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Chapter Four: Determinants

Definition

Geometry of Determinants

Other Formulas

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1.1 Example

In this permutation expansion

$$
\begin{align*}
  \begin{vmatrix}
              t_{1,1}  &t_{1,2}  &t_{1,3}  \\
              t_{2,1}  &t_{2,2}  &t_{2,3}  \\
              t_{3,1}  &t_{3,2}  &t_{3,3}
           \end{vmatrix}             
  &=
\begin{aligned}[t]
     &t_{1,1}t_{2,2}t_{3,3}\begin{vmatrix}
                             1  &0  &0  \\
                             0  &1  &0  \\
                             0  &0  &1
                           \end{vmatrix}
      +t_{1,1}t_{2,3}t_{3,2}\begin{vmatrix}
                              1  &0  &0  \\
                              0  &0  &1  \\
                              0  &1  &0
                            \end{vmatrix}           \\
      &\hbox{}\quad\hbox{}
         +t_{1,2}t_{2,1}t_{3,3}\begin{vmatrix}
                                 0  &1  &0  \\
                                 1  &0  &0  \\
                                 0  &0  &1
                                \end{vmatrix}
         +t_{1,2}t_{2,3}t_{3,1}\begin{vmatrix}
                                 0  &1  &0  \\
                                 0  &0  &1  \\
                                 1  &0  &0
                               \end{vmatrix}        \\
        &\hbox{}\quad\hbox{}       
         +t_{1,3}t_{2,1}t_{3,2}\begin{vmatrix}
                                 0  &0  &1  \\
                                 1  &0  &0  \\
                                 0  &1  &0
                                \end{vmatrix}
         +t_{1,3}t_{2,2}t_{3,1}\begin{vmatrix}
                                 0  &0  &1  \\
                                 0  &1  &0  \\
                                 1  &0  &0
                               \end{vmatrix}  
    \end{aligned}
\end{align*}
$$

we can, for instance, factor out the entries from the first row

$$
\begin{align*}
  &=t_{1,1}\cdot \left[t_{2,2}t_{3,3}\begin{vmatrix}
                                 1  &0  &0  \\
                                 0  &1  &0  \\
                                 0  &0  &1
                                \end{vmatrix}
                 +t_{2,3}t_{3,2}\begin{vmatrix}
                                  1  &0  &0  \\
                                  0  &0  &1  \\
                                  0  &1  &0
                                \end{vmatrix}\,\right]    \\
         &\hbox{}\quad\hbox{}
          +t_{1,2}\cdot \left[t_{2,1}t_{3,3}\begin{vmatrix}
                                        0  &1  &0  \\
                                        1  &0  &0  \\
                                        0  &0  &1
                                       \end{vmatrix}
                        +t_{2,3}t_{3,1}\begin{vmatrix}
                                         0  &1  &0  \\
                                         0  &0  &1  \\
                                         1  &0  &0
                                        \end{vmatrix}\,\right]  \\
        &\hbox{}\quad\hbox{}
         +t_{1,3}\cdot \left[t_{2,1}t_{3,2}\begin{vmatrix}
                                        0  &0  &1  \\
                                        1  &0  &0  \\
                                        0  &1  &0
                                       \end{vmatrix}
                        +t_{2,2}t_{3,1}\begin{vmatrix}
                                         0  &0  &1  \\
                                         0  &1  &0  \\
                                         1  &0  &0
                                        \end{vmatrix}\,\right]
\end{align*}
$$

and swap rows in the permutation matrices to get this.

$$
\begin{align*}
  &=t_{1,1}\cdot \left[t_{2,2}t_{3,3}\begin{vmatrix}
                                 1  &0  &0  \\
                                 0  &1  &0  \\
                                 0  &0  &1
                                \end{vmatrix}
                +t_{2,3}t_{3,2}\begin{vmatrix}
                                 1  &0  &0  \\
                                 0  &0  &1  \\
                                 0  &1  &0
                                \end{vmatrix}\,\right]    \\
         &\hbox{}\quad\hbox{}
          -t_{1,2}\cdot \left[t_{2,1}t_{3,3}\begin{vmatrix}
                                         1  &0  &0  \\
                                         0  &1  &0  \\
                                         0  &0  &1
                                        \end{vmatrix}
                        +t_{2,3}t_{3,1}\begin{vmatrix}
                                         1  &0  &0  \\
                                         0  &0  &1  \\
                                         0  &1  &0
                                        \end{vmatrix}\,\right]  \\
         &\hbox{}\quad\hbox{}
          +t_{1,3}\cdot \left[t_{2,1}t_{3,2}\begin{vmatrix}
                                         1  &0  &0  \\
                                         0  &1  &0  \\
                                         0  &0  &1
                                        \end{vmatrix}
                        +t_{2,2}t_{3,1}\begin{vmatrix}
                                         1  &0  &0  \\
                                         0  &0  &1  \\
                                         0  &1  &0
                                        \end{vmatrix}\,\right] 
\end{align*}
$$

