5. Cross Product

a. Determinants

The most common method of computing a determinant is expansion on a row or column, which does generalize to any size matrix.

2. Expansion by Minors on a Row or Column

The formula for the determinant using the method of diagonals given on the previous page can be rewritten as follows. Notice that each entry on the first row occurs in exactly two terms. Pair up those two terms and factor out the entry from the first row. Then factor a minus out of the second term: \[\begin{aligned} \begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix} &=aei+dhc+gbf-gec-ahf-dbi \\ &=a(ei-hf)+d(hc-bi)+g(bf-ec) \\ &=a(ei-hf)-d(bi-hc)+g(bf-ec) \end{aligned}\] (The reason for the minus on the second term will explained shortly.) Look at the \(3\) quantities in parentheses. They are \(2\times2\) determinants! \[ \begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix} =a\begin{vmatrix} e & h \\ f & i \end{vmatrix} -d\begin{vmatrix} b & h \\ c & i \end{vmatrix} +g\begin{vmatrix} b & e \\ c & f \end{vmatrix} \qquad \text{(*)} \] This formula is called the expansion by minors on the \(1^\text{st}\) row. We can generalize this formula to any row or column by following these steps:

• In a \(3\times3\) matrix, choose a row or column to expand on. Write down the three entries in that row or column. For example in the expansion given above, we write the three entries in the first row: \[ a \qquad\qquad d \qquad\qquad g \]
• For each entry in that row or column, cross out the row and column of that entry, and take the determinant of the remaining \(2\times2\) matrix. (These matrices are called the minors of the original matrix and their determinants are called the minor determinants.) Multiply each entry on the selected row or column by its minor determinant. In the example given above, for each entry in the first row, its minor is in bold and the product of the entry and its minor determinant is shown below: \[ \begin{vmatrix} a\hspace{-8pt}\boldsymbol{\bigcirc} & d\hspace{-5pt}\boldsymbol{\backslash} & g\hspace{-5pt}\boldsymbol{\backslash} \\ b\hspace{-5pt}\boldsymbol{\backslash} & \boldsymbol{e} & \boldsymbol{h} \\ c\hspace{-5pt}\boldsymbol{\backslash} & \boldsymbol{f} & \boldsymbol{i} \\ \end{vmatrix} \qquad \begin{vmatrix} a\hspace{-5pt}\boldsymbol{\backslash} & d\hspace{-8pt}\boldsymbol{\bigcirc} & g\hspace{-5pt}\boldsymbol{\backslash} \\ \boldsymbol{b} & e\hspace{-5pt}\boldsymbol{\backslash} & \boldsymbol{h} \\ \boldsymbol{c} & f\hspace{-5pt}\boldsymbol{\backslash} & \boldsymbol{i} \\ \end{vmatrix} \qquad \begin{vmatrix} a\hspace{-5pt}\boldsymbol{\backslash} & d\hspace{-5pt}\boldsymbol{\backslash} & g\hspace{-8pt}\boldsymbol{\bigcirc} \\ \boldsymbol{b} & \boldsymbol{e} & h\hspace{-5pt}\boldsymbol{\backslash} \\ \boldsymbol{c} & \boldsymbol{f} & i\hspace{-5pt}\boldsymbol{\backslash} \\ \end{vmatrix} \] \[ a(ei-hf) \qquad \quad d(bi-hc) \qquad \quad g(bf-ec) \]
• Finally you need to put a plus or minus in front of each term before adding them. The plus or minus is taken from the following checker board pattern: \[ \begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix} \] So the expansion on the \(1^\text{st}\) row, given above, has a \(+,-,+\) giving the formula in (*). Similarly, an expansion on the \(2^\text{nd}\) column would have a \(-,+,-\) and look like: \[ \begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix} =-d\begin{vmatrix} b & h \\ c & i \end{vmatrix} +e\begin{vmatrix} a & g \\ c & i \end{vmatrix} -f\begin{vmatrix} a & g \\ b & h \end{vmatrix} \] If you expand the three \(2\times2\) determinants, you can check this gives the same \(3\times3\) determinant.

Compute \(\begin{vmatrix} 2 & 4 & 1 \\ 3 & 1 & 5 \\ -2 & 1 & 4 \end{vmatrix}\) by expanding on the third column.

