Determinants

Two by Two Determinant

The determinant of a 2x2 matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ is given by $det(A) = ad - bc$. This determinant is crucial for testing the invertibility of a matrix. It also relates to how areas and volumes change under a linear transformation represented by the matrix.

• A matrix (A) is invertible if and only if $det(A) \neq 0$.

• A matrix (A) is not invertible if and only if $det(A) = 0$.

Applications of Determinants

1. Testing Matrix Invertibility: The primary use, as noted above.

2. Theoretical Development: Properties of determinants facilitate various theoretical developments.

3. Area and Volume Measurement: Determinants relate to how areas and volumes transform.

Determinants are computationally intensive for large matrices, making them less suitable for large-scale computational problems. Numerical stability issues, such as the division step in matrix inversion, further limit their applicability in large matrix settings.

Justification of the Two by Two Determinant Formula

Consider a 2x2 matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$.

Case 1: a is non-zero

1. Divide the first row by (a).

2. Subtract (c) copies of the first row from the second row:
   $\begin{bmatrix} a & b \ c & d \end{bmatrix} \rightarrow \begin{bmatrix} 1 & \frac{b}{a} \ 0 & d - \frac{bc}{a} \end{bmatrix} = \begin{bmatrix} 1 & \frac{b}{a} \ 0 & \frac{ad - bc}{a} \end{bmatrix}$

The bottom row will be all zeros if and only if $ad - bc = 0$.

• If $ad - bc = 0$, the matrix is singular (not invertible).

• If $ad - bc \neq 0$, the matrix is invertible.

Case 2: a = 0 and c is non-zero

Swap rows one and two and perform a similar calculation, leading to the same conclusion.

Case 3: Both a and c are zero

If both $a = 0$ and $c = 0$, the matrix has a column of zeros and is not invertible. In this case, $ad - bc = 0$.

Examples of Calculating Two by Two Determinants

Example 1:

Calculate the determinant of $A = \begin{bmatrix} 2 & 1 \ 1 & -1 \end{bmatrix}$.

$det(A) = (2)(-1) - (1)(1) = -2 - 1 = -3$

This determinant indicates that the plane is flipped (orientation reversal) and areas are scaled by a factor of 3.

Example 2:

Calculate the determinant of $A = \begin{bmatrix} 1 & -3 \ -1 & 3 \end{bmatrix}$.

$det(A) = (1)(3) - (-1)(-3) = 3 - 3 = 0$

Since the determinant is zero, the matrix is not invertible. The rows (or columns) are linearly dependent (one is a multiple of the other).

Example 3:

Calculate the determinant of $A = \begin{bmatrix} -1 & 5 \ -3 & 3 \end{bmatrix}$.

$det(A) = (-1)(3) - (-3)(5) = -3 + 15 = 12$

Example 4:

Calculate the determinant of $A = \begin{bmatrix} 5 & 3 \ -1 & -1 \end{bmatrix}$.

$det(A) = (5)(-1) - (-1)(3) = -5 + 3 = -2$

Three by Three Determinant

Row reduction on a 3x3 matrix, assuming appropriate terms are non-zero, leads to a determinant formula:

$det(A) = a<em>{11}(a</em>{22}a<em>{33} - a</em>{23}a<em>{32}) - a</em>{12}(a<em>{21}a</em>{33} - a<em>{23}a</em>{31}) + a<em>{13}(a</em>{21}a<em>{32} - a</em>{22}a_{31})$

This formula determines whether a matrix is invertible. If the determinant is zero, the matrix is singular and not invertible. If non-zero, the matrix is invertible.

Remembering the 3x3 Determinant Formula

To remember the formula, repeat the first two columns of the matrix:

$\begin{bmatrix} a<em>{11} &amp; a</em>{12} &amp; a<em>{13} \ a</em>{21} &amp; a<em>{22} &amp; a</em>{23} \ a<em>{31} &amp; a</em>{32} &amp; a<em>{33} \end{bmatrix} \rightarrow \begin{bmatrix} a</em>{11} &amp; a<em>{12} &amp; a</em>{13} &amp; a<em>{11} &amp; a</em>{12} \ a<em>{21} &amp; a</em>{22} &amp; a<em>{23} &amp; a</em>{21} &amp; a<em>{22} \ a</em>{31} &amp; a<em>{32} &amp; a</em>{33} &amp; a<em>{31} &amp; a</em>{32} \end{bmatrix}$

Multiply along the diagonals. The positive terms are obtained by going diagonally down from the top left to the bottom right. The negative terms are obtained by going down the other direction.

Cofactor Expansion

The 3x3 determinant can be expressed in terms of cofactors:

$det(A) = a<em>{11}C</em>{11} + a<em>{12}C</em>{12} + a<em>{13}C</em>{13}$,

where,

• $C_{11}$ is the determinant of the 2x2 matrix obtained by crossing out the first row and first column of $A$.

• $C_{12}$ is the negative of the determinant of the 2x2 matrix obtained by crossing out the first row and second column of $A$.

• $C_{13}$ is the determinant of the 2x2 matrix obtained by crossing out the first row and third column of $A$.

Examples of Calculating Three by Three Determinants

Example 1:

Calculate the determinant of:

$\begin{vmatrix} 3 &amp; -1 &amp; 0 \ -2 &amp; 3 &amp; 4 \ 5 &amp; -2 &amp; -4 \end{vmatrix}$

Using cofactor expansion along the first row:

$det(A) = 3 \begin{vmatrix} 3 &amp; 4 \ -2 &amp; -4 \end{vmatrix} - (-1) \begin{vmatrix} -2 &amp; 4 \ 5 &amp; -4 \end{vmatrix} + 0 \begin{vmatrix} -2 &amp; 3 \ 5 &amp; -2 \end{vmatrix}$

$= 3((3)(-4) - (4)(-2)) + 1((-2)(-4) - (4)(5)) + 0$
$= 3(-12 + 8) + (8 - 20) = 3(-4) + (-12) = -12 - 12 = -24$

Example 2:

Calculate the determinant of:

$\begin{vmatrix} 1 &amp; -2 &amp; -1 \ 5 &amp; -1 &amp; 0 \ -3 &amp; -1 &amp; 0 \end{vmatrix}$

Using cofactor expansion along the first row:

$det(A) = 1 \begin{vmatrix} -1 &amp; 0 \ -1 &amp; 0 \end{vmatrix} - (-2) \begin{vmatrix} 5 &amp; 0 \ -3 &amp; 0 \end{vmatrix} + (-1) \begin{vmatrix} 5 &amp; -1 \ -3 &amp; -1 \end{vmatrix}$

$= 1((-1)(0) - (0)(-1)) + 2((5)(0) - (0)(-3)) - 1((5)(-1) - (-1)(-3))$
$= 1(0) + 2(0) - 1(-5 - 3) = 0 + 0 - 1(-8) = 8$

N by N Matrix

Here's the determinant of an n by n matrix:

$det(A) = a<em>{11}C</em>{11} - a<em>{12}C</em>{12} + a<em>{13}C</em>{13} …$

Where C is a matrix, formed by crossing out a row and column from the original matrix to get a (n-one) by (n-one) matrix.