det3.tex

% Chapter 4, Section 3 _Linear Algebra_ Jim Hefferon
%  http://joshua.smcvt.edu/linearalgebra
%  2001-Jun-12
\section{Laplace's Formula}
% \textit{This section is optional.
%   Later sections do not depend on this material.}
Determinants are a font of interesting and amusing formulas.
Here is one that is often used
to compute determinants by hand.



\subsectionoptional{Laplace's Expansion}
\index{Laplace determinant expansion|(}

% We can compute 
% the determinant of an~$\nbyn{n}$ matrix as a combination
% of determinants of $\nbyn{(n-1)}$ mtrices.
The example shows a $\nbyn{3}$ case but the approach works for
any size~$n>1$.

\begin{example}  \label{ex:ExpThreeFirstRow}
%<*LaplaceExample0>
Consider the permutation expansion.
\begin{align*}
  \begin{vmat}
              t_{1,1}  &t_{1,2}  &t_{1,3}  \\
              t_{2,1}  &t_{2,2}  &t_{2,3}  \\
              t_{3,1}  &t_{3,2}  &t_{3,3}
           \end{vmat}             
  &=\begin{aligned}[t]
     &t_{1,1}t_{2,2}t_{3,3}\begin{vmat}[r]
                             1  &0  &0  \\
                             0  &1  &0  \\
                             0  &0  &1
                           \end{vmat}
      +t_{1,1}t_{2,3}t_{3,2}\begin{vmat}[r]
                              1  &0  &0  \\
                              0  &0  &1  \\
                              0  &1  &0
                            \end{vmat}           \\
      &\hbox{}\quad\hbox{}
         +t_{1,2}t_{2,1}t_{3,3}\begin{vmat}[r]
                                 0  &1  &0  \\
                                 1  &0  &0  \\
                                 0  &0  &1
                                \end{vmat}
         +t_{1,2}t_{2,3}t_{3,1}\begin{vmat}[r]
                                 0  &1  &0  \\
                                 0  &0  &1  \\
                                 1  &0  &0
                               \end{vmat}        \\
        &\hbox{}\quad\hbox{}       
         +t_{1,3}t_{2,1}t_{3,2}\begin{vmat}[r]
                                 0  &0  &1  \\
                                 1  &0  &0  \\
                                 0  &1  &0
                                \end{vmat}
         +t_{1,3}t_{2,2}t_{3,1}\begin{vmat}[r]
                                 0  &0  &1  \\
                                 0  &1  &0  \\
                                 1  &0  &0
                               \end{vmat}  
    \end{aligned}
\end{align*}
Pick a row or column and factor out its entries; here we 
do the entries in the first row.
% $t_{1,1}$, $t_{1,2}$, $t_{1,3}$.
%</LaplaceExample0>
%<*LaplaceExample1>
\begin{align*}
  &=t_{1,1}\cdot \left[t_{2,2}t_{3,3}\begin{vmat}[r]
                                 1  &0  &0  \\
                                 0  &1  &0  \\
                                 0  &0  &1
                                \end{vmat}
                 +t_{2,3}t_{3,2}\begin{vmat}[r]
                                  1  &0  &0  \\
                                  0  &0  &1  \\
                                  0  &1  &0
                                \end{vmat}\,\right]    \\
         &\hbox{}\quad\hbox{}
          +t_{1,2}\cdot \left[t_{2,1}t_{3,3}\begin{vmat}[r]
                                        0  &1  &0  \\
                                        1  &0  &0  \\
                                        0  &0  &1
                                       \end{vmat}
                        +t_{2,3}t_{3,1}\begin{vmat}[r]
                                         0  &1  &0  \\
                                         0  &0  &1  \\
                                         1  &0  &0
                                        \end{vmat}\,\right]  \\
        &\hbox{}\quad\hbox{}
         +t_{1,3}\cdot \left[t_{2,1}t_{3,2}\begin{vmat}[r]
                                        0  &0  &1  \\
                                        1  &0  &0  \\
                                        0  &1  &0
                                       \end{vmat}
                        +t_{2,2}t_{3,1}\begin{vmat}[r]
                                         0  &0  &1  \\
                                         0  &1  &0  \\
                                         1  &0  &0
                                        \end{vmat}\,\right]
\end{align*}
In those permutation matrices, swap to get the first rows into place.
This requires one swap to each of the permutation matrices on
the second line, and two swaps to each on the third line.
(Recall that row swaps change the sign of the determinant.)
%</LaplaceExample1>
%<*LaplaceExample2>
\begin{align*}
  &=t_{1,1}\cdot \left[t_{2,2}t_{3,3}\begin{vmat}[r]
                                 1  &0  &0  \\
                                 0  &1  &0  \\
                                 0  &0  &1
                                \end{vmat}
                +t_{2,3}t_{3,2}\begin{vmat}[r]
                                 1  &0  &0  \\
                                 0  &0  &1  \\
                                 0  &1  &0
                                \end{vmat}\,\right]    \\
         &\hbox{}\quad\hbox{}
          -t_{1,2}\cdot \left[t_{2,1}t_{3,3}\begin{vmat}[r]
                                         1  &0  &0  \\
                                         0  &1  &0  \\
                                         0  &0  &1
                                        \end{vmat}
                        +t_{2,3}t_{3,1}\begin{vmat}[r]
                                         1  &0  &0  \\
                                         0  &0  &1  \\
                                         0  &1  &0
                                        \end{vmat}\,\right]  \\
         &\hbox{}\quad\hbox{}
          +t_{1,3}\cdot \left[t_{2,1}t_{3,2}\begin{vmat}[r]
                                         1  &0  &0  \\
                                         0  &1  &0  \\
                                         0  &0  &1
                                        \end{vmat}
                        +t_{2,2}t_{3,1}\begin{vmat}[r]
                                         1  &0  &0  \\
                                         0  &0  &1  \\
                                         0  &1  &0
                                        \end{vmat}\,\right] 
\end{align*}
%</LaplaceExample2>
%<*LaplaceExample3>
On each line the terms in square brackets involve only the second and
third row and column, and
simplify to a $\nbyn{2}$ determinant.
\begin{equation*}
  =t_{1,1}\cdot \begin{vmat}
            t_{2,2}  &t_{2,3}  \\
            t_{3,2}  &t_{3,3}
          \end{vmat}
   -t_{1,2}\cdot \begin{vmat}
             t_{2,1}  &t_{2,3}  \\
             t_{3,1}  &t_{3,3}
           \end{vmat}
   +t_{1,3}\cdot \begin{vmat}
             t_{2,1}  &t_{2,2}  \\
             t_{3,1}  &t_{3,2}
           \end{vmat}
\end{equation*}
%</LaplaceExample3>
%This gives a $\nbyn{3}$ determinant as a combination of 
%$\nbyn{2}$ determinants.
The formula given in \nearbytheorem{th:LaPlaceExp},
which generalizes this example,
is a \definend{recurrence}\index{recurrence} \Dash 
the determinant is expressed as a combination of determinants.
This formula isn't circular because it gives the $\nbyn{n}$
case in terms of smaller ones.
\end{example}

