Determinants of Order Two and Three

Introduction to Determinants

A determinant is a scalar value that can be computed from the elements of a square matrix. It's a fundamental concept in linear algebra with numerous applications in solving systems of linear equations, finding eigenvalues, and calculating areas and volumes. For a square matrix, the determinant provides crucial information about its properties, such as invertibility.

We will focus on determinants of order two and three, which are the most commonly encountered in introductory linear algebra and in exams like JEE Main.

Determinant of a Matrix of Order Two

Consider a general square matrix of order two:

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The determinant of matrix A, denoted as $|A|$ or $\det(A)$, is calculated by subtracting the product of the elements on the minor diagonal from the product of the elements on the main diagonal.

The formula is:

$|A| = ad - bc$

Example:

Let $A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}$.

The determinant is:

$|A| = (4 \times 3) - (2 \times 1) = 12 - 2 = 10$.

Determinant of a Matrix of Order Three

Consider a general square matrix of order three:

$A = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$

The determinant of a 3x3 matrix can be calculated using several methods. One common method is the cofactor expansion, often visualized using Sarrus's rule for 3x3 matrices.

Method 1: Cofactor Expansion (along the first row)

The determinant is given by:

$|A| = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Now, we expand the 2x2 determinants:

$|A| = a(ei - fh) - b(di - fg) + c(dh - eg)$

Expanding further:

$|A| = aei - afh - bdi + bfg + cdh - ceg$

Method 2: Sarrus's Rule

To use Sarrus's rule, we rewrite the first two columns of the matrix to the right of the third column:

$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \begin{matrix} a & b \\ d & e \\ g & h \end{matrix}$

Then, we sum the products of the diagonals going from top-left to bottom-right and subtract the sum of the products of the diagonals going from top-right to bottom-left.

Products of the main diagonals (top-left to bottom-right):

$(a \times e \times i) + (b \times f \times g) + (c \times d \times h)$

Products of the anti-diagonals (top-right to bottom-left):

$(c \times e \times g) + (a \times f \times h) + (b \times d \times i)$

The determinant is the difference between these two sums:

$|A| = (aei + bfg + cdh) - (ceg + afh + bdi)$

This formula yields the same result as the cofactor expansion.

Example:

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 2 & 0 & 5 \end{bmatrix}$.

Using Sarrus's Rule:

Rewrite: $\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 2 & 0 & 5 \end{vmatrix} \begin{matrix} 1 & 2 \\ 0 & 1 \\ 2 & 0 \end{matrix}$

Sum of main diagonal products:

$(1 \times 1 \times 5) + (2 \times 4 \times 2) + (3 \times 0 \times 0) = 5 + 16 + 0 = 21$

Sum of anti-diagonal products:

$(3 \times 1 \times 2) + (1 \times 4 \times 0) + (2 \times 0 \times 5) = 6 + 0 + 0 = 6$

Determinant: $|A| = 21 - 6 = 15$.

Properties of Determinants

Understanding the properties of determinants can greatly simplify calculations and is essential for solving many problems.

Property 1: Interchange of Rows or Columns

If two rows or two columns of a determinant are interchanged, the determinant is multiplied by -1.

Example:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $|A| = ad - bc$.

Let $B$ be the matrix obtained by interchanging the rows of $A$: $B = \begin{bmatrix} c & d \\ a & b \end{bmatrix}$.

$|B| = cb - da = -(ad - bc) = -|A|$.

Property 2: Identical Rows or Columns

If any two rows or any two columns of a determinant are identical, the value of the determinant is zero.

Example:

Let $A = \begin{bmatrix} a & b \\ a & b \end{bmatrix}$.

$|A| = ab - ba = 0$.

Similarly, if two columns are identical, the determinant is zero.

Property 3: Multiplication of a Row or Column by a Scalar

If each element of any row or any column of a determinant is multiplied by a scalar $k$, then the determinant is multiplied by $k$.

Example:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $|A| = ad - bc$.

Let $B$ be the matrix obtained by multiplying the first row by $k$: $B = \begin{bmatrix} ka & kb \\ c & d \end{bmatrix}$.

$|B| = (ka)d - (kb)c = k(ad - bc) = k|A|$.

This implies that if a row (or column) has a common factor, it can be taken out of the determinant.

Property 4: Adding a Multiple of One Row/Column to Another

The value of a determinant remains unchanged if to any one row (or to any one column), is added $k$ times the corresponding elements of any other row (or column).

This property is crucial for simplifying determinants using row/column operations, similar to Gaussian elimination.

