Numbers Between Two Vertical Bars
Here are two arrangements of numbers. Neither of them sits inside square brackets. Each one is closed in by a pair of straight vertical lines.
\[ |6| \]
\[ \begin{vmatrix} 2 & 4 \\ 5 & 9 \end{vmatrix} \]
The first arrangement holds only one number. It has one row and one column, so it is the number arrangement of a square matrix of order \( 1 \times 1 \). The second has two rows and two columns, so it is the number arrangement of a square matrix of order \( 2 \times 2 \).
The two arrangements have two things in common. Both are square, with the same number of rows as columns. Both are written between two vertical lines. Placing a square arrangement of numbers between two vertical lines is the sign of a determinant.
| What we write | How it is enclosed | What it is |
| \( A = \begin{bmatrix} 2 & 4 \\ 5 & 9 \end{bmatrix} \) | Square brackets | A matrix: an arrangement of numbers |
| \( |A| = \begin{vmatrix} 2 & 4 \\ 5 & 9 \end{vmatrix} \) | Vertical bars | A determinant: it has one value |
What a Determinant Is
Determinant: numbers arranged in rows and columns in a square shape and written between two vertical lines form a determinant. Only a square matrix has a determinant. A determinant is a scalar quantity. If A is a square matrix, its determinant is written \( |A| \).
A scalar is a single number, the same as the plain numbers you use in arithmetic. A matrix is a whole table of numbers, but its determinant comes down to one value. That is the most useful thing about a determinant. It lets us describe a matrix with one number.
The determinant of a matrix A is written \( |A| \) and is sometimes written D. We read \( |A| \) as "determinant of A". Only a square matrix has one. A matrix of order \( 2 \times 3 \) has two rows and three columns, which is not a square arrangement, so it has no determinant at all.
A matrix and its determinant are different things. \( [C] \) with square brackets is a matrix, and \( |C| \) with vertical bars is a number. Use bars whenever you mean the determinant.
Determinant of Order One Matrix
A square matrix of order \( 1 \times 1 \) holds only one element. Its determinant is that element itself, with its sign unchanged.
\[ A = [8] \Rightarrow |A| = |8| = 8 \]
\[ B = [-7] \Rightarrow |B| = |-7| = -7 \]
Determinant of order one matrix: if a square matrix has one row and one column, its determinant is called the determinant of an order one matrix. Its value is equal to the element itself.
Example 1. Find \( |C| \) if \( C = [-12] \).
Step 1: The matrix given is:
\[ C = [-12] \]
Step 2: C is of order \( 1 \times 1 \), so its determinant is its only element:
\[ |C| = |-12| = -12 \]
So \( |C| = -12 \). The minus sign stays.
A determinant is not an absolute value, even though both use vertical bars. The absolute value of \( -7 \) is 7, because absolute value measures distance from zero and is never negative. The determinant of the matrix \( [-7] \) is \( -7 \). When the bars enclose the element of a matrix, keep its sign.
Determinant of Order Two Matrix
Take a general square matrix of order \( 2 \times 2 \), \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). Its determinant is written with bars in place of the brackets, and its value is:
\[ |A| = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc \]
Here a, b, c and d are called the elements of the determinant, and \( ad - bc \) is called the expansion of \( |A| \). The line from the top left element to the bottom right element, through a and d, is the main diagonal. The line from the top right to the bottom left, through b and c, is the secondary diagonal.
Determinant of order two matrix: if a square matrix has two rows and two columns, its determinant is called the determinant of an order two matrix. Its value is found by multiplying the elements of the main diagonal and of the secondary diagonal separately, then subtracting the second product from the first.
To find the value of any determinant of order two:
- Multiply the two elements on the main diagonal.
- Multiply the two elements on the secondary diagonal.
- Subtract the second product from the first.
Example 2. Find the determinant of \( B = \begin{bmatrix} 5 & 3 \\ 2 & 4 \end{bmatrix} \).