\begin{definition} \label{df:Minor}
%<*df:Minor>
For any \( \nbyn{n} \) matrix \( T \), the \( \nbyn{(n-1)} \) matrix formed by
deleting row~\( i \) and column~\( j \) of \( T \) is the
\( i,j \) \definend{minor}\index{minor, of a matrix}\index{determinant!minor}%
\index{matrix!minor} 
of \( T \).
The \( i,j \) \definend{cofactor}\index{cofactor}\index{determinant!using cofactors}%
% \index{matrix!cofactor} 
\( T_{i,j} \) of \( T \) is
\( (-1)^{i+j} \) times the determinant of the \( i,j \) minor of \( T \).
%</df:Minor>
\end{definition}

\begin{example}
The $1,2$ cofactor of the matrix from \nearbyexample{ex:ExpThreeFirstRow}
is the negative of the second $\nbyn{2}$ determinant.
\begin{equation*}
  T_{1,2}=
  -1\cdot\begin{vmat}
    t_{2,1}  &t_{2,3}  \\
    t_{3,1}  &t_{3,3}
  \end{vmat}
\end{equation*}
\end{example}

\begin{example}
Where
\begin{equation*}
   T=
   \begin{mat}[r]
      1  &2  &3  \\
      4  &5  &6  \\
      7  &8  &9
   \end{mat}
\end{equation*}
these are the  \( 1,2 \) and \( 2,2 \) cofactors.
\begin{equation*}
   T_{1,2}=
   (-1)^{1+2}\cdot\begin{vmat}[r]
                4  &6  \\
                7  &9
             \end{vmat}=6
  \qquad
   T_{2,2}=
   (-1)^{2+2}\cdot\begin{vmat}[r]
                1  &3  \\
                7  &9
             \end{vmat}=-12
\end{equation*}
\end{example}

\begin{theorem}[Laplace Expansion of Determinants]
\label{th:LaPlaceExp}
\index{determinant!Laplace expansion}%
% \index{Laplace expansion!computes determinant}
%<*th:LaPlaceExp>
Where \( T \) is an \( \nbyn{n} \) matrix, we can find
the determinant by expanding by cofactors on any
row~$i$ or column~$j$.
\begin{align*}
   \deter{T}
   &=t_{i,1}\cdot T_{i,1}+t_{i,2}\cdot T_{i,2}+\cdots+t_{i,n}\cdot T_{i,n}  \\
   &=t_{1,j}\cdot T_{1,j}+t_{2,j}\cdot T_{2,j}+\cdots+t_{n,j}\cdot T_{n,j}
\end{align*}
%</th:LaPlaceExp>
\end{theorem}

\begin{proof}
\nearbyexercise{exer:LaplaceProof}.
\end{proof}

\begin{example} \label{ex:ExpLaPlace}
We can compute the determinant
\begin{equation*}
   \deter{T}=
   \begin{vmat}[r]
     1  &2  &3  \\
     4  &5  &6  \\
     7  &8  &9
   \end{vmat}
\end{equation*}
by expanding along the first row, as in
\nearbyexample{ex:ExpThreeFirstRow}.
\begin{equation*}
   \deter{T}
   =1\cdot(+1)\begin{vmat}[r]
                5  &6  \\
                8  &9
              \end{vmat}
   +2\cdot(-1)\begin{vmat}[r]
                4  &6  \\
                7  &9
              \end{vmat}
   +3\cdot(+1)\begin{vmat}[r]
                4  &5  \\
                7  &8
              \end{vmat}     
  =-3+12-9     
  =0
\end{equation*}
Or, we could expand down the second column.
\begin{equation*}
   \deter{T}
   =2\cdot(-1)\begin{vmat}[r]
                4  &6  \\
                7  &9
              \end{vmat}
   +5\cdot(+1)\begin{vmat}[r]
                1  &3  \\
                7  &9
              \end{vmat}
   +8\cdot(-1)\begin{vmat}[r]
                1  &3  \\
                4  &6
              \end{vmat}     
  =12-60+48   
  =0
\end{equation*}
\end{example}

\begin{example}
A row or column with many zeroes suggests a Laplace expansion.
\begin{equation*}
  \begin{vmat}[r]
    1 &5  &0  \\
    2 &1  &1  \\
    3 &-1 &0
  \end{vmat}
   =
   0\cdot(+1)\begin{vmat}[r]
               2  &1  \\
               3  &-1
             \end{vmat}+
   1\cdot(-1)\begin{vmat}[r]
               1  &5  \\
               3  &-1
             \end{vmat}+
   0\cdot(+1)\begin{vmat}[r]
               1  &5  \\
               2  &1
             \end{vmat}
  =16
\end{equation*}
\end{example}

