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.
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.
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
\[\begin{array}{rcl}
\det (A)&=&\sum_{j=1}^3 (-1)^{1+j}a_{1j}\det (A_{1j})\\ & =& (-1)^{1+1}a_{11}\det (A_{11}) + (-1)^{1+2}a_{12}\det (A_{12}) + (-1)^{1+3}a_{13}\det (A_{13})\\ & =& -(-10)+ 0\cdot\det (A_{12})+ 2\cdot 5 \\ &=& 20\end{array}
\]
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: \[A=\matrix{0 & -4 & 1 \\ 5 & -1 & -1 \\ -2 & 6 & 6 \\ }\]
Calculate the determinant of #A# by expansion along a row or column.
\(\det (A)= \) \(140\)
The # (1,1)#-entry of #A# equals #0#. Therefore we expand along the first row.
\[\begin{array}{rcl}\det(A) &=& (-1)^{1+1}\cdot a_{1 1}\cdot \det(A_{1 1}) + (-1)^{1+2}\cdot a_{1 2}\cdot \det(A_{1 2}) + (-1)^{1+3}\cdot a_{1 3}\cdot \det(A_{1 3})\\
&=& 0\cdot\left|\begin{array}{cc}-1&-1\\6&6\end{array}\right|
+4\cdot\left|\begin{array}{cc}5&-1\\-2&6\end{array}\right|
+1\cdot\left|\begin{array}{cc}5&-1\\-2&6\end{array}\right|\\
&=& 0+4\cdot28 + 1\cdot28\\
&=& 140
\end{array}\]
Row and column expansion
