute a determinant:

Theorem: If \(A\) is a square matrix, then we have: if \(B\) is the matrix that arises from \(A\) by

1. adding a multiple of a row of \(A\) to another row, then: \(\det(B)=\det(A)\);


2. interchanging two rows of \(A\), then: \(\det(B)=-\det(A)\);


3. multiplying one row of \(A\) by a scalar \(k\), then: \(\det(B)=k\cdot\det(A)\).

Theorem: If \(A\) is a square matrix, then we have: \(\det(A^T)=\det(A)\).

This implies that the row operations in the previous theorem may also be applied to the columns of the matrix in order to compute the determinant.

Examples:

1)

\[\begin{vmatrix}101&102&103\\201&203&205\\302&305&309\end{vmatrix}=\begin{vmatrix}101&102&103\\-1&-1&-1\\-1&-1&0\end{vmatrix} =-\begin{vmatrix}-1&-1&0\\-1&-1&-1\\101&102&103\end{vmatrix}=-\begin{vmatrix}-1&-1&0\\0&0&-1\\0&1&103\end{vmatrix} =\begin{vmatrix}-1&-1&0\\0&1&103\\0&0&-1\end{vmatrix}=(-1)\cdot1\cdot(-1)=1.\]

2)

\[\begin{vmatrix}101&102&103\\201&203&205\\302&305&309\end{vmatrix}=\begin{vmatrix}101&1&2\\201&-1&4\\302&3&7\end{vmatrix} =\begin{vmatrix}101&1&0\\201&2&0\\302&3&1\end{vmatrix}=\begin{vmatrix}101&1\\201&2\end{vmatrix}=\begin{vmatrix}0&1\\-1&2\end{vmatrix}=0-(-1)=1.\]

Theorem: If \(A\) and \(B\) are \(n\times n\) matrices, then we have: \(\det(AB)=\det(A)\cdot\det(B)\).

Example: Consider \(A=\begin{pmatrix}1&-2\\3&4\end{pmatrix}\) and \(B=\begin{pmatrix}-1&1\\2&1\end{pmatrix}\), then we have: \(\det(A)=4+6=10\) and \(\det(B)=-1-2=-3\). Furthermore, we have:

\[AB=\begin{pmatrix}1&-2\\3&4\end{pmatrix}\begin{pmatrix}-1&1\\2&1\end{pmatrix}=\begin{pmatrix}-5&-1\\5&7\end{pmatrix} \quad\Longrightarrow\quad\det(AB)=-35+5=-30\]

and

\[BA=\begin{pmatrix}-1&1\\2&1\end{pmatrix}\begin{pmatrix}1&-2\\3&4\end{pmatrix}=\begin{pmatrix}2&6\\5&0\end{pmatrix} \quad\Longrightarrow\quad\det(BA)=0-30=-30.\]

Although the matrices do not commute (\(BA\neq AB\)), we see that \(\det(AB)=\det(A)\cdot\det(B)=\det(B)\cdot\det(A)=\det(BA)\).

Corollary: If \(A\) is an invertible matrix, then we have: \(\det(A^{-1})=\displaystyle\frac{1}{\det(A)}\).

Proof: Since \(A\) is invertible, we have: \(\det(A)\neq0\) and \(AA^{-1}=I\). This implies that \(1=\det(I)=\det(AA^{-1}) =\det(A)\cdot\det(A^{-1})\) and therefore \(\det(A^{-1})=\displaystyle\frac{1}{\det(A)}\).

Corollary: If \(A\) is an \(n\times n\) matrix, then we have: \(\det(\lambda A)=\lambda^n\det(A)\) for every \(\lambda\in\mathbb{R}\).

Proof: Since \(\lambda A=(\lambda I)A\) we conclude that \(\det(\lambda A)=\det(\lambda I)\cdot\det(A)=\lambda^n\det(A)\).

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Last modified on March 22, 2021