We finish by applying Laplace's expansion to
derive a new formula for the inverse of a matrix.
With \nearbytheorem{th:LaPlaceExp}, we can calculate 
the determinant of a matrix 
by taking linear combinations of entries from a row with
their associated cofactors.
\begin{equation*}
  t_{i,1}\cdot T_{i,1}+t_{i,2}\cdot T_{i,2}+\dots+t_{i,n}\cdot T_{i,n}
   =\deter{T}  
\tag*{($*$)}\end{equation*}
Recall that a matrix with two identical rows has a zero determinant.
Thus,
weighing the cofactors by entries from 
row~$k$ with $k\neq i$ gives zero
\begin{equation*}
  t_{i,1}\cdot T_{k,1}+t_{i,2}\cdot T_{k,2}+\dots+t_{i,n}\cdot T_{k,n}=0
\tag*{($**$)}\end{equation*}
because it represents the expansion along the row~$k$ of a matrix with 
row~\( i \) equal to row~\( k \).
This summarizes ($*$) and~($**$).
\begin{equation*}
 \generalmatrix{t}{n}{n}
 \begin{mat}
   T_{1,1}  &T_{2,1}  &\ldots  &T_{n,1}  \\
   T_{1,2}  &T_{2,2}  &\ldots  &T_{n,2}  \\
            &\vdots   &        &         \\
   T_{1,n}  &T_{2,n}  &\ldots  &T_{n,n}
 \end{mat}                                  
 =\begin{mat}
     |T|      &0        &\ldots  &0        \\
     0        &|T|      &\ldots  &0        \\
              &\vdots   &        &         \\
     0        &0        &\ldots  &|T|
   \end{mat} 
\end{equation*}
Note that the order of the subscripts in the matrix of cofactors
is opposite to the order of subscripts in the other matrix; e.g.,
along the first row of the matrix of cofactors 
the subscripts are $1,1$ then $2,1$, etc.

\begin{definition}  \label{df:Adjoint}
%<*df:Adjoint>
The matrix \definend{adjoint}\index{matrix!adjoint}\index{adjoint matrix}
to the square matrix \( T \) is
\begin{equation*}
  \adj(T)=
    \begin{mat}
      T_{1,1}  &T_{2,1}  &\ldots  &T_{n,1}  \\
      T_{1,2}  &T_{2,2}  &\ldots  &T_{n,2}  \\
               &\vdots   &        &         \\
      T_{1,n}  &T_{2,n}  &\ldots  &T_{n,n}
    \end{mat}
\end{equation*}
where \( T_{j,i} \) is the \( j,i \) cofactor.
%</df:Adjoint>
\end{definition}

\begin{theorem} \label{th:MatTimesAdjEqDiagDets}
%<*th:MatTimesAdjEqDiagDets>
Where \( T \) is a square matrix,
$T\cdot \adj(T)=\adj(T)\cdot T=\deter{T}\cdot I$.
Thus if $T$ has an inverse, if $\deter{T}\neq 0$, then
$T^{-1}=(1/\deter{T})\cdot\adj(T)$.\index{matrix!inverse}\index{inverse!matrix}
%</th:MatTimesAdjEqDiagDets>
\end{theorem}

\begin{proof}
Equations~($*$) and~($**$).
\end{proof}

\begin{example}    \label{ex:MatTimesAdjEqDets}
If
\begin{equation*}
  T=\begin{mat}[r]
        1  &0  &4  \\
        2  &1  &-1 \\
        1  &0  &1
      \end{mat}
\end{equation*}
then $\adj(T)$ is
\begin{equation*}
    \begin{mat}
      T_{1,1}  &T_{2,1}  &T_{3,1} \\
      T_{1,2}  &T_{2,2}  &T_{3,2} \\
      T_{1,3}  &T_{2,3}  &T_{3,3}   
    \end{mat}
    \!\!=\!\!\begin{mat}
      \begin{vmat}[r]
           1  &-1 \\
           0  &1
         \end{vmat}
      &-\begin{vmat}[r]
             0  &4  \\
             0  &1
           \end{vmat}
      &\begin{vmat}[r]
             0  &4  \\
             1  &-1
         \end{vmat}             \\[2.1ex]
      -\begin{vmat}[r]
           2  &-1 \\
           1  &1
         \end{vmat}
      &\begin{vmat}[r]
             1  &4  \\
             1  &1
           \end{vmat}
      &-\begin{vmat}[r]
             1  &4  \\
             2  &-1
           \end{vmat}            \\[2.1ex]
      \begin{vmat}[r]
           2  &1  \\
           1  &0
         \end{vmat}
      &-\begin{vmat}[r]
             1  &0  \\
             1  &0
           \end{vmat}
      &\begin{vmat}[r]
             1  &0  \\
             2  &1
           \end{vmat}
    \end{mat}
    \!\!=\!                 
    \begin{mat}[r]
       1  &0  &-4  \\
       -3 &-3 &9  \\
       -1 &0  &1
    \end{mat}         
\end{equation*}
and taking the product with $T$ gives the diagonal matrix $\deter{T}\cdot I$.
\begin{equation*}
  \begin{mat}[r]
    1  &0  &4  \\
    2  &1  &-1 \\
    1  &0  &1
  \end{mat}
  \begin{mat}[r]
    1  &0  &-4  \\
    -3 &-3 &9  \\
    -1 &0  &1
   \end{mat}         
  =\begin{mat}[r]
     -3  &0  &0  \\
      0  &-3 &0  \\
      0  &0  &-3
   \end{mat}
\end{equation*}
% \end{example}