Notation: $R_i \rightarrow R_i + kR_j$ or $C_i \rightarrow C_i + kC_j$.

Example:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.

If we perform $R_1 \rightarrow R_1 + kR_2$, the new matrix is $\begin{bmatrix} a+kc & b+kd \\ c & d \end{bmatrix}$.

The determinant is $(a+kc)d - (b+kd)c = ad + kcd - bc - kcd = ad - bc = |A|$.

Property 5: Determinant of a Product of Matrices

For two square matrices A and B of the same order, $|AB| = |A||B|$.

Property 6: Determinant of the Transpose

The determinant of the transpose of a matrix is equal to the determinant of the original matrix: $|A^T| = |A|$.

This means properties involving rows also hold true for columns, and vice versa.

Property 7: Determinant of Scalar Multiple of a Matrix

If A is an $n \times n$ matrix and $k$ is a scalar, then $|kA| = k^n |A|$.

Note the power $n$, which is the order of the matrix. For a 2x2 matrix, $|kA| = k^2|A|$. For a 3x3 matrix, $|kA| = k^3|A|$.

Minors and Cofactors

The concepts of minors and cofactors are fundamental to understanding the cofactor expansion method for calculating determinants and are also used in finding the adjoint and inverse of a matrix.

Minor (M_ij)

The minor of an element $a_{ij}$ (the element in the $i$-th row and $j$-th column) of a square matrix is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column.

For a 3x3 matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$:

The minor of $a_{11}$ is $M_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} = a_{22}a_{33} - a_{23}a_{32}$.

The minor of $a_{12}$ is $M_{12} = \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} = a_{21}a_{33} - a_{23}a_{31}$.

The minor of $a_{23}$ is $M_{23} = \begin{vmatrix} a_{11} & a_{12} \\ a_{31} & a_{32} \end{vmatrix} = a_{11}a_{32} - a_{12}a_{31}$.

Cofactor (C_ij)

The cofactor of an element $a_{ij}$ is defined as $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of $a_{ij}$.

The term $(-1)^{i+j}$ determines the sign of the cofactor. It follows a checkerboard pattern:

$\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$

So, $C_{ij} = M_{ij}$ if $i+j$ is even, and $C_{ij} = -M_{ij}$ if $i+j$ is odd.

Example:

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 2 & 0 & 5 \end{bmatrix}$.

Let's find the cofactor of $a_{23}$ (which is 4).

First, find the minor $M_{23}$: Delete the 2nd row and 3rd column.

$M_{23} = \begin{vmatrix} 1 & 2 \\ 2 & 0 \end{vmatrix} = (1 \times 0) - (2 \times 2) = 0 - 4 = -4$.

Now, find the cofactor $C_{23}$: $i=2, j=3$. So $i+j = 5$ (odd).

$C_{23} = (-1)^{2+3} M_{23} = (-1)^5 (-4) = (-1)(-4) = 4$.

Determinant using Cofactors

The determinant of a matrix can be calculated by summing the product of elements of any row (or column) with their corresponding cofactors.

Expansion along the $i$-th row: $|A| = a_{i1}C_{i1} + a_{i2}C_{i2} + \dots + a_{in}C_{in}$.

Expansion along the $j$-th column: $|A| = a_{1j}C_{1j} + a_{2j}C_{2j} + \dots + a_{nj}C_{nj}$.

Example:

Using the same matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 2 & 0 & 5 \end{bmatrix}$.

Let's expand along the first column (j=1):

$a_{11}=1, a_{21}=0, a_{31}=2$.

We need $C_{11}, C_{21}, C_{31}$.

$M_{11} = \begin{vmatrix} 1 & 4 \\ 0 & 5 \end{vmatrix} = 5 - 0 = 5$. $C_{11} = (-1)^{1+1}M_{11} = 1 \times 5 = 5$.

$M_{21} = \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = 10 - 0 = 10$. $C_{21} = (-1)^{2+1}M_{21} = -1 \times 10 = -10$.

$M_{31} = \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = 8 - 3 = 5$. $C_{31} = (-1)^{3+1}M_{31} = 1 \times 5 = 5$.

$|A| = a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31} = (1)(5) + (0)(-10) + (2)(5) = 5 + 0 + 10 = 15$.

This matches the result obtained using Sarrus's rule.

Applications of Determinants

Determinants have wide-ranging applications, especially in solving systems of linear equations.

