Determinants

Student project for MA265

## 1. Introduction:

The determinants are directly associated with a square matrix A when the determinant is A. Determinants arose in the solution of linear systems and are really usefull for the study of linear transformations. Lets S = {1,2,...,n} be the set of integers from 1 to n, arranged in ascending order. A rearrangement j1,j2,...,jn of the elements ofS is called a permutation of S. We can consider a permutation of S to be a one-to-one mapping of S onto itself.

If A is a square matrix then the determinant function is denoted by det and det(A)

For an instance we have a 2 x 2 matrix denominated A, therefore:

$det(A)=\left(\begin{array}{cccc}a11&a12\\a21&a22\end{array}\right)$

As we already defined the determinant function we can write some formulas. The formulas for any 2 x 2 and 3 x 3 matrix will be:

The determinant function for a 2 x 2 matrix is:

$det(A)=\left(\begin{array}{cccc}a11&a12\\a21&a22\end{array}\right)$

= (a11 * a22) - (a12 * a21 )

The determinant function for a 3 x 3 matrix is:

$det(A)=\left(\begin{array}{cccc}a11&a12&a13\\a21&a22&a23\\a31&a32&a33\end{array}\right)$

= (a11 * a22 * a33) + (a12 * a23 * a31) + (a13 * a21 * a32) - (a12 * a21 * a33) - (a11 * a23 * a32) - (a13 * a22 * a31

Example: Solving for a determinant

$det(A)=\left(\begin{array}{cccc}1&2&3\\2&1&3\\3&1&2\end{array}\right)= (1)(1)(2) + (2)(3)(3) + (3)(2)(1) - (1)(3)(1) - (2)(2)(2) - (3)(1)(3) = 6 = det(A)$

## 2. Properties of Determinants:

Theorem 1: Let A be an n x n matrix then; det(A) = det(At)

Example:

$A^t =\left(\begin{array}{cccc}1&2&3\\2&1&1\\3&3&2\end{array}\right) |A| = (1)(1)(2) + (2)(1)(3) + (3)(2)(3) - (1)(1)(3) - (2)(2)(2) - (3)(1)(3) = 6 = |A|$

Theorem 2: If a matrix B results from matrix A by interchanging two different rows (columns) of A, then; det(B) = - det(A)

Example:

$|A|=\left(\begin{array}{cccc}2&-1\\3&2\end{array}\right)=-\left(\begin{array}{cccc}3&2\\2&-1\end{array}\right)=\left(\begin{array}{cccc}2&3\\-1&2\end{array}\right)=|A^t|=7$

Theorem 3: If two rows (columns) of A are equal, then; det(A) = 0

Example:

$\left(\begin{array}{cccc}1&2&3\\-1&0&7\\1&2&3\end{array}\right)=0$

Theorem 4: If a row (column) of A consists entirely of zeros, then; det(A) = 0

Example:

$\left(\begin{array}{cccc}1&2&3\\4&5&6\\0&0&0\end{array}\right)=0$

Theorem 5: If B obtained from A by multiplying a row (column) of A by a real number k, then;det(B) = kdet(A)

Example:

$\left(\begin{array}{cccc}2&6\\1&12\end{array}\right)=2\left(\begin{array}{cccc}1&3\\1&12\end{array}\right)=(2)(3)\left(\begin{array}{cccc}1&1\\1&4\end{array}\right)=6(4-1)=18$

Theorem 6: If B = [bij] is obained from A = [aij] by adding to each element of the rth row (column) of A, k times the corresponding element of the sth row (column), r not equal s, of A, then; det(B) = det(A)

Example:

$\left(\begin{array}{cccc}1&2&3\\2&-1&3\\1&0&1\end{array}\right)=\left(\begin{array}{cccc}5&0&9\\2&-1&3\\1&0&1\end{array}\right)$

Theorem 7: If a matrix A = [aij] is upper (lower) triangular, then; det(A) = a11*a12...ann ; tha is, the determinant of a triangular matrix is the product of the element on themain diagonal.