% \begin{corollary} \label{cor:InvFromAdj}
% \index{matrix!inverse}\index{inverse!matrix}
% %<*co:InvFromAdj>
% If \( \deter{T}\neq 0 \) then
% $T^{-1}=(1/\deter{T})\cdot\adj(T)$.
% %</co:InvFromAdj>
% \end{corollary}

% \begin{example}
\noindent The inverse of $T$ 
% The inverse of the matrix from \nearbyexample{ex:MatTimesAdjEqDets}
is $(1/-3)\cdot\adj(T)$.
\begin{equation*}
  T^{-1}
  =\begin{mat}[r]  % braces make the - a negative sign?
     1/\hbox{$-3$}  &0/\hbox{$-3$}  &-4/\hbox{$-3$}  \\
    -3/\hbox{$-3$}  &-3/\hbox{$-3$} &9/\hbox{$-3$}   \\
    -1/\hbox{$-3$}  &0/\hbox{$-3$}  &1/\hbox{$-3$}
   \end{mat}
  =\begin{mat}[r]
     -1/3  &0  &4/3  \\
      1    &1  &-3   \\
      1/3  &0  &-1/3
   \end{mat}
\end{equation*}
\end{example}

The formulas from this subsection are often used for by-hand
calculation and are sometimes useful with special types of matrices.
However, for generic matrices they are
not the best choice 
because they require more arithmetic than, for instance, the 
Gauss-Jordan method.




