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\section*{Lecture 19: Determinant formulas and cofactors}

Now that we know the properties of the determinant, it's time to learn
some (rather messy) formulas for computing it.

\subsection*{Formula for the determinant}

We know that the determinant has the following three properties:
\begin{enumerate}
\item $\det \mb I = 1$
\item Exchanging rows reverses the sign of the determinant.
\item The determinant is linear in each row separately.
\end{enumerate}

Last class we listed seven consequences of these properties.  We can use these ten properties to find a formula for the determinant of a 2
by 2 matrix:
\begin{eqnarray*}
\twobytwodet abcd &=& \twobytwodet a0cd + \twobytwodet 0bcd\\
&=& \twobytwodet a0c0 + \twobytwodet a00d + \twobytwodet 0bc0 + \twobytwodet 0b0d\\
&=& 0 + ad + (-cb) + 0\\
&=& ad-bc.
\end{eqnarray*}
By applying property 3 to separate the individual entries of each row we could get a formula for any other square
matrix.  However, for a 3 by 3 matrix we'll have to add the determinants of twenty seven
different matrices!  Many of those determinants are zero.  The
non-zero pieces are:
\begin{small}\begin{eqnarray*}
\threebythreedet {a_{11}}{a_{12}}{a_{13}}{a_{21}}{a_{22}}{a_{23}}{a_{31}}{a_{32}}{a_{33}}
&=&
\threebythreedet {a_{1 1}}{0}{0}{0}{a_{2 2}}{0}{0}{0}{a_{3 3}} +
\threebythreedet {a_{1 1}}{0}{0}{0}{0}{a_{2 3}}{0}{a_{3 2}}{0} + 
\threebythreedet {0}{a_{12}}{0}{a_{21}}{0}{0}{0}{0}{a_{33}}\\
&& 
 + \threebythreedet {0}{a_{12}}{0}{0}{0}{a_{23}}{a_{31}}{0}{0} + 
\threebythreedet {0}{0}{a_{13}}{a_{21}}{0}{0}{0}{a_{32}}{0} + 
\threebythreedet {0}{0}{a_{13}}{0}{a_{22}}{0}{a_{31}}{0}{0}\\
&=& a_{11}a_{22}a_{33} - a_{11}a_{23}a_{33} - a_{12}a_{21}a_{33}\\
&& + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31}.
\end{eqnarray*}\end{small}
Each of the non-zero pieces has one entry from each row in each
column, as in a permutation matrix.  Since the determinant of a
permutation matrix is either 1 or -1, we can again use property 3 to find
the determinants of each of these summands and obtain our formula.

One way to remember this formula is that the positive terms are
products of entries going down and to the right in our original
matrix, and the negative terms are products going down and to the
left.  This rule of thumb doesn't work for matrices larger than 3 by 3.

\bigskip

The number of parts with non-zero determinants was 2 in the 2
by 2 case, 6 in the 3 by 3 case, and will be $24=4!$ in the 4 by 4
case.  This is because there are $n$ ways to choose an element from
the first row (i.e. a value for $\alpha$), after which there are only
$n-1$ ways to choose an element from the second row that avoids a zero determinant.  Then there are $n-2$ choices from
the third row, $n-3$ from the fourth, and so on.

The big formula for computing the determinant of any square matrix is:
$$\det \mb A = \sum_{n! \text{ terms}} \pm a_{1 \alpha} a_{2\beta}
a_{3\gamma} ... a_{n\omega}$$ where $(\alpha, \beta, \gamma,
... \omega)$ is some permutation of $(1, 2, 3, ..., n)$.  If we test
this on the identity matrix, we find that all the terms are zero
except the one corresponding to the trivial permutation $\alpha = 1$,
$\beta = 2$, ..., $\omega = n$.  This agrees with the first property:
$\det \mb I = 1$.  It's possible to check all the other properties
as well, but we won't do that here.

\bigskip

Applying the method of elimination and multiplying the diagonal
entries of the result (the pivots) is another good way to find the determinant of a
matrix.