Example:

Compute det(A)

$A=\left(\begin{array}{cccc}4&3&2\\3&-2&5\\2&4&6\end{array}\right) = det(A) = 2det(A)$

$=2det\left(\begin{array}{cccc}4&3&2\\3&-2&5\\1&2&3\end{array}\right)$

$=(-1)2det\left(\begin{array}{cccc}4&3&2\\3&-2&5\\1&2&3\end{array}\right)$

$=-2det\left(\begin{array}{cccc}1&2&3\\3&-2&5\\4&3&2\end{array}\right)$

$=-2det\left(\begin{array}{cccc}1&2&3\\0&-8&-4\\0&-5&-10\end{array}\right)$

$=-2det\left(\begin{array}{cccc}1&2&3\\0&-8&-4\\0&0&-30/4\end{array}\right) = det(A)= -2(1)(-8)(-30/4)=-120$

Theorem 8: If A is an n x n matrix, then A is nonsingular if and only ifdet(A) not equal 0

Theorem 9: If A and B are n x n matrices, then; det(AB) = det(A)det(B)

Example:

Let

$A=\left(\begin{array}{cccc}1&2\\3&4\end{array}\right)$ and $A=\left(\begin{array}{cccc}2&-1\\1&2\end{array}\right)$

then,

|A| = -2   and   |B| = 5

while,

$AB=\left(\begin{array}{cccc}4&3\\10&5\end{array}\right)$,

and |AB| = -10 = |A||B|

NOTE; At this point of learning we have shown the following:

1. A is invertible.
2. The only solution to the system Ax = 0 is the trivial solution.
3. A is row equivalent to In .
4. A is expressible as a product of elementary matrices.
5. Ax = b has exactly one solution for every n ¥1 matrix b.
6. Ax = b is consistent for every n ¥1 matrix b.
7. det(A)10

## 3. Cofactor Expansion:

The cofactor expansion is a method for evaluating the determinant of an n xn matrix that reduces the problem to the evaluation of determinants of matrices of order n - 1. We should repeat the proces of (n-1) x (n-1) until we have a 2 x 2 matrices.

Let A = [aij] be an n x n matrix. Let Mij be the (n-1) x (n-1) submatrix of A obtained by deleting the ith row and jth row column of A. The determinant det(Mij) is called the minor aij. Also, Let A = [aij] be an n x n matrix. The cofactor Aij of aij is defined as Aij = (-1)i+j det(Mij)

Theorem 10: Let A = [aij] be an n x n matrix. then;

det(A) = ai1Ai1+ai2Ai2+...+ainAin                             and                        det(A)=a1jA1j+a2jA2j+...+anjAnj

[expansion of det(A) along the ith row]                                                [expansion of det(A) along the jth column]

Example:

$\left(\begin{array}{cccc}1&2&-3&4\\-4&2&1&3\\3&0&0&-3\\2&0&-2&3\end{array}\right)$

$=(-1)^{3+1}(3)\left(\begin{array}{cccc}2&-3&4\\2&1&3\\0&-2&3\end{array}\right)+(-1)^{3+2}(0)\left(\begin{array}{cccc}1&-3&4\\-4&1&3\\2&-2&3\end{array}\right)$
$+(-1)^{3+3}(0)\left(\begin{array}{cccc}1&2&4\\-4&2&3\\2&0&3\end{array}\right)+(-1)^{3+4}(-3)\left(\begin{array}{cccc}1&2&-3\\-4&2&1\\2&0&-2\end{array}\right)$

= ( + 1)(3)(20) + 0 + 0 + ( − 1)( − 3)( − 4) = 48

## 4. Inverse of a Matrix:

Theorem 11: If A = [aij] is an n x nmatrix, then;

ai1Akl+ai2Ak2+...+ainAkn = 0    for i not equal k    ;    a1jA1k+a2jA2k+...+anjAnk    for j not equal k Example:

Let, $A=\left(\begin{array}{cccc}1&2&3\\-2&3&1\\4&5&-2\end{array}\right)$, then

$A21=(-1)^{2+1}\left(\begin{array}{cccc}2&3\\5&-2\end{array}\right)=19$

$A22=(-1)^{2+2}\left(\begin{array}{cccc}1&3\\4&-2\end{array}\right)=-14$

$A23=(-1)^{2+3}\left(\begin{array}{cccc}1&2\\4&5\end{array}\right)=3$

Now

a31A21+a32A22+a33+A23 = (4)(19) + (5)(-14) + (-2)(3) = 0,

and

a11A21+a12A22+a13A23 = (1)(19)+(2)(-14)+(3)(3) = 0.