\begin{exercises}
  \recommended \item 
    Find the cofactor.
    \begin{equation*}
      T=\begin{mat}[r]
          1  &0  &2  \\
         -1  &1  &3  \\
          0  &2  &-1
        \end{mat}
    \end{equation*}
    \begin{exparts*}
       \partsitem \( T_{2,3} \)
       \partsitem \( T_{3,2} \)
       \partsitem \( T_{1,3} \)
    \end{exparts*}
    \begin{answer}
      \begin{exparts*}
       \partsitem \( (-1)^{2+3}\begin{vmat}[r]
                            1  &0  \\
                            0  &2
                          \end{vmat}=-2  \)
       \partsitem \( (-1)^{3+2}\begin{vmat}[r]
                            1  &2  \\
                           -1  &3
                          \end{vmat}=-5  \)
       \partsitem \( (-1)^{4}\begin{vmat}[r]
                           -1  &1  \\
                            0  &2
                          \end{vmat}=-2  \)
      \end{exparts*}  
    \end{answer}
  \recommended \item 
    Find the determinant
    by expanding
    \begin{equation*}
      \begin{vmat}[r]
         3  &0  &1  \\
         1  &2  &2  \\
        -1  &3  &0
      \end{vmat}
    \end{equation*}
    \begin{exparts*}
      \partsitem on the first row
      \partsitem on the second row
      \partsitem on the third column.
    \end{exparts*}
    \begin{answer}
      \begin{exparts}
         \partsitem \( 3\cdot(+1)\begin{vmat}[r]
                          2  &2  \\
                          3  &0
                       \end{vmat}
                 +0\cdot (-1)\begin{vmat}[r]
                          1  &2  \\
                         -1  &0
                       \end{vmat}
                 +1\cdot (+1)\begin{vmat}[r]
                          1  &2  \\
                         -1  &3
                       \end{vmat} =-13 \)
         \partsitem \( 1\cdot (-1)\begin{vmat}[r]
                          0  &1  \\
                          3  &0
                       \end{vmat}
                 +2\cdot (+1)\begin{vmat}[r]
                          3  &1  \\
                         -1  &0
                       \end{vmat}
                 +2\cdot (-1)\begin{vmat}[r]
                          3  &0  \\
                         -1  &3
                       \end{vmat} =-13 \)
         \partsitem \( 1\cdot (+1)\begin{vmat}[r]
                          1  &2  \\
                         -1  &3
                       \end{vmat}
                 +2\cdot (-1)\begin{vmat}[r]
                          3  &0  \\
                         -1  &3
                       \end{vmat}
                 +0\cdot (+1)\begin{vmat}[r]
                          3  &0  \\
                          1  &2
                       \end{vmat} =-13 \)
       \end{exparts}  
     \end{answer}
  \item 
    Find the adjoint of the matrix in \nearbyexample{ex:ExpLaPlace}.
    \begin{answer}
      This is $\adj(T)$.
      \begin{align*}
         \begin{mat}
           T_{1,1}  &T_{2,1}  &T_{3,1} \\
           T_{1,2}  &T_{2,2}  &T_{3,2} \\
           T_{1,3}  &T_{2,3}  &T_{3,3}   
         \end{mat}
         &=\begin{mat}
            \begin{vmat}
               5  &6  \\ 8  &9
            \end{vmat}
            &-\begin{vmat}
               2  &3  \\  8  &9
            \end{vmat}
            &+\begin{vmat}
               2  &3  \\  5  &6
            \end{vmat}        \\[2.5ex]
            -\begin{vmat}
               4  &6  \\  7  &9
            \end{vmat}
            &+\begin{vmat}
               1  &3  \\  7  &9
            \end{vmat}
            &-\begin{vmat}
               1  &3  \\  4  &6
            \end{vmat}        \\[2.5ex]
            +\begin{vmat}
               4  &5  \\  7  &8
            \end{vmat}
            &-\begin{vmat}
               1  &2  \\  7  &8
            \end{vmat}
            &+\begin{vmat}
               1  &2  \\  4  &5
            \end{vmat}
         \end{mat}                                   \\
         &=\begin{mat}[r]
          -3  &6   &-3 \\
           6  &-12 &6  \\
          -3  &6   &-3
         \end{mat}
      \end{align*}
    \end{answer}
  \recommended \item 
    Find the matrix adjoint to each.
    \begin{exparts*}
      \partsitem \( \begin{mat}[r]
                 2   &1  &4  \\
                -1   &0  &2  \\
                 1   &0  &1
               \end{mat}  \)
      \partsitem \( \begin{mat}[r]
                 3  &-1  \\
                 2  &4
               \end{mat}  \)
      \partsitem \( \begin{mat}[r]
                 1   &1  \\
                 5   &0
               \end{mat}  \)
      \partsitem \( \begin{mat}[r]
                 1   &4  &3  \\
                -1   &0  &3  \\
                 1   &8  &9
               \end{mat}  \)
    \end{exparts*}
    \begin{answer}
      \begin{exparts}
        \partsitem This is the adjoint. 
          \begin{align*}
          \begin{mat}
            T_{1,1}  &T_{2,1}  &T_{3,1} \\ 
            T_{1,2}  &T_{2,2}  &T_{3,2} \\ 
            T_{1,3}  &T_{2,3}  &T_{3,3} 
          \end{mat}
          &=\begin{mat}
            \begin{vmat}[r]
              0  &2  \\
              0  &1 
            \end{vmat}
            &-\begin{vmat}[r]
               1  &4  \\
               0  &1
            \end{vmat}
            &\begin{vmat}[r]
               1  &4  \\
               0  &2 
            \end{vmat}        \\[2.5ex]
            -\begin{vmat}[r]
              -1  &2  \\
               1  &1
            \end{vmat}
            &\begin{vmat}[r]
               2  &4  \\
               1  &1
            \end{vmat}
            &-\begin{vmat}[r]
               2  &4  \\ 
              -1  &2
            \end{vmat}         \\[2.5ex]
            \begin{vmat}[r]
              -1  &0  \\
               1  &0
            \end{vmat} 
            &-\begin{vmat}[r]
               2  &1  \\
               1  &0
            \end{vmat}
            &\begin{vmat}[r]
               2  &1  \\
              -1  &0
            \end{vmat}
          \end{mat}                         \\
          &=\begin{mat}[r]
             0  &-1  &2  \\
             3  &-2  &-8 \\
             0  &1   &1            
          \end{mat}
          \end{align*}
        \partsitem The minors are $\nbyn{1}$.
          $\begin{mat}
             T_{1,1}  &T_{2,1} \\
             T_{1,2}  &T_{2,2}  
          \end{mat}
          =
          \begin{mat}
            \begin{vmat}
              4              
            \end{vmat}
            &-\begin{vmat}
              -1
            \end{vmat}        \\[1.5ex]
            -\begin{vmat}
               2              
            \end{vmat}
            &\begin{vmat}
               3
            \end{vmat}
          \end{mat}
          =
          \begin{mat}[r]
            4  &1  \\
           -2  &3
          \end{mat}$
        \partsitem 
          $\begin{mat}[r]
               0  &-1 \\
              -5  &1
          \end{mat}$
        \partsitem The minors are $\nbyn{2}$.
          \begin{align*}
          \begin{mat}
            T_{1,1}  &T_{2,1}  &T_{3,1} \\ 
            T_{1,2}  &T_{2,2}  &T_{3,2} \\ 
            T_{1,3}  &T_{2,3}  &T_{3,3} 
          \end{mat}
          &=\begin{mat}
            \begin{vmat}[r]
              0  &3  \\
              8  &9
            \end{vmat}
            &-\begin{vmat}[r]
              4  &3  \\
              8  &9
            \end{vmat}
            &\begin{vmat}[r]
              4  &3  \\
              0  &3 
            \end{vmat}       \\[2.5ex]
            -\begin{vmat}[r]
              -1  &3  \\
               1  &9
            \end{vmat}
            &\begin{vmat}[r]
               1  &3  \\
               1  &9              
            \end{vmat}
            &-\begin{vmat}[r]
               1  &3 \\
              -1  &3              
            \end{vmat}        \\[2.5ex]
            \begin{vmat}[r]
               -1  &0  \\
                1  &8              
            \end{vmat}
            &-\begin{vmat}[r]
                1  &4 \\
                1  &8              
            \end{vmat}
            &\begin{vmat}[r]
                1  &4  \\
               -1  &0              
            \end{vmat}
          \end{mat}                       \\
          &=
          \begin{mat}[r]
            -24  &-12  &12  \\
             12  &6    &-6   \\
             -8  &-4   &4
          \end{mat}
        \end{align*}
      \end{exparts}
    \end{answer}
\recommended \item
    Find the inverse of each matrix in the prior question with
    \nearbytheorem{th:MatTimesAdjEqDiagDets}.
    \begin{answer}
      \begin{exparts}
        \partsitem $(1/3)\cdot 
           \begin{mat}[r]
             0  &-1  &2  \\
             3  &-2  &-8 \\
             0  &1   &1            
          \end{mat}          
          =
          \begin{mat}[r]
            0  &-1/3  &2/3  \\
            1  &-2/3  &-8/3 \\ 
            0  &1/3   &1/3
          \end{mat}$
        \partsitem $(1/14)\cdot 
          \begin{mat}[r]
            4  &1  \\
           -2  &3
          \end{mat}
          =
          \begin{mat}[r]
            2/7  &1/14  \\
           -1/7  &3/14
          \end{mat}$
        \partsitem $(1/-5)\cdot 
          \begin{mat}[r]
               0  &-1 \\
              -5  &1
          \end{mat}
          =
          \begin{mat}[r]
            0  &1/5 \\
            1  &-1/5
          \end{mat}$
        \partsitem The matrix has a zero determinant, and so
           has no inverse.
      \end{exparts}
    \end{answer}
  \item 
    Find the matrix adjoint to this one.
    \begin{equation*}
      \begin{mat}[r]
        2  &1  &0  &0  \\
        1  &2  &1  &0  \\
        0  &1  &2  &1  \\
        0  &0  &1  &2
      \end{mat}
    \end{equation*}
    \begin{answer}
      $\begin{mat}
        T_{1,1}  &T_{2,1}  &T_{3,1}  &T_{4,1}  \\ 
        T_{1,2}  &T_{2,2}  &T_{3,2}  &T_{4,2}  \\ 
        T_{1,3}  &T_{2,3}  &T_{3,3}  &T_{4,3}  \\ 
        T_{1,4}  &T_{2,4}  &T_{3,4}  &T_{4,4}  
      \end{mat}
      =
      \begin{mat}[r]
        4  &-3  &2  &-1  \\
       -3  &6   &-4 &2   \\
        2  &-4  &6  &-3  \\
        -1 &2  &-3  &4
      \end{mat}$
    \end{answer}
  \recommended \item
    Expand across the first row to derive the formula for the determinant
    of a \( \nbyn{2}  \) matrix.
    \begin{answer}
       The determinant
       \begin{equation*}
         \begin{vmat}
           a  &b  \\
           c  &d
         \end{vmat}
       \end{equation*} 
       expanded on the first row gives    
       \( a\cdot (+1)\deter{d}+b\cdot (-1)\deter{c}=ad-bc \)
       (note the two $\nbyn{1}$ minors).
     \end{answer}
  \recommended \item
    Expand across the first row to derive the formula for the determinant
    of a \( \nbyn{3} \) matrix.
    \begin{answer}
      The determinant of
      \begin{equation*}
        \begin{mat}
          a  &b  &c  \\
          d  &e  &f  \\
          g  &h  &i
        \end{mat}
      \end{equation*}
      is this.
      \begin{equation*}
        a\cdot \begin{vmat}
            e  &f  \\
            h  &i
         \end{vmat}
        -b\cdot \begin{vmat}
            d  &f  \\
            g  &i
         \end{vmat}
        +c\cdot \begin{vmat}
            d  &e  \\
            g  &h
         \end{vmat}
        =a(ei-fh)-b(di-fg)+c(dh-eg)
      \end{equation*}  
    \end{answer}
  \recommended \item 
   \begin{exparts}
     \partsitem Give a formula for the adjoint of a \( \nbyn{2} \) matrix.
     \partsitem Use it to derive the formula for the inverse.
   \end{exparts} 
    \begin{answer}
      \begin{exparts}
        \partsitem
          $\begin{mat}
             T_{1,1}  &T_{2,1}  \\
             T_{1,2}  &T_{2,2}
            \end{mat}
            =\begin{mat}
            \begin{vmat}
              t_{2,2}
            \end{vmat}
           &-\begin{vmat}
               t_{1,2}
             \end{vmat}       \\
           -\begin{vmat}
               t_{2,1}
            \end{vmat}
           &\begin{vmat}
              t_{1,1} 
            \end{vmat}
          \end{mat}
          =\begin{mat}
             t_{2,2}  &-t_{1,2}  \\
            -t_{2,1} &t_{1,1}
           \end{mat}$
        \partsitem
          $(1/t_{1,1}t_{2,2}-t_{1,2}t_{2,1})\cdot
          \begin{mat}
             t_{2,2}  &-t_{1,2}  \\
            -t_{2,1} &t_{1,1}
           \end{mat}$
      \end{exparts}
    \end{answer}
  \recommended \item
    Can we compute a determinant by expanding down the diagonal?
    \begin{answer}
      No.
      Here is a determinant whose value
      \begin{equation*}
        \begin{vmat}[r]
          1  &0  &0  \\
          0  &1  &0  \\
          0  &0  &1
        \end{vmat}=1
      \end{equation*}
      doesn't equal the result of
      expanding down the diagonal.
      \begin{equation*}
        1\cdot (+1)\begin{vmat}[r]
               1  &0  \\
               0  &1
             \end{vmat}
       +1\cdot (+1)\begin{vmat}[r]
               1  &0  \\
               0  &1
             \end{vmat}
       +1\cdot (+1)\begin{vmat}[r]
               1  &0  \\
               0  &1
             \end{vmat}=3
      \end{equation*}  
    \end{answer}
  \item 
    Give a formula for the adjoint of a diagonal matrix.
    \begin{answer}
      Consider this diagonal matrix.
      \begin{equation*}
        D=
        \begin{mat}
          d_1  &0   &0   &\ldots    \\
          0    &d_2 &0   &          \\
          0    &0   &d_3            \\
               &    &    &\ddots    \\
               &    &    &      &d_n   
        \end{mat}
      \end{equation*}
      If $i\neq j$ then the $i,j$~minor is an $\nbyn{(n-1)}$ matrix
      with only $n-2$ nonzero entries, because we have deleted
      both $d_i$ and $d_j$.
      Thus, at least one row or column of the minor is all zeroes, and
      so the cofactor $D_{i,j}$ is zero.
      If $i=j$ then the minor is the diagonal matrix with entries
      $d_1$, \ldots, $d_{i-1}$, $d_{i+1}$, \ldots, $d_n$.
      Its determinant is obviously $(-1)^{i+j}=(-1)^{2i}=1$ 
      times the product of those.
      \begin{equation*}
        \adj(D)
        =
        \begin{mat}
          d_2\cdots d_n    &0                 &      &0    \\
          0                &d_1d_3\cdots d_n  &      &0    \\
                           &                  &\ddots      \\
                           &                  &       &d_1\cdots d_{n-1} 
        \end{mat}
      \end{equation*}

