### 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: \[A=\matrix{5 & 5 & 3 \\ -6 & 3 & -4 \\ 2 & 0 & -3 \\ }\]

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

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

\(\det (A)= \) \(-193\)

The # (3,2)#-entry of #A# equals #0#. Therefore we expand along the third row.

\[\begin{array}{rcl}\det(A) &=& (-1)^{3+1}\cdot a_{3 1}\cdot \det(A_{3 1}) + (-1)^{3+2}\cdot a_{3 2}\cdot \det(A_{3 2}) + (-1)^{3+3}\cdot a_{3 3}\cdot \det(A_{3 3})\\

&=& 2\cdot\left|\begin{array}{cc}5&3\\3&-4\end{array}\right| -0\cdot\left|\begin{array}{cc}5&3\\-6&-4\end{array}\right| -3\cdot\left|\begin{array}{cc}5&5\\-6&3\end{array}\right|\\

&=& 2\cdot(-29) +0 -3\cdot45\\

&=& -193

\end{array}\]

The # (3,2)#-entry of #A# equals #0#. Therefore we expand along the third row.

\[\begin{array}{rcl}\det(A) &=& (-1)^{3+1}\cdot a_{3 1}\cdot \det(A_{3 1}) + (-1)^{3+2}\cdot a_{3 2}\cdot \det(A_{3 2}) + (-1)^{3+3}\cdot a_{3 3}\cdot \det(A_{3 3})\\

&=& 2\cdot\left|\begin{array}{cc}5&3\\3&-4\end{array}\right| -0\cdot\left|\begin{array}{cc}5&3\\-6&-4\end{array}\right| -3\cdot\left|\begin{array}{cc}5&5\\-6&3\end{array}\right|\\

&=& 2\cdot(-29) +0 -3\cdot45\\

&=& -193

\end{array}\]

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