"idea"

$a, b, c \rightarrow \text { scalars }$

$\frac{a \times(b+c)}{2}$ (1 multiplication 1 addition)

$=\frac{a \times b+a \times c}{1}$ (2 multiplication 1 addition)

same quantity, but different

Property 1. If the entire row (or column ) of a square matrix is zero, then the value of its determinant is also zero

Example:

$\left[\begin{array}{ll}0 & 0 \\c & d\end{array}\right]$

$\rightarrow \frac{0 \times()}{0} + 0 \times( )$

Property 2: For a diagonal matrix, the determinant is the product of the diagonal entries

Example:

$\left|\left[\begin{array}{lll}a_{11} & 0 & 0 \\0 & a_{22} & 0 \\0 & 0 & a_{33}\end{array}\right]\right|$

$=a_{11}\left|\begin{array}{cc}a_{22} & 0 \\0 & a_{33}\end{array}\right|+0+0$

$=\underline{a_{11} a_{12} a_{13} }$

Property 3: For a triangular matrix, the determinant is the product of diagonal entries

Example:

a) $\left|\left[\begin{array}{cc}-a & b \\ 0 & c\end{array}\right]\right|=a c$, just the product of the diagonal entries

b) $\left|\left[\begin{array}{lll}a & 0 & 0 \\ \cdots & b & 0 \\ \cdots & \cdots & c\end{array}\right]\right|$

$=a\left|\begin{array}{ll}b & 0 \\ \cdot & c\end{array}\right|$

$=a b c$, again the product of diagonal

Note :

$\left[\left(\begin{array}{cc}0 & 0 \\ a & b\end{array}\right)\right]$

$\therefore$ area enclosed by the column vectors is 0

Consistent with the determinant being 0

Property 4. Interchanging the row and column of a square matrix, IT does not change the value of determinant

$\operatorname{det}(A)=\operatorname{det}\left(A^{\top}\right)$

Example

${\left[\begin{array}{ll}a & b \\c & d\end{array}\right] \text { transpose }\left[\begin{array}{ll}a & c \\b & d\end{array}\right]}$

$a d-b c \stackrel{\text { same }}{\longleftrightarrow} a d-b c$

Property 5: If any two rows (or columns) are interchanged, then the sign of the determinant changes

Example:

${\left.\begin{array}{ll} R_1 \rightarrow \\ R_2 \rightarrow \end{array}\right.} {\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]}$

${R_ 1 \leftrightarrow R_2 \Rightarrow\left[\begin{array}{ll} c & d \\ a & b \end{array}\right]}$

$\begin{aligned} & a d-b c , \quad c b-a d \ & =-(a d-b c) \ & \end{aligned}$

Property 6. If two rows (or columns) of a matrix are identical, then the value of its determinant is zero

Proof: Consider square matrix $A$. with two identical rows $R_i R_j$

$R_i \leftrightarrow R_j$ will give same matrix $A$

But, from previous property, sign of determinant changes

$-\operatorname{det}(A)=\operatorname{det}(\underbrace{A^{R_i \rightarrow R_j}}_A)$

Note : Consider $3 \times 3$ matrix

$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{array}\right]$

$A_{i j}$ is cofactor of $a_{i j}$

$d e t=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$

$a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=?$

$a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23}=0$

$=a_{11}(-1)^{2+1}\left|\begin{array}{ll}a_{12} & a_{13} \\ a_{32} & a_{33}\end{array}\right|+a_{12}(-1)^{2+2}\left|\begin{array}{ll}a_{11} & a_{13} \\ a_{31} & a_{33}\end{array}\right|$ $+ a_{13}(-1)^{2+3}\left|\begin{array}{ll}a_{11} & a_{12} \\ a_{31} & a_{32}\end{array}\right|$

Property 7: If each element of a row (or a column) is multiplied by a constant ' $k$ ' 'then the value of the determinant gets multiplied by '$k$'

Example:

${\left[\begin{array}{ll}a & b \\c & d\end{array}\right] \xrightarrow{R_1 \rightarrow k R_1}\left[\begin{array}{rr}k a & k b\\c & d\end{array}\right]}$

$\quad \downarrow$ $\qquad \qquad \qquad \qquad \downarrow$

$ad -bc \quad \quad \quad k a(d)-k b(c) =k\left[a(d)-b (c)\right]$

$\Rightarrow$ det scales by 'k'.

Property 8: If some or all elements of a new (or column)can be written as a sum of two terms, then the determinant can be expressed as the sum of determinants of matrices obtained by separating the origin matrix

Example:

$\left|\left[\begin{array}{cc}a+x & b+y \\ c & d\end{array}\right]\right|=\left|\begin{array}{ll}a & b \\c & d\end{array}\right|+\left|\begin{array}{ll}x & y \\c & d\end{array}\right|$

$(a+x) (d)+(b+y) (-c)$

$={a(d)+b (-c)} +{x(d)+y (-c)}$

$= {det}\left[\begin{array}{ll}a & b \\c & d\end{array}\right] + {det}\left[\begin{array}{ll}x & y \\c & d\end{array}\right]$

Property 9: If each element of a row(or column) is replaced by a sum of that element and an element of another row or column), then the value of the determinant remains the some

Example:

$\left[\begin{array}{ll}a & b \\c & d\end{array}\right] \xrightarrow{R_ 1 \rightarrow R_1+R_2} \left[\begin{array}{cc}a+c & b+d \\c & d\end{array}\right]$

$\quad \quad \downarrow$ $\qquad \qquad \qquad \qquad \qquad \downarrow$

$a d-b c$ $\qquad \qquad \qquad$ $\left|\begin{array}{ll}a & b \\c & d\end{array}\right|+ \left|\begin{array}{cc}c & d \\c & d\end{array}\right| = \left|\begin{array}{ll}a & b \\c & d\end{array}\right| + 0$