      By the way, \nearbytheorem{th:MatTimesAdjEqDiagDets} provides
      a slicker way to derive this conclusion.
    \end{answer}
  \recommended \item
    Prove that the transpose of the adjoint is the adjoint of the transpose.
    \begin{answer}
      Just note that if $S=\trans{T}$ then the cofactor
      $S_{j,i}$ equals the cofactor $T_{i,j}$ because $(-1)^{j+i}=(-1)^{i+j}$
      and because the minors are the transposes of each other (and the
      determinant of a transpose equals the determinant of the matrix).
    \end{answer}
  \item 
    Prove or disprove: \( \adj(\adj(T))=T \).
    \begin{answer}
      It is false; here is an example.
      \begin{equation*}
         T=\begin{mat}[r]
           1  &2   &3  \\
           4  &5   &6  \\
           7  &8   &9
         \end{mat}
         \qquad
         \adj(T)=\begin{mat}[r]
          -3  &6   &-3 \\
           6  &-12 &6  \\
          -3  &6   &-3
         \end{mat}
         \qquad
         \adj(\adj(T))=\begin{mat}[r]
           0  &0   &0  \\
           0  &0   &0  \\
           0  &0   &0
         \end{mat}
      \end{equation*}
    \end{answer}
\item
    A square matrix is \definend{upper triangular}\index{matrix!triangular} if
    each \( i,j \) entry is zero in the part above the diagonal,
    that is, when \( i>j \).
    \begin{exparts}
      \partsitem
        Must the adjoint of an upper triangular matrix be upper triangular?
        Lower triangular?
      \partsitem Prove that the inverse of a upper triangular matrix
        is upper triangular, if an inverse exists.
    \end{exparts}
    \begin{answer}
      \begin{exparts}
        \partsitem An example 
          \begin{equation*}
             M=
            \begin{mat}[r]
              1  &2  &3  \\
              0  &4  &5  \\
              0  &0  &6
            \end{mat}
          \end{equation*}
          suggests the right answer.
          \begin{align*}
            \adj(M)=
            \begin{mat}
             M_{1,1}  &M_{2,1}  &M_{3,1}  \\
             M_{1,2}  &M_{2,2}  &M_{3,2}  \\
             M_{1,3}  &M_{2,3}  &M_{3,3}  
            \end{mat}
            &=\begin{mat}
              \begin{vmat}
                4  &5 \\ 0 &6
              \end{vmat}
              &-\begin{vmat}
                2  &3  \\  0  &6
              \end{vmat}
              &\begin{vmat}
                2  &3  \\  4  &5 
              \end{vmat}          \\[1.5ex]
              -\begin{vmat}
                0  &5  \\  0  &6
              \end{vmat}
              &\begin{vmat}
                1  &3  \\  0  &6
              \end{vmat}
              &-\begin{vmat}
                1  &3  \\  0  &5  
              \end{vmat}           \\[1.5ex]
              \begin{vmat}
                0  &4  \\  0  &0
              \end{vmat}
              &-\begin{vmat}
                1  &2  \\  0  &0
              \end{vmat}
              &\begin{vmat}
                1  &2  \\  0  &4
              \end{vmat}
            \end{mat}                                \\
            &=
            \begin{mat}[r]
              24  &-12 &-2 \\
               0  &6   &-5  \\
               0  &0   &4
            \end{mat}
          \end{align*}
          The result is indeed upper triangular.

