### 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.

Expansion along a row or column

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})

\]

Let \(A\) be the following \((3\times3)\)-matrix: \[A=\matrix{3 & 4 & 0 \\ -4 & 5 & -4 \\ 4 & -5 & -1 \\ }\]

Calculate the determinant of #A# by expansion along a row or column.

Calculate the determinant of #A# by expansion along a row or column.

\(\det (A)= \) \(-155\)

The # (1,3)#-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})\\

&=& 3\cdot\left|\begin{array}{cc}5&-4\\-5&-1\end{array}\right| -4\cdot\left|\begin{array}{cc}-4&-4\\4&-1\end{array}\right| +0\cdot\left|\begin{array}{cc}-4&5\\4&-5\end{array}\right|\\

&=& 3\cdot(-25) - 4\cdot20 +0\\

&=& -155

\end{array}\]

The # (1,3)#-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})\\

&=& 3\cdot\left|\begin{array}{cc}5&-4\\-5&-1\end{array}\right| -4\cdot\left|\begin{array}{cc}-4&-4\\4&-1\end{array}\right| +0\cdot\left|\begin{array}{cc}-4&5\\4&-5\end{array}\right|\\

&=& 3\cdot(-25) - 4\cdot20 +0\\

&=& -155

\end{array}\]

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