Let A = [aij] be an n x n matrix. Then n xn adj A, called the adjoint of A, is the matrix whose (i,j)th entry is the cofactor Aji of aji. Thus;

$adj A=\left(\begin{array}{cccc}A11&A21&...&An1\\A12&A22&...&An2\\...&...&...&...\\A1n&A2n&...&Ann\end{array}\right)$

Theorem 12: If A = [aij] is an n x n matrix, then; A(adj A) = (adj A)A = det(A)In.

Example:

Let $A=\left(\begin{array}{cccc}1&2&3\\-2&3&1\\4&5&-2\end{array}\right)$

$\left(\begin{array}{cccc}3&-2&1\\5&6&2\\1&0&-3\end{array}\right) \left(\begin{array}{cccc}-18&-6&-10\\17&-10&-1\\-6&-2&28\end{array}\right)=\left(\begin{array}{cccc}-94&0&0\\0&-94&0\\0&0&-94\end{array}\right)=-94\left(\begin{array}{cccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

and

$\left(\begin{array}{cccc}-18&-6&-10\\17&-10&-1\\-6&-2&28\end{array}\right) \left(\begin{array}{cccc}3&-2&1\\5&6&2\\1&0&-3\end{array}\right)=-94\left(\begin{array}{cccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

## 5. Other applications of Determinants:

To obtain another method for solving a linear system of n equations in n unknowns is known as the Cramer's Rule.

Theorem 13: Cramer's Rule

Let;

a11x1 + a12x2 + ... + a1nxn = b1

a21x1 + a22x2 + ... + a2nxn = b2

...

an1x1 + an2x2 + ... + annxn = bn

be a linear system of n equations in n unknowns, and let A = [aij] be the coefficient matrix so that we can write the given system as Ax = b, where

$b=\left(\begin{array}{cccc}b1\\b2\\...\\bn\end{array}\right)$

If det(A) not equal 0, then the system has the unique solutions

x1 = det(A1)/det(A),      x2 = det(A2)/det(A),      ...,      xn = det(An)/det(A),

where Ai is the matrix obtained from A by replacing the ith column of A by b.

Example:

Consider the following linear system:

-2x1 + 3x2 - x3 = 1

x1 + 2x2 - x3 = 4

-2x1  -  x2  - x3 = -3

We have                                                        Then,

$|A|=\left(\begin{array}{cccc}-2&3&-1\\1&2&-1\\-2&-1&1\end{array}\right)=-2$

$x1={\left(\begin{array}{cccc}1&3&-1\\4&2&-1\\-3&-1&1\end{array}\right)}/{|A|}=-4/-2=2$,

$x2={\left(\begin{array}{cccc}-2&1&-1\\1&4&-1\\-2&-3&1\end{array}\right)}/{|A|}=-6/-2=3$,

and

$x3={\left(\begin{array}{cccc}-2&3&1\\1&2&4\\-2&-1&-3\end{array}\right)}/{|A|}=-8/-2=4$.

We note that Cramer's rule is only applicable when we have n equations in n unknowns and the coefficient matrix A is nonsingular. If we are facing a linear system of n equations in n unknowns whose coefficient matrix is singular, we must use the Gaussian elimination or Gauss-Jordan reduction methods.

NOTE; At this point of learning we have shown the following:

1. A is nonsingular

2. Ax = 0 has only the trival solution

3. A is row (column) equivalent to In

4. The linear system Ax = b has a unique solution for every n x 1 matrix b.

5. A is a product of elementray matrices

6. det(A) not equal 0

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