          This check is detailed but not hard.
          The entries in the upper triangle of the adjoint are 
          $M_{a,b}$ where $a>b$.
          We need to verify that the cofactor $M_{a,b}$ is zero if  $a>b$.
          With $a>b$, row~$a$ and column~$b$ of $M$, 
          \begin{equation*}
            \begin{mat}
              m_{1,1} &\ldots   &m_{1,b}  &\ldots       &        \\
              m_{2,1} &\ldots   &m_{2,b}  &       &        \\
              \vdots  &         &\vdots   &       &        \\
              m_{a,1} &\ldots   &m_{a,b}  &\ldots &m_{a,n} \\
              \vdots        &         &\vdots   &       &        \\
                      &         &m_{n,b}  &       &                   
            \end{mat}
          \end{equation*}
          when deleted, leave an upper triangular minor,
          because entry~$i,j$ of the minor is either entry~$i,j$ of
          $M$ (this happens if $a>i$ and~$b>j$;
          in this case $i<j$ implies that the entry is zero)
          or it is entry~$i,j+1$ of $M$ (this happens if $i<a$ and
          $j>b$; in this case, $i<j$ implies that $i<j+1$, which implies
          that the entry is zero), or it is entry~$i+1,j+1$ of $M$
          (this last case happens when $i>a$ and~$j>b$; obviously here
          $i<j$ implies that $i+1<j+1$ and so the entry is zero). 
          Thus the determinant of the minor is the product down the
          diagonal.
          Observe that the $a-1,a$ entry of $M$ is the
          $a-1,a-1$~entry of the minor (it doesn't get
          deleted because the relation $a>b$ is strict).
          But this entry is zero because $M$ is upper triangular and 
          $a-1<a$.
          Therefore the cofactor is zero, and the adjoint is upper triangular.
          (The lower triangular case is similar.)  
        \partsitem This is immediate from the prior part, by
          \nearbytheorem{th:MatTimesAdjEqDiagDets}.
      \end{exparts}
    \end{answer}
  \item 
    \label{exer:LaplaceProof}
    \textit{This question requires material from the optional Determinants
       Exist subsection.}
    Prove \nearbytheorem{th:LaPlaceExp}
    by using the permutation expansion.
    \begin{answer}
      We will show that each determinant can be expanded along 
      row~\( i \).
      The argument for column~$j$ is similar.

