Determinant

Notation: For an $n \times n$ matrix $A,$ by $A({\alpha}\vert \beta),$ we mean the submatrix $B$ of $A,$ which is obtained by deleting the ${\alpha}^{\mbox{th}}$ row and $\beta^{\mbox{th}}$ column.

EXAMPLE 2.6.1   Consider a matrix $A = \begin{bmatrix}1 & 2 & 3\\ 1 &3&2 \\ 2 & 4&7 \end{bmatrix}.$ Then $A(1\vert 2) = \begin{bmatrix}1 & 2 \\ 2 & 7 \end{bmatrix},$ $A(1\vert 3) = \begin{bmatrix}1 & 3 \\ 2 & 4 \end{bmatrix},$ and $A(1,2\vert 1,3) = [4].$

DEFINITION 2.6.2 (Determinant of a Square Matrix)   Let $A$ be a square matrix of order $n.$ With $A,$ we associate inductively (on $n$ ) a number, called the determinant of $A,$ written $\det (A)$ (or $\vert A\vert$ ) by

$$\det(A) = \left \{ \begin{array}{lc} a & {\mbox{if }} A = [a]... ...l(A(1\vert j)\bigr), & {\mbox{ otherwise}}. \end{array} \right.$$

EXAMPLE 2.6.3  

1. Let $A= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}.$ Then,
   $$\det(A) = \vert A\vert = a_{11} \det(A{(1\vert 1)}) - a_{12} \det(A{(1\vert 2)}) = a_{11} a_{22} - a_{12} a_{21}.$$

   For example, for $A = \begin{bmatrix}1 & 2 \\ 3 & 5 \end{bmatrix},$ $\; \det(A) = \begin{vmatrix}1 & 2 \\ 3 & 5 \end{vmatrix} = 1\cdot 5 - 2 \cdot 3 = -1.$

2. Let $A= \begin{bmatrix}a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}.$ Then,
   

   $$\det(A) = \vert A\vert = a_{11} \det(A{(1\vert 1)}) - a_{12} \det(A{(1\vert 2)}) + a_{13} \det(A{(1\vert 3)})$$

   $$= a_{11} \begin{vmatrix}a_{22} & a_{23} \\ \nonumber a_{32} & a_{33... ..._{13} \begin{vmatrix}a_{21} & a_{22} \\ \nonumber a_{31} & a_{32} \end{vmatrix}$$

   $$= a_{11} ( a_{22} a_{33} - a_{23} a_{32} ) - a_{12} ( a_{21} a_{33} - a_{31} a_{23} ) + a_{13} ( a_{21} a_{32} - a_{31} a_{22} )$$

   $$= a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} - a_{12} a_{21} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21}a_{32} - a_{13} a_{22} a_{31}$$ (2.6.1)

   
   

   For example, if $A = \begin{bmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & 2 & 2 \end{bmatrix}$ then
   $\det(A) = \begin{vmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \\ 1 & 2 & 2 \end{vmatrix} = 1... ...x} + 3 \cdot \begin{vmatrix}2 & 3 \\ 1 & 2 \end{vmatrix} = 4 - 2(3) + 3(1) = 1.$

EXERCISE 2.6.4  

1. Find the determinant of the following matrices.
   $i) \; \begin{bmatrix}1 & 2 & 7 & 8 \\ 0 & 4 & 3 & 2 \\ 0 & 0 & 2 & 3\\ 0 & 0 ... ...ii) \; \begin{bmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{bmatrix}.$
2. Show that the determinant of a triangular matrix is the product of its diagonal entries.

DEFINITION 2.6.5   A matrix $A$ is said to be a singular matrix if $\det (A) = 0.$ It is called non-singular if $\det (A) \neq 0.$

The proof of the next theorem is omitted. The interested reader is advised to go through Appendix 14.3.

THEOREM 2.6.6   Let $A$ be an $n \times n$ matrix. Then

1. if $B$ is obtained from $A$ by interchanging two rows, then $\det (B) = - \det (A)$ ,
2. if $B$ is obtained from $A$ by multiplying a row by $c$ then $\det (B) = c \det (A)$ ,
3. if all the elements of one row or column of $A$ are 0 then $\det (A) = 0$ ,
4. if $B$ is obtained from $A$ by replacing the $j$ th row by itself plus $k$ times the $i$ th row, where $i \neq j$ then $\det (B) = \det (A)$ ,
5. if $A$ is a square matrix having two rows equal then $\det (A) = 0$ .

Remark 2.6.7  

1. Many authors define the determinant using ``Permutations." It turns out that THE WAY WE HAVE DEFINED DETERMINANT is usually called the expansion of the determinant along the first row.
2. Part 1 of Lemma 2.6.6 implies that ``one can also calculate the determinant by expanding along any row." Hence, for an $n \times n$ matrix $A,$ for every $k, \; 1 \le k \le n$ , one also has
   $$\det (A) = \sum_{j=1}^n (-1)^{k+j} a_{kj} \det\bigl(A(k\vert j)\bigr).$$

