Row and column expansion

Matrix calculus: Determinants

Row and column expansion

We will now focus on the actual calculation of determinants. There is a range of useful mathematical rules. We will focus here on the so-called expansion of a determinant along a row or column. Below we discuss the role of row and column operations on matrices.

The gist of expansion along a row or column is that the calculation of a determinant is reduced to a calculation of smaller determinants. Here is the exact wording.

The determinant of an $(n\times n)$ -matrix $A$ can be calculated by expansion along a row or column. By this we mean the following equations, where $A_{ij}$ is the $((n-1)\times(n-1))$ -matrix obtained from $A$ by deleting the $i$ -th row and the $j$ -th column.

Expansion along the $i$ -th row:
$\det (A)=\sum_{j=1}^n (-1)^{i+j}a_{ij}\det (A_{ij})$
Expansion along the $j$ -th column:
$\det (A)=\sum_{i=1}^n (-1)^{i+j}a_{ij}\det (A_{ij})$

Consider the $(n\times n)$ -matrix

$A=\left(\,\begin{array}{ccc} a_{11} & \ldots & a_{1n}\\ a_{21} & & \vdots\\ \vdots & & \vdots\\ a_{n1} & \ldots & a_{nn} \end{array}\,\right)$ All terms of the sum formula for the determinant have factors from the first column. First look at a term that has the factor $a_{11}$ . All of its other factors do not come from the first column or first row, and thus come from the matrix $A_{11}$ . Now look at a term containing the factor $a_{21}$ . The other factors of such a term do not lie in the second row or the first column, so these come from $A_{21}$ . And so on. A close examination yields

$\det (A)=a_{11}\det (A_{11})-a_{21}\det (A_{21})+\ldots +(-1)^{n+1}a_{n1}\det (A_{n1})$ This formula is known as expansion along the first column. We may also expand along other columns and rows.

Selection of row or column

The above expansions are of course very useful when a row or column has a lot of zeros: expansion in that row or column then provides a sum of a few determinants, which are smaller.

Example Consider

$A=\left(\,\begin{array}{rrr} -1 & 0 & 2\\ 2 & 1 & 3\\ 1 & 3 & -1 \end{array}\,\right)$ Then

$A_{12}=\left(\,\begin{array}{rr} 2 & 3\\ 1 & -1 \end{array}\,\right)\hbox{ and } A_{33}=\left(\,\begin{array}{rr} -1 & 0\\ 2 & 1 \end{array}\,\right)$ In order to calculate the determinant of $A$ by expansion along the first row, the next two subdeterminants are needed:

$\begin{array}{rcl}\det( A_{11})&=&\left|\,\begin{array}{rr} 1 & 3\\ 3 & -1 \end{array}\,\right| =-10\\ \det( A_{13})&=&\left|\,\begin{array}{rr} 2 & 1\\ 1 & 3 \end{array}\,\right| = 5 \end{array}$ The formula for expansion along the first row yields

$\begin{array}{rcl} \det (A)&=&\sum_{j=1}^3 (-1)^{1+j}a_{1j}\det (A_{1j})\\ & =& (-1)^{1+1}a_{11}\det (A_{11}) + (-1)^{1+2}a_{12}\det (A_{12}) + (-1)^{1+3}a_{13}\det (A_{13})\\ & =& -(-10)+ 0\cdot\det (A_{12})+ 2\cdot 5 \\ &=& 20\end{array}$

Upper triangular matrix

A fairly extreme example can be found in the determination of the determinant of a so-called triangular matrix: above (or below) the diagonal are all zeros. The answer is clear from Determinants of some special matrices; if a term has a factor above the diagonal, then there is also a factor in this term below the diagonal, which is $0$ ; therefore, such a term equals $0$ . The only term that remains is the product of the diagonal elements. Here we derive this result again, but now by repeated expansion along the first column:

$\begin{array}{l l} \det (A) & =\left|\,\begin{array}{ccccc} a_{11} & \ast & \ldots & \ldots & \ast\\ 0 & a_{22} & & & \vdots\\ \vdots & \ddots & \ddots & & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\ 0 & \ldots & \ldots & 0 & a_{nn} \end{array}\,\right|\ \\\\ & =\ a_{11}\cdot \left|\,\begin{array}{ccccc} a_{22} & \ast & \ldots & \ldots & \ast\\ 0 & \ddots & & & \vdots\\ \vdots & \ddots & \ddots & & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\ 0 & \ldots & \ldots & 0 & a_{nn} \end{array}\,\right|\\\\ & =\ a_{11}\cdot a_{22}\cdot \left|\,\begin{array}{ccc} a_{33} & \ldots & \ast\\ & \ddots & \vdots\\ 0 & & a_{nn} \end{array}\,\right|\\\\ &=\ a_{11}\cdot a_{22}\cdots\ \left|\,\begin{array}{cc} a_{n-1,n-1} & \ast\\ 0 & a_{nn} \end{array}\,\right|\\\\ & =\ a_{11}\cdot a_{22}\,\cdots\, a_{n-1,n-1}\cdot a_{nn} \end{array}$ Suggestion from Ernst:

$\begin{array}{l l} \det (A) &=\left|\,\begin{array}{ccccc} a_{11} & \ast & \ldots & \ldots & \ast\\ 0 & a_{22} & & & \vdots\\ \vdots & \ddots & \ddots & & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\ 0 & \ldots & \ldots & 0 & a_{nn} \end{array}\,\right|\ \\\\ & =\ a_{11}\cdot \left|\,\begin{array}{ccccc} a_{22} & \ast & \ldots & \ldots & \ast\\ 0 & a_{33} & & & \vdots\\ \vdots & \ddots & \ddots & & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\ 0 & \ldots & \ldots & 0 & a_{nn} \end{array}\,\right|\ \\\\ & =\ a_{11}\cdot a_{22}\cdot \left|\,\begin{array}{ccccc} a_{33} & \ast & \ldots & \ldots & \ast\\ 0 & a_{44} & & & \vdots\\ \vdots & \ddots & \ddots & & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n-1} & \ast\\ 0 & \ldots & \ldots & 0 & a_{nn} \end{array}\,\right|\ \\\\&=\ a_{11}\cdot a_{22}\cdots a_{n-2,n-2}\cdot\left|\,\begin{array}{cc} a_{n-1,n-1} & \ast\\ 0 & a_{nn} \end{array}\,\right|\\\\ & =\ a_{11}\cdot a_{22}\,\cdots\, a_{n-2,n-2}\cdot a_{n-1,n-1}\cdot a_{nn} \end{array}$

Let $A$ be the following $(3\times3)$ -matrix:

$A=\matrix{5 & 0 & -5 \\ 6 & -2 & 4 \\ -1 & 2 & -3 \\ }$
Calculate the determinant of $A$ by expansion along a row or column.

$\det (A)=$ $-60$

The $(1,2)$ -entry of $A$ equals $0$ . Therefore we expand along the first row.

$\begin{array}{rcl}\det(A) &=& (-1)^{1+1}\cdot a_{1 1}\cdot \det(A_{1 1}) + (-1)^{1+2}\cdot a_{1 2}\cdot \det(A_{1 2}) + (-1)^{1+3}\cdot a_{1 3}\cdot \det(A_{1 3})\\ &=& 5\cdot\left|\begin{array}{cc}-2&4\\2&-3\end{array}\right| -0\cdot\left|\begin{array}{cc}6&4\\-1&-3\end{array}\right| -5\cdot\left|\begin{array}{cc}6&-2\\-1&2\end{array}\right|\\ &=& 5\cdot(-2) +0 -5\cdot10\\ &=& -60 \end{array}$

New example