      Each term in the permutation expansion contains one and
      only one entry from each row.
      As in \nearbyexample{ex:ExpThreeFirstRow},
      factor out each row~$i$ entry to get
      \( \deter{T}
           =t_{i,1}\cdot\hat{T}_{i,1}+\dots+t_{i,n}\cdot\hat{T}_{i,n} \),
      where each \( \hat{T}_{i,j} \) is a sum of terms not containing any
      elements of row \( i \).
      We will show that \( \hat{T}_{i,j} \) is the \( i,j \) cofactor.

      Consider the \( i,j=n,n \) case first:
      \begin{equation*}
        t_{n,n}\cdot\hat{T}_{n,n}
        =t_{n,n}\cdot
            \sum_{\phi}t_{1,\phi(1)}t_{2,\phi(2)}\dots\,t_{n-1,\phi(n-1)}
                      \sgn(\phi)
      \end{equation*}
      where the sum is over all \( n \)-permutations \( \phi \) such that
      \( \phi(n)=n \).
      To show that 
      \( \hat{T}_{i,j} \) is the minor \( T_{i,j} \), we need only show
      that if \( \phi \) is an \( n \)-permutation such that 
      \( \phi(n)=n \) and
      \( \sigma \) is an \( n-1 \)-permutation with
      \( \sigma(1)=\phi(1) \), \ldots, \( \sigma(n-1)=\phi(n-1) \)
      then \( \sgn(\sigma)=\sgn(\phi) \).
      But that's true because $\phi$ and $\sigma$ 
      have the same number of inversions.

      Back to the general \( i,j \) case.
      Swap adjacent rows until the \( i \)-th is last and swap adjacent
      columns until the \( j \)-th is last.
      Observe that the determinant of the \( i,j \)-th minor is not affected by
      these adjacent
      swaps because inversions are preserved (since the minor has the
      \( i \)-th row and \( j \)-th column omitted).
      On the other hand, the sign of \( \deter{T} \) and \( \hat{T}_{i,j} \)
      changes \( n-i \) plus \( n-j \) times.
      Thus \( \hat{T}_{i,j}=(-1)^{n-i+n-j}\deter{T_{i,j}}
                =(-1)^{i+j}\deter{T_{i,j}} \).  
    \end{answer}
  \item 
    Prove that the determinant of a matrix equals the determinant of its
    transpose using Laplace's expansion and induction on the size 
    of the matrix.
    \begin{answer}
      This is obvious for the \( \nbyn{1} \) base case.

      For the inductive case, 
      assume that the determinant of a matrix equals the determinant of
      its transpose for all \( \nbyn{1} \), \ldots, \( \nbyn{(n-1)} \)
      matrices.
      Expanding on row~\( i \) gives
      \( \deter{T}=t_{i,1}T_{i,1}+\dots\,+t_{i,n}T_{i,n} \)
      and expanding on column~\( i \) gives
      \( \deter{\trans{T}}=t_{1,i}(\trans{T})_{1,i}
           +\dots+t_{n,i}(\trans{T})_{n,i} \)
      Since \( (-1)^{i+j}=(-1)^{j+i} \) the signs are the same in the
      two summations.
      Since the \( j,i \) minor of \( \trans{T} \) is the transpose
      of the \( i,j \) minor of \( T \), the inductive hypothesis
      gives \( \deter{(\trans{T})_{i,j}}=\deter{T_{i,j}} \).  
    \end{answer}
  \puzzle \item 
    Show that
    \begin{equation*}
      F_n=
      \begin{vmat}
        1  &-1  &1  &-1  &1  &-1  &\ldots  \\
        1  &1   &0  &1   &0  &1   &\ldots  \\
        0  &1   &1  &0   &1  &0   &\ldots  \\
        0  &0   &1  &1   &0  &1   &\ldots  \\
        .  &.   &.  &.   &.  &.   &\ldots
      \end{vmat}
    \end{equation*}
    where \( F_n \) is the \( n \)-th term of
    \( 1,1,2,3,5,\dots,x,y,x+y,\ldots\, \), the Fibonacci sequence,
    and the determinant is of order \( n-1 \).
    \cite{Monthly49p409}
    \begin{answer}
      \answerasgiven %
      Denoting the above determinant by \( D_n \), it is seen that
      \( D_2=1 \), \( D_3=2 \).
      It remains to show that \( D_n=D_{n-1}+D_{n-2},\; n\geq 4 \).
      In \( D_n \) subtract the \( (n-3) \)-th column from the \( (n-1) \)-th,
      the \( (n-4) \)-th from the \( (n-2) \)-th, \ldots, the first from
      the third, obtaining
      \begin{equation*}
        F_n=
        \begin{vmat}
          1  &-1  &0  &0   &0  &0   &\ldots  \\
          1  &1   &-1 &0   &0  &0   &\ldots  \\
          0  &1   &1  &-1  &0  &0   &\ldots  \\
          0  &0   &1  &1   &-1 &0   &\ldots  \\
          .  &.   &.  &.   &.  &.   &\ldots
        \end{vmat}.
      \end{equation*}
      By expanding this determinant with reference to the first row, there
      results the desired relation.  
    \end{answer}
%  \recommended \item 
%    Prove that a square matrix is singular if and only if its adjoint is
%    singular.
%  \item 
%    Prove that if \( T \) is an \( \nbyn{n} \) matrix then
%    \( \det(\adj(T))=(\det(T))^{n-1} \).
%    \begin{answer}
%      By \nearbytheorem{th:MatTimesAdjEqDiagDets}, 
%      $\deter{\adj(T)\cdot T}=\deter{T}^n$ because 
%      $\deter{T}\cdot I$ is a diagonal matrix.
%      Then $\deter{T}^n=\deter{\adj(T)\cdot T}
%                         =\deter{\adj(T)}\cdot \deter{T}$.
%    \end{answer}
\end{exercises}
\index{Laplace determinant expansion|)}