The point of the swapping (one swap to each of the permutation matrices on the second line and two swaps to each on the third line) is that the three lines simplify to three terms.

$$
\begin{equation*}
  =t_{1,1}\cdot \begin{vmatrix}
            t_{2,2}  &t_{2,3}  \\
            t_{3,2}  &t_{3,3}
          \end{vmatrix}
   -t_{1,2}\cdot \begin{vmatrix}
             t_{2,1}  &t_{2,3}  \\
             t_{3,1}  &t_{3,3}
           \end{vmatrix}
   +t_{1,3}\cdot \begin{vmatrix}
             t_{2,1}  &t_{2,2}  \\
             t_{3,1}  &t_{3,2}
           \end{vmatrix}
\end{equation*}
$$

The formula given in nearbytheorem, which generalizes this example, is a recurrence — the determinant is expressed as a combination of determinants. This formula isn’t circular because, as here, the determinant is expressed in terms of determinants of matrices of smaller size.

1.2 Definition

For any \nbyn{n} matrix T, the \nbyn{(n-1)} matrix formed by deleting row i and column j of T is the (1)j minor %

of T. The (2)j cofactor %

T_{i,j} of T is
(-1)^{i+j} times the determinant of the (3)j minor of T.

1.3 Example

The 1,2 cofactor of the matrix from nearbyexample is the negative of the second \nbyn{2} determinant.

$$
\begin{equation*}
  T_{1,2}=
  -1\cdot\begin{vmatrix}
    t_{2,1}  &t_{2,3}  \\
    t_{3,1}  &t_{3,3}
  \end{vmatrix}
\end{equation*}
$$

1.4 Example

Where

$$
\begin{equation*}
   T=
   \begin{pmatrix}
      1  &2  &3  \\
      4  &5  &6  \\
      7  &8  &9
   \end{pmatrix}
\end{equation*}
$$

these are the 1,2 and 2,2 cofactors.

$$
\begin{equation*}
   T_{1,2}=
   (-1)^{1+2}\cdot\begin{vmatrix}
                4  &6  \\
                7  &9
             \end{vmatrix}=6
  \qquad
   T_{2,2}=
   (-1)^{2+2}\cdot\begin{vmatrix}
                1  &3  \\
                7  &9
             \end{vmatrix}=-12
\end{equation*}
$$

1.5 Theorem[Laplace Expansion of Determinants]

Where T is an \nbyn{n} matrix, the determinant can be found by expanding by cofactors on row i or column j.

$$
\begin{align*}
   \deter{T}
   &=t_{i,1}\cdot T_{i,1}+t_{i,2}\cdot T_{i,2}+\cdots+t_{i,n}\cdot T_{i,n}  \\
   &=t_{1,j}\cdot T_{1,j}+t_{2,j}\cdot T_{2,j}+\cdots+t_{n,j}\cdot T_{n,j}
\end{align*}
$$

PROOF:

nearbyexercise. QED

1.6 Example

We can compute the determinant

$$
\begin{equation*}
   \deter{T}=
   \begin{vmatrix}
     1  &2  &3  \\
     4  &5  &6  \\
     7  &8  &9
   \end{vmatrix}
\end{equation*}
$$

by expanding along the first row, as in nearbyexample.

$$
\begin{equation*}
   \deter{T}
   =1\cdot(+1)\begin{vmatrix}
                5  &6  \\
                8  &9
              \end{vmatrix}
   +2\cdot(-1)\begin{vmatrix}
                4  &6  \\
                7  &9
              \end{vmatrix}
   +3\cdot(+1)\begin{vmatrix}
                4  &5  \\
                7  &8
              \end{vmatrix}     
  =-3+12-9     
  =0
\end{equation*}
$$

Alternatively, we can expand down the second column.

$$
\begin{equation*}
   \deter{T}
   =2\cdot(-1)\begin{vmatrix}
                4  &6  \\
                7  &9
              \end{vmatrix}
   +5\cdot(+1)\begin{vmatrix}
                1  &3  \\
                7  &9
              \end{vmatrix}
   +8\cdot(-1)\begin{vmatrix}
                1  &3  \\
                4  &6
              \end{vmatrix}     
  =12-60+48   
  =0
\end{equation*}
$$

1.7 Example

A row or column with many zeroes suggests a Laplace expansion.

$$
\begin{equation*}
  \begin{vmatrix}
    1 &5  &0  \\
    2 &1  &1  \\
    3 &-1 &0
  \end{vmatrix}
   =
   0\cdot(+1)\begin{vmatrix}
               2  &1  \\
               3  &-1
             \end{vmatrix}+
   1\cdot(-1)\begin{vmatrix}
               1  &5  \\
               3  &-1
             \end{vmatrix}+
   0\cdot(+1)\begin{vmatrix}
               1  &5  \\
               2  &1
             \end{vmatrix}
  =16
\end{equation*}
$$