\subsubsection*{Example}

In a matrix with many zero entries, many terms in the formula
are zero.  We can compute the determinant of:
$$\left[ \begin{array}{rrrr} 0&0&1&1\\0&1&1&0\\1&1&0&0\\1&0&0&1
\end{array}\right]$$
by choosing a non-zero
entry from each row and column, multiplying those entries, giving the product
the appropriate sign, then adding the results.

The permutation corresponding to the diagonal running from $a_{14}$
to $a_{41}$ is $(4, 3, 2, 1)$.  This contributes $1$ to the
determinant of the matrix; the contribution is positive because it takes two row
exchanges to convert the permutation $(4, 3, 2, 1)$ to the identity
$(1, 2, 3, 4)$.

Another non-zero term of $\sum \pm a_{1 \alpha} a_{2\beta} a_{3\gamma}
a_{4\omega}$ comes from the permutation $(3, 2,1,4)$.  This
contributes $-1$ to the sum, because one exchange (of the first and third
rows) leads to the identity.

These are the only two non-zero terms in the sum, so the determinant
is $0$.  We can confirm this by noting that row 1 minus row 2 plus row 3 minus row 4
equals zero.

\subsection*{Cofactor formula}

The cofactor formula rewrites the big formula for the determinant of an $n$ by $n$ matrix in terms of the determinants of smaller matrices.

In the $3 \times 3$ case, the formula looks like:
\begin{eqnarray*}
\det \mb A &=& a_{11}(a_{22}a_{33} - a_{23}a_{32}) + a_{12}(-a_{21}a_{33} + a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})\\
&=& \threebythreedet {a_{11}} 0 0 0 {a_{22}} {a_{23}} 0 {a_{32}} {a_{33}}
+  \threebythreedet 0 {a_{12}} 0 {a_{21}} 0 {a_{23}} {a_{31}} 0 {a_{33}}
+  \threebythreedet 0 0 {a_{13}} {a_{21}} {a_{22}} 0 {a_{31}} {a_{32}} 0
\end{eqnarray*}
This comes from grouping all the multiples of $a_{ij}$ in the big formula.  Each element is multiplied
by the {\em cofactors} in the parentheses following it.  Note that
each cofactor is (plus or minus) the determinant of a two by two
matrix.  That determinant is made up of products of elements in the
rows and columns NOT containing $a_{1j}$.

In general, the cofactor $C_{ij}$ of $a_{ij}$ can be found by looking
at all the terms in the big formula that contain $a_{ij}$.  $C_{ij}$
equals $(-1)^{i+j}$ times the determinant of the $n-1$ by $n-1$ square
matrix obtained by removing row $i$ and column $j$.  ($C_{ij}$ is
positive if $i+j$ is even and negative if $i+j$ is odd.)

\bigskip

For $n \times n$ matrices, the cofactor formula is:
$$\boxed{\det A = a_{11}C_{11} +a_{12}C_{12} + \cdots + a_{1n}C_{1n}.}$$
Applying this to a $2 \times 2$ matrix gives us:
$$\twobytwodet abcd = ad + b(-c).$$

\subsection*{Tridiagonal matrix}

A {\em tridiagonal matrix} is one for which the only non-zero entries
lie on or adjacent to the diagonal.  For example, the $4 \times 4$
tridiagonal matrix of $1$'s is: 

$$A_4 = \left[ \begin{array}{rrrr} 1&1&0&0\\1&1&1&0\\0&1&1&1\\0&0&1&1\end{array}\right].$$


What is the determinant of an $n \times n$ tridiagonal
matrix of $1$'s?
$$|A_1| = 1, |A_2| =\twobytwodet 1111 = 0, |A_3| = \threebythreedet 110111011 = -1$$
$$|A_4| = 1\threebythreedet 110111011 - 1\threebythreedet 110011011 = |A_3| - 1 |A_2| = -1$$
In fact, $|A_n| = |A_{n-1}| - |A_{n-2}|$.
We get a sequence which repeats every six terms:
$$|A_1| = 1, |A_2| = 0, |A_3| = -1, |A_4| = -1, |A_5| = 0, |A_6| = 1, |A_7| = 1.$$

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