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

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

EXAMPLE 2.6.3  
  1. Let $ A= \begin{bmatrix}a_{11} & a_{12} \\ a_{21}
& a_{22}
\end{bmatrix}.$ Then,

    $\displaystyle \det(A) = \vert A\vert = a_{11} \det(A{(1\vert 1)}) - a_{12} \det(A{(1\vert 2)}) =
a_{11} a_{22} - a_{12} a_{21}.$

    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,
    $\displaystyle \det(A)$ $\displaystyle =$ $\displaystyle \vert A\vert = a_{11} \det(A{(1\vert 1)}) - a_{12} \det(A{(1\vert 2)}) + a_{13} \det(A{(1\vert 3)})$  
      $\displaystyle =$ $\displaystyle a_{11} \begin{vmatrix}a_{22} & a_{23} \\ \nonumber
a_{32} & a_{33...
..._{13} \begin{vmatrix}a_{21} & a_{22} \\ \nonumber
a_{31} & a_{32} \end{vmatrix}$  
      $\displaystyle =$ $\displaystyle a_{11} ( a_{22} a_{33} - a_{23} a_{32} ) - a_{12} ( a_{21} a_{33} - a_{31}
a_{23} ) + a_{13} ( a_{21} a_{32} - a_{31} a_{22} )$  
      $\displaystyle =$ $\displaystyle a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} - a_{12}
a_{21} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21}a_{32}
- a_{13} a_{22} a_{31}$ (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

    $\displaystyle \det (A) = \sum_{j=1}^n
(-1)^{k+j} a_{kj} \det\bigl(A(k\vert j)\bigr).$

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

    $\displaystyle {\mbox{Claim: }} \hspace{.5in} {\mbox{Area}} \; (PQRS) = \vert u_1 v_2 - u_2 v_1\vert
= \begin{vmatrix}u_1 & v_1 \\ u_2 & v_2 \end{vmatrix}.$

    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

    $\displaystyle \cos(\theta) = \frac{ {\mathbf u}\bullet {\mathbf v}}{\ell({\mathbf u}) \ell({\mathbf v})}.$

    Which tells us,
    $\displaystyle {\mbox{Area}}(PQRS)$ $\displaystyle =$ $\displaystyle \ell({\mathbf u}) \ell({\mathbf v}) \sin(\theta) =
\ell({\mathbf ...
...{\mathbf u}\bullet {\mathbf v}}{\ell({\mathbf u}) \ell({\mathbf v})}
\right)^2}$  
      $\displaystyle =$ $\displaystyle \sqrt{ \ell({\mathbf u})^2 + \ell(v)^2 - ({\mathbf u}\bullet{\mathbf v})^2} =
\sqrt{(u_1 v_2 - u_2 v_1)^2}$  
      $\displaystyle =$ $\displaystyle \vert u_1 v_2 - u_2 v_1\vert.$  

    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,

    $\displaystyle {\mathbf u}\times {\mathbf v}= (u_2 v_3 - u_3 v_2, u_3 v_1 - u_1 v_3, u_1 v_2 - u_2 v_1).$

    Note here that if $ A = [ {\mathbf u}^t, {\mathbf v}^t, {\mathbf w}^t],$ then

    $\displaystyle \det(A) = \begin{vmatrix}u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\
u...
...hbf w}\times {\mathbf u}) = {\mathbf w}\bullet ({\mathbf u}\times {\mathbf v}).$

    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,

    $\displaystyle {\mbox{volume }} (P) = \vert {\mathbf w}\bullet ({\mathbf u}\times {\mathbf v}) \vert.$

    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.



Subsections
A K Lal 2007-09-12