Cramer's Rule

Cramer's Rule provides a method for solving a system of linear equations with a unique solution using determinants. Consider a system of $n$ linear equations in $n$ variables:

$a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n = b_1$
$a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n = b_2$
...
$a_{n1}x_1 + a_{n2}x_2 + \dots + a_{nn}x_n = b_n$

This system can be written in matrix form as $AX = B$, where $A$ is the coefficient matrix, $X$ is the column vector of variables, and $B$ is the column vector of constants.

If the determinant of the coefficient matrix, $|A|$, is non-zero ($|A| \neq 0$), then the system has a unique solution. Cramer's Rule states that the solution is given by:

$x_i = \frac{|A_i|}{|A|}$ for $i = 1, 2, \dots, n$.

Here, $|A_i|$ is the determinant of the matrix formed by replacing the $i$-th column of $A$ with the column vector $B$.

Example (2x2 system):

Consider the system:

$2x + 3y = 7$
$x - y = 1$

The coefficient matrix is $A = \begin{bmatrix} 2 & 3 \\ 1 & -1 \end{bmatrix}$.

The constant vector is $B = \begin{bmatrix} 7 \\ 1 \end{bmatrix}$.

Calculate $|A|$: $|A| = (2)(-1) - (3)(1) = -2 - 3 = -5$. Since $|A| \neq 0$, a unique solution exists.

To find $x$, replace the first column of $A$ with $B$ to get $A_1 = \begin{bmatrix} 7 & 3 \\ 1 & -1 \end{bmatrix}$.

$|A_1| = (7)(-1) - (3)(1) = -7 - 3 = -10$.

$x = \frac{|A_1|}{|A|} = \frac{-10}{-5} = 2$.

To find $y$, replace the second column of $A$ with $B$ to get $A_2 = \begin{bmatrix} 2 & 7 \\ 1 & 1 \end{bmatrix}$.

$|A_2| = (2)(1) - (7)(1) = 2 - 7 = -5$.

$y = \frac{|A_2|}{|A|} = \frac{-5}{-5} = 1$.

The solution is $x=2, y=1$.

Geometric Interpretation

For a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the absolute value of its determinant, $| |A| |$, represents the area of the parallelogram formed by the row vectors (or column vectors) of the matrix when viewed as vectors in a 2D plane originating from the origin.

For a 3x3 matrix, the absolute value of its determinant represents the volume of the parallelepiped formed by the three row vectors (or column vectors) in 3D space.

Area of a Triangle

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be calculated using determinants:

Area $= \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

The absolute value is taken because area must be positive. If the determinant is zero, it means the three points are collinear.

Determinants of Special Matrices

Determinants of certain types of matrices have specific, often simpler, calculation methods.

Diagonal Matrix

A diagonal matrix has non-zero elements only on the main diagonal. All off-diagonal elements are zero.

Example: $D = \begin{bmatrix} d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix}$

The determinant of a diagonal matrix is the product of its diagonal elements: $|D| = d_1 d_2 d_3$.

Triangular Matrix (Upper or Lower)

A triangular matrix has all elements above (upper triangular) or below (lower triangular) the main diagonal equal to zero.

Example (Upper Triangular): $U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}$

The determinant of a triangular matrix (both upper and lower) is also the product of its diagonal elements: $|U| = u_{11} u_{22} u_{33}$.

Skew-Symmetric Matrix

A skew-symmetric matrix $A$ satisfies $A^T = -A$. For an $n \times n$ skew-symmetric matrix:

If $n$ is odd, then $|A| = 0$.

If $n$ is even, then $|A|$ is the square of a polynomial in the matrix entries.

Example (3x3 skew-symmetric): $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$

$|A| = 0(0 - (-c^2)) - a(0 - (-bc)) + b(ac - 0) = -a(bc) + b(ac) = -abc + abc = 0$.

This confirms that for odd order $n=3$, the determinant is 0.

Practice Problems Strategy

When faced with a determinant problem in an exam:

1. Examine the Matrix: Look for patterns, zeros, or identical rows/columns.
2. Apply Properties: Use properties to simplify the determinant. Try to create zeros in a row or column using elementary row/column operations ($R_i \rightarrow R_i + kR_j$). Aim for a row/column with at least two zeros.
3. Choose Expansion Method: If simplification leads to a 2x2 or 3x3 determinant, calculate it directly. If simplification results in a matrix with zeros, use cofactor expansion along the row/column with the most zeros.
4. Check for Special Cases: Is it a diagonal, triangular, symmetric, or skew-symmetric matrix? This can drastically simplify the calculation.
5. Verify using Properties: If the problem involves variables, check if the determinant is zero or non-zero based on conditions (e.g., for unique solutions in Cramer's rule, or for linear dependence).