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Determinant I

For each $n$, there is a function called the determinant (of order $n$) which maps each $n\times n$ matrix over $K$ to an element of $K$. In other words, $$ \det \nolimits_n \colon M_n(K) \to K. $$ We simply write $\det$ for $\det_n$ if $n$ is understood or can be arbitrary.

One can/should view $\det$ as a function on the rows (or the columns) of a matrix.

When $n =2, 3$ the formula for determinants are given by:

$$\det\left[\begin{matrix} a &b \\ c&d \end{matrix}\right] = ad-bc$$

and $$\det \left[\begin{matrix} a & b &c \\ d & e & f \\ g & h & i \end{matrix}\right] = aei + bfg + cdh - gec- hfa - dbi $$

Checkpoint. Find the determinant of

$$A = \left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & -2 & -1 \\ 0 & 2 & 2 \end{array}\right] \tag{1} $$

We can ask SAGE to spit out these formulas.

var('a,b,c,d,e,f,g,h,i')
A2 = matrix([[a,b],[c,d]]); A2,det(A2)

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[\begin{array}{rr} a & b \\ c & d \end{array}\right], -b c + a d\right)$$

A3 = matrix([[a,b,c],[d,e,f],[g,h,i]]); A3,A3.det() #another way of calling determinant.

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[\begin{array}{rrr} a & b & c \\ d & e & f \\ g & h & i \end{array}\right], -{\left(f h - e i\right)} a + {\left(c h - b i\right)} d - {\left(c e - b f\right)} g\right)$$

det(A3).expand() #this is the criss-cross formula for 3x3 determinant.

$$\newcommand{\Bold}[1]{\mathbf{#1}}-c e g + b f g + c d h - a f h - b d i + a e i$$

The order 3 and the order 2 determinants are related. More precisely, let $A$ be a $3\times 3$ matrix, then $$ \det A = a_{11}\det A_{11} - a_{12}\det A_{12} + a_{13}\det A_{13} $$ where $A_{ij}$ is the submatrix of $A$ obtained by deleting the $i$-th row and $j$-th column of $A$.

The (i,j) minor of $A$ ( or the minor of $A$ at $(i,j)$ ), denoted by $m(A)_{ij}$ (or simply $m_{ij}$ if $A$ is understood), is defined to be $\det(A_{ij})$.

The (i,j) cofactor of $A$ (or the cofactor of $A$ at $(i,j)$, denoted by $c_{ij}=c(A)_{ij}$, is defined to be

$$c_{ij} = (-1)^{i+j}m_{ij}.$$

One can think of cofactor as a signed version of minor.

def cross_out(A,i,j):
    return A.delete_rows([i-1]).delete_columns([j-1])
#write a little program so we don't need to remember offsetting the indices everytime.

cross_out(A3,1,1), cross_out(A3,1,3) #this is A_11

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[\begin{array}{rr} e & f \\ h & i \end{array}\right], \left[\begin{array}{rr} d & e \\ g & h \end{array}\right]\right)$$

A3.minors(2) #all 2x2 minors of A.

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left[-b d + a e, -c d + a f, -c e + b f, -b g + a h, -c g + a i, -c h + b i, -e g + d h, -f g + d i, -f h + e i\right]$$

In fact a much more general fact is true:

For each $1 \le i \le n$, we have the following expansion by the $i$-th row formula for determinant. $$ \det(A) = \sum_{j=1}^n a_{ij}c_{ij} = \sum_{j=1}^n (-1)^{i+j}a_{ij}m_{ij} $$

Likewise, for each $1 \le j \le n$, the expansion by the $j$-th column formula for determinant is $$ \det(A) = \sum_{i=1}^n a_{ij}c_{ij} = \sum_{i=1}^n (-1)^{i+j}a_{ij}m_{ij} $$.

Checkpoint. Compute the determinant of the matrix $A$ in the checkpoint by expanding

1. the 1st row
2. the 2nd row
3. the 3rd column.

The adjugate (or adjoint) of $A$, denoted by adj$A$ is the transpose of the matrix formed by $A$'s cofactors, i.e. $$ \text{adj}(A) = [c_{ij}]^T = [c_{ji}] $$

A2, A2.adjugate()

$$\newcommand{\Bold}[1]{\mathbf{#1}}\left(\left[\begin{array}{rr} a & b \\ c & d \end{array}\right], \left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right]\right)$$

Proposition. (adj$A$)$A = \det(A)I$.

Corollary. $A$ is invertible if and only if $\det(A) \neq 0$. And in the case $\det(A) \neq 0$, $$A^{-1} = \dfrac{1}{\det(A)}\text{adj}(A)$$

Checkpoint. Compute adj$(A)$ by hand for the $A$ in the first checkpoint and hence $A^{-1}$.