We finish by applying this result to derive a new formula for the inverse of a matrix. With nearbytheorem, the determinant of an \nbyn{n} matrix T can be calculated by taking linear combinations of entries from a row and 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, for any matrix T, weighing the cofactors by entries from the wrong row — 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 equation summarizes (*) and (**).

$$
\begin{equation*}
 \generalmatrix{t}{n}{n}
 \begin{pmatrix}
   T_{1,1}  &T_{2,1}  &\ldots  &T_{n,1}  \\
   T_{1,2}  &T_{2,2}  &\ldots  &T_{n,2}  \\
            &\vdots   &        &         \\
   T_{1,n}  &T_{2,n}  &\ldots  &T_{n,n}
 \end{pmatrix}                                  
 =\begin{pmatrix}
     |T|      &0        &\ldots  &0        \\
     0        &|T|      &\ldots  &0        \\
              &\vdots   &        &         \\
     0        &0        &\ldots  &|T|
   \end{pmatrix} 
\end{equation*}
$$

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.

1.8 Definition

The matrix adjoint to the square matrix T is

$$
\begin{equation*}
  \adj(T)=
    \begin{pmatrix}
      T_{1,1}  &T_{2,1}  &\ldots  &T_{n,1}  \\
      T_{1,2}  &T_{2,2}  &\ldots  &T_{n,2}  \\
               &\vdots   &        &         \\
      T_{1,n}  &T_{2,n}  &\ldots  &T_{n,n}
    \end{pmatrix}
\end{equation*}
$$

where T_{j,i} is the (4)i cofactor.

1.9 Theorem

Where T is a square matrix,

T\cdot \adj(T)=\adj(T)\cdot T=\deter{T}\cdot I.

PROOF:

Equations (*) and (**). QED

1.10 Example

If

$$
\begin{equation*}
  T=\begin{pmatrix}
        1  &0  &4  \\
        2  &1  &-1 \\
        1  &0  &1
      \end{pmatrix}
\end{equation*}
$$

then the adjoint \adj(T) is

$$
\begin{equation*}
    \begin{pmatrix}
      T_{1,1}  &T_{2,1}  &T_{3,1} \\
      T_{1,2}  &T_{2,2}  &T_{3,2} \\
      T_{1,3}  &T_{2,3}  &T_{3,3}   
    \end{pmatrix}
    \!\!=\!\!\begin{pmatrix}
      \begin{vmatrix}
           1  &-1 \\
           0  &1
         \end{vmatrix}
      &-\begin{vmatrix}
             0  &4  \\
             0  &1
           \end{vmatrix}
      &\begin{vmatrix}
             0  &4  \\
             1  &-1
         \end{vmatrix}             \\[2.1ex]
      -\begin{vmatrix}
           2  &-1 \\
           1  &1
         \end{vmatrix}
      &\begin{vmatrix}
             1  &4  \\
             1  &1
           \end{vmatrix}
      &-\begin{vmatrix}
             1  &4  \\
             2  &-1
           \end{vmatrix}            \\[2.1ex]
      \begin{vmatrix}
           2  &1  \\
           1  &0
         \end{vmatrix}
      &-\begin{vmatrix}
             1  &0  \\
             1  &0
           \end{vmatrix}
      &\begin{vmatrix}
             1  &0  \\
             2  &1
           \end{vmatrix}
    \end{pmatrix}
    \!\!=\!                 
    \begin{pmatrix}
       1  &0  &-4  \\
       -3 &-3 &9  \\
       -1 &0  &1
    \end{pmatrix}         
\end{equation*}
$$

and taking the product with T gives the diagonal matrix \deter{T}\cdot I.

$$
\begin{equation*}
  \begin{pmatrix}
    1  &0  &4  \\
    2  &1  &-1 \\
    1  &0  &1
  \end{pmatrix}
  \begin{pmatrix}
    1  &0  &-4  \\
    -3 &-3 &9  \\
    -1 &0  &1
   \end{pmatrix}         
  =\begin{pmatrix}
     -3  &0  &0  \\
      0  &-3 &0  \\
      0  &0  &-3
   \end{pmatrix}
\end{equation*}
$$

1.11 Corollary

If \deter{T}\neq 0 then

T^{-1}=(1/\deter{T})\cdot\adj(T).

1.12 Example

The inverse of the matrix from nearbyexample is (1/-3)\cdot\adj(T).

$$
\begin{equation*}
  T^{-1}
  =\begin{pmatrix}  % braces make the - a negative sign?
     1/\hbox{{$-3$}}  &0/\hbox{{$-3$}}  &-4/\hbox{{$-3$}}  \\
    -3/\hbox{{$-3$}}  &-3/\hbox{{$-3$}} &9/\hbox{{$-3$}}   \\
    -1/\hbox{{$-3$}}  &0/\hbox{{$-3$}}  &1/\hbox{{$-3$}}
   \end{pmatrix}
  =\begin{pmatrix}
     -1/3  &0  &4/3  \\
      1    &1  &-3   \\
      1/3  &0  &-1/3
   \end{pmatrix}
\end{equation*}
$$

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

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