Expand on the third column: \[\begin{aligned} \begin{vmatrix} 2 & 4 & 1 \\ 3 & 1 & 5 \\ -2 & 1 & 4 \end{vmatrix} &=1\begin{vmatrix} 3 & 1 \\ -2 & 1 \end{vmatrix} -5\begin{vmatrix} 2 & 4 \\ -2 & 1 \end{vmatrix} +4\begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix} \\ &=1(3+2)-5(2+8)+4(2-12)=-85 \end{aligned}\] This is the same answer as we got in an example on the previous page.

It is advantageous to expand on the row or column with the most zeros.

Compute \(\begin{vmatrix} 2 & 4 & 1 \\ 3 & 5 & 9 \\ 0 & 0 & 2 \end{vmatrix}\).

Expand on the third row: \[ \begin{vmatrix} 2 & 4 & 1 \\ 3 & 5 & 9 \\ 0 & 0 & 2 \end{vmatrix} =0\begin{vmatrix} * & * \\ * & * \end{vmatrix} -0\begin{vmatrix} * & * \\ * & * \end{vmatrix} +2\begin{vmatrix} 2 & 4 \\ 3 & 5 \end{vmatrix} =-4 \] In this example, you can skip the computation of the determinants of the first two matrices because they are mutiplied by \(0\).

Generalization to Any Square Matrix

This method is called Expansion by Minors and generalizes to \(4\times4\), \(5\times5\), and \(n\times n\) matrices. In summary: Pick a row or column. Multiply each entry by the determinant of its minor and a plus or minus according to a checker board pattern and add them up. The general checkerboard pattern is: \[ \begin{pmatrix} + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ + & - & + & - & \cdots \\ - & + & - & + & \cdots \\[4pt] \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} \] always starting with a plus in the top left entry.

Notice that a \(4\times4\) determinant is the sum of \(4\) terms each of which has a \(3\times3\) determinant which has \(6\) terms. So all together there are \(24\) terms, as mentioned on the previous page. Similarly, an \(n\times n\) determinant is the sum of \(n\) terms each of which has an \((n-1)\times(n-1)\) determinant which all together adds up to \(n!\) terms.

Compute the determinant of the \(4\times4\) matrix \(\begin{pmatrix} 5 & 0 & 2 & 1 \\ 0 & 0 & 3 & 0 \\ 2 & 3 & 1 & 3 \\ 0 & 4 & 4 & 2 \end{pmatrix}\) using the method of expansion by minors on the second row.

The \(2^\text{nd}\) row is chosen for the expansion because it contains three zeros. Thus, upon expansion we get one \(3\times3\) minor. The contents of the other minors are irrelevant because they are all multiplied by zero: \[\begin{aligned} \begin{vmatrix} 5 & 0 & 2 & 1 \\ 0 & 0 & 3 & 0 \\ 2 & 3 & 1 & 3 \\ 0 & 4 & 4 & 2 \end{vmatrix} &=-0\begin{vmatrix} * & * & * \\ * & * & * \\ * & * & * \end{vmatrix} +0\begin{vmatrix} * & * & * \\ * & * & * \\ * & * & * \end{vmatrix} -3\begin{vmatrix} 5 & 0 & 1 \\ 2 & 3 & 3 \\ 0 & 4 & 2 \end{vmatrix} +0\begin{vmatrix} * & * & * \\ * & * & * \\ * & * & * \end{vmatrix} \\ &=-3\begin{vmatrix} 5 & 0 & 1 \\ 2 & 3 & 3 \\ 0 & 4 & 2 \end{vmatrix} \end{aligned}\] Finally we choose to expand upon the \(1^\text{st}\) column because it contains a zero: (The \(2^\text{nd}\) column or \(1^\text{st}\) or \(3^\text{rd}\) row would be just as good.) \[\begin{aligned} -3\begin{vmatrix} 5 & 0 & 1 \\ 2 & 3 & 3 \\ 0 & 4 & 2 \end{vmatrix} &=-3\left( 5\begin{vmatrix} 3 & 3 \\ 4 & 2 \end{vmatrix} -2\begin{vmatrix} 0 & 1 \\ 4 & 2 \end{vmatrix} +0\begin{vmatrix} * & * \\ * & * \end{vmatrix} \right) \\ &=-3(5(6-12)-2(0-4)+0) \\ &=-3(-30+8+0)=66 \end{aligned}\] Thus the determinant of the matrix is \(66\).