Remark 2.6.8  

1. Let ${\mathbf u}^t = (u_1, u_2)$ and ${\mathbf v}^t = (v_1, v_2)$ be two vectors in ${\mathbb{R}}^2.$ Then consider the parallelogram, $PQRS,$ formed by the vertices $P = (0,0)^t, Q = {\mathbf u}, R={\mathbf u}+{\mathbf v}$ and $S= {\mathbf v}.$ We
   
   $${\mbox{Claim: }} \hspace{.5in} {\mbox{Area}} \; (PQRS) = \vert u_1 v_2 - u_2 v_1\vert = \begin{vmatrix}u_1 & v_1 \\ u_2 & v_2 \end{vmatrix}.$$
   Recall that the dot product, ${\mathbf u}\bullet {\mathbf v}= u_1 v_1 + u_2 v_2,$ and $\sqrt{{\mathbf u}\bullet{\mathbf u}} = \sqrt{(u_1^2 + u_2^2)},$ is the length of the vector ${\mathbf u}.$ We denote the length by $\ell({\mathbf u}).$ With the above notation, if $\theta$ is the angle between the vectors ${\mathbf u}$ and ${\mathbf v},$ then
   $$\cos(\theta) = \frac{ {\mathbf u}\bullet {\mathbf v}}{\ell({\mathbf u}) \ell({\mathbf v})}.$$
   Which tells us,
   

   $${\mbox{Area}}(PQRS) = \ell({\mathbf u}) \ell({\mathbf v}) \sin(\theta) = \ell({\mathbf ... ...{\mathbf u}\bullet {\mathbf v}}{\ell({\mathbf u}) \ell({\mathbf v})} \right)^2}$$

   $$= \sqrt{ \ell({\mathbf u})^2 + \ell(v)^2 - ({\mathbf u}\bullet{\mathbf v})^2} = \sqrt{(u_1 v_2 - u_2 v_1)^2}$$

   $$= \vert u_1 v_2 - u_2 v_1\vert.$$

   
   Hence, the claim holds. That is, in ${\mathbb{R}}^2,$ the determinant is $\pm$ times the area of the parallelogram.
2. Let ${\mathbf u}= (u_1, u_2, u_3), {\mathbf v}= (v_1, v_2, v_3)$ and ${\mathbf w}= (w_1, w_2, w_3)$ be three elements of ${\mathbb{R}}^3.$ Recall that the cross product of two vectors in ${\mathbb{R}}^3$ is,
   $${\mathbf u}\times {\mathbf v}= (u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1).$$
   Note here that if $A = [ {\mathbf u}^t, {\mathbf v}^t, {\mathbf w}^t],$ then
   $$\det(A) = \begin{vmatrix}u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u... ...hbf w}\times {\mathbf u}) = {\mathbf w}\bullet ({\mathbf u}\times {\mathbf v}).$$
   Let $P$ be the parallelopiped formed with $(0,0,0)$ as a vertex and the vectors ${\mathbf u}, {\mathbf v}, {\mathbf w}$ as adjacent vertices. Then observe that ${\mathbf u}\times {\mathbf v}$ is a vector perpendicular to the plane that contains the parallelogram formed by the vectors ${\mathbf u}$ and ${\mathbf v}.$ So, to compute the volume of the parallelopiped $P,$ we need to look at $\cos (\theta),$ where $\theta$ is the angle between the vector ${\mathbf w}$ and the normal vector to the parallelogram formed by ${\mathbf u}$ and ${\mathbf v}.$ So,
   $${\mbox{volume }} (P) = \vert {\mathbf w}\bullet ({\mathbf u}\times {\mathbf v}) \vert.$$

   Hence, $\vert\det (A)\vert = {\mbox{volume }}\; ( P).$

3. Let ${\mathbf u}_1, {\mathbf u}_2,\ldots, {\mathbf u}_n \in {\mathbb{R}}^{n \times 1}$ and let $A = [ {\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n]$ be an $n \times n$ matrix. Then the following properties of $\det (A)$ also hold for the volume of an $n$ -dimensional parallelopiped formed with ${\mathbf 0}\in {\mathbb{R}}^{n\times 1}$ as one vertex and the vectors ${\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n$ as adjacent vertices:
   1. If ${\mathbf u}_1 = (1,0,\ldots, 0)^t, {\mathbf u}_2 = (0,1,0,\ldots, 0)^t, \ldots,$ and ${\mathbf u}_n = (0,\ldots, 0, 1)^t,$ then $\det(A) = 1.$ Also, volume of a unit $n$ -dimensional cube is $1.$
   2. If we replace the vector ${\mathbf u}_i$ by ${\alpha}{\mathbf u}_i,$ for some ${\alpha}\in {\mathbb{R}},$ then the determinant of the new matrix is ${\alpha}\cdot \det(A)$ . This is also true for the volume, as the original volume gets multiplied by ${\alpha}.$
   3. If ${\mathbf u}_1 = {\mathbf u}_i$ for some $i, \; 2 \leq i \leq n,$ then the vectors ${\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n$ will give rise to an $(n-1)$ -dimensional parallelopiped. So, this parallelopiped lies on an $(n-1)$ -dimensional hyperplane. Thus, its $n$ -dimensional volume will be zero. Also, $\vert\det(A)\vert = \vert 0\vert = 0.$

   In general, for any $n \times n$ matrix $A,$ it can be proved that $\vert\det (A)\vert$ is indeed equal to the volume of the $n$ -dimensional parallelopiped. The actual proof is beyond the scope of this book.