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Note:  In ch. 3 all matrices are SQUARE.
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3.1 Defn:  $det A = \Sigma \pm a_{1 j_1}a_{2j_2} ... a_{nj_n}$

$2\times 2$ short-cut: 
$det \left[\matrix{  a_{11} & a_{12} \cr
                a_{21} & a_{22} \cr }\right]$ = 

$3\times 3$ short-cut: 
$det \left[\matrix{  a_{11} & a_{12} & a_{13}\cr
                a_{21} & a_{22} & a_{23} \cr 
                a_{31} & a_{32} & a_{33} \cr }\right]$

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$\matrix{  a_{11} & a_{12} \cr
                a_{21} & a_{22}  \cr 
                a_{31} & a_{32} \cr }$

Note there is no short-cut for $n \times n$ matrices when $n >
3$.

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Definition of Determinant using cofactor expansion
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Defn:  $A_{ij}$ is the matrix obtained from $A$ by deleting the ith row 
and
the jth column.

Defn:  Let $A = (a_{ij})$ by an $n \times n$ square matrix. The 
determinant of
A is

\hskip 10pt 1.) If $n = 1$, $det A = a_{11}$. 

\hskip 10pt 2.) If $n > 1$,
$det A = \Sigma_{k = 1}^n
(-1)^{1 + k}a_{1k} det A_{1k}  $
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\rightline{= $a_{11}det A_{11} -
a_{12}det A_{12} + ... + (-1)^{1 + n}a_{1n} det A_{1n}$}

Note the above definition is an inductive or recursive definition.

Thm:  Let $A = (a_{ij})$ by an $n \times n$ square matrix, $n >1$.  Then
expanding along row $i$,
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$$det A = \Sigma_{k = 1}^n (-1)^{i + k}a_{ik} det
A_{ik}.$$



Or expanding along column $j$,
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$$det A = 
\Sigma_{k = 1}^n (-1)^{k + j}a_{kj} det
A_{kj}.$$

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Defn:  $det A_{ij}$ is the i, j-minor of $A$.

$(-1)^{i+j}det A_{ij}$ is the i, j-cofactor of $A$.

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3.2:  Properties of Determinants



Thm:  If $A ~{\overrightarrow{~_{R_i \rightarrow cR_i~}}} B$, then $det B
=
c (det
A)$.

Warning note:  $det(cA) = c^n det A$.

Thm:  If $A ~{\overrightarrow{~_{R_i \leftrightarrow R_j~}}}
B$, then $det B =
- (det A)$.

Thm:  If $A ~{\overrightarrow{~_{R_i + cR_j \rightarrow R_i~}}}
B$, then $det
B = det A$.
\vfil
\eject
Some Shortcuts:

Thm: If A is an $n \times n$ matrix which is either lower triangular or
upper
triangular, then $det A = a_{11}a_{22} ...a_{nn}$, the product of the 
entries
along the main diagonal.

Cor:  $det(I_n) = 1$.

Thm: If a square matrix has a row or column containing all zeros, its 
determinant is
zero.

Thm:  If some row (column) of a square matrix $A$ is a scalar multiple of
another row (column), then $det A = 0$.



Thm:  A square matrix is invertible if and only if $det A \not= 0$.

Thm:  Let $A$ be a square matrix.  Then the linear system $Ax = b$ has a
unique solution for every $b$ if and only if $det A \not= 0$.

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Thm:  $det AB = (det A)(det B)$.

Cor:  $det A^{-1} = {1 \over det A}$.

$det (A + B) \not= det A + det B$.

Thm:  $det A^T = det A.$ 

Proof of thm $det AB = (det A)(det B)$:

Lemma 1:  
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Let $M$ be a square matrix, and let $E$ be an elementary matrix of
the same order.  Then $det (EM) = (det E) (det M)$.

Lemma 2:  Let $M$ be a square matrix, and let $E_1, E_2, ..., E_k $ be
elementary matrices of the same order as $M$.  Then $det (E_1 E_2 ... E_k 
M) =
(det E_1) (det E_2) . . . (det E_k) (det M)$.

Lemma 3:  
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Let $E_1, E_2, ..., E_k $ be elementary matrices of the same order.  
Then $det (E_1 E_2 ... E_k ) = (det E_1) (det E_2) . . . (det E_k)$.

\end