Step 1: Writing the determinant of B:
\[ |B| = \begin{vmatrix} 5 & 3 \\ 2 & 4 \end{vmatrix} \]
Step 2: Taking the main diagonal product minus the secondary diagonal product:
\[ |B| = 5 \times 4 - 3 \times 2 \]
Step 3: Working out the two products and subtracting:
\[ |B| = 20 - 6 = 14 \]
Negative elements need more care. Example 3. Find the determinant of \( E = \begin{bmatrix} 3 & -2 \\ 4 & -5 \end{bmatrix} \).
Step 1: Writing the determinant of E:
\[ |E| = \begin{vmatrix} 3 & -2 \\ 4 & -5 \end{vmatrix} \]
Step 2: Taking the main diagonal product minus the secondary diagonal product, with each negative number kept in brackets:
\[ |E| = 3 \times (-5) - (-2) \times 4 \]
Step 3: Working out each product:
\[ |E| = -15 - (-8) \]
Step 4: Subtracting a negative number is the same as adding it:
\[ |E| = -15 + 8 = -7 \]
The minus sign in \( ad - bc \) always stays in the formula, whatever the signs of the elements are. If the product bc is itself negative, minus a negative becomes plus. Put every negative element in brackets before you multiply, or the sign gets lost.
Singular Matrix
Some square matrices have a determinant equal to zero. Take \( A = \begin{bmatrix} 2 & 6 \\ 1 & 3 \end{bmatrix} \).
Step 1: Writing the determinant of A:
\[ |A| = \begin{vmatrix} 2 & 6 \\ 1 & 3 \end{vmatrix} \]
Step 2: Taking the main diagonal product minus the secondary diagonal product:
\[ |A| = 2 \times 3 - 6 \times 1 \]
Step 3: Both products are 6, so they cancel:
\[ |A| = 6 - 6 = 0 \]
Singular matrix: a square matrix whose determinant is zero is called a singular matrix. If A is a square matrix and \( |A| = 0 \), then A is a singular matrix.
Look closely at A. Its second column, 6 and 3, is exactly three times its first column, 2 and 1. So the two diagonal products are always equal and cancel. Whenever one row of a \( 2 \times 2 \) matrix is a multiple of the other row, or one column is a multiple of the other column, the matrix is singular.
Non-singular Matrix
Most matrices do not give zero. Take \( M = \begin{bmatrix} 4 & 3 \\ 5 & 6 \end{bmatrix} \).
Step 1: Taking the main diagonal product minus the secondary diagonal product:
\[ |M| = 4 \times 6 - 3 \times 5 \]
Step 2: Working out the products and subtracting:
\[ |M| = 24 - 15 = 9 \]
Since \( 9 \neq 0 \), the determinant of M is not zero.
Non-singular matrix: a square matrix whose determinant is not zero is called a non-singular matrix. It is also called a regular matrix. If A is a square matrix and \( |A| \neq 0 \), then A is a non-singular matrix.
| Singular matrix | Non-singular matrix | |
| Value of the determinant | \( |A| = 0 \) | \( |A| \neq 0 \) |
| Other name | None | Regular matrix |
| Example | \( \begin{bmatrix} 2 & 6 \\ 1 & 3 \end{bmatrix} \) | \( \begin{bmatrix} 4 & 3 \\ 5 & 6 \end{bmatrix} \) |
To decide whether a square matrix is singular or non-singular, find its determinant. Zero means singular. Any other value, positive or negative, means non-singular.
Worked Example: Trigonometric Elements
The elements of a matrix can be trigonometric ratios. The method does not change. Find \( |A| \) if \( A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \).
Step 1: Writing the determinant of A:
\[ |A| = \begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \]
Step 2: Taking the main diagonal product minus the secondary diagonal product:
\[ |A| = \cos\theta \times \cos\theta - (-\sin\theta) \times \sin\theta \]
Step 3: Minus a negative becomes plus:
\[ |A| = \cos^2\theta + \sin^2\theta \]
Step 4: Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[ |A| = 1 \]
The answer does not depend on \( \theta \). For every angle the determinant is 1, which is never zero, so this matrix is always non-singular.
Worked Example: Finding an Unknown Element
If a question tells you a matrix is singular, it has given you an equation: the determinant equals zero. Find the value of x if \( A = \begin{bmatrix} -1 & x \\ -3 & 6 \end{bmatrix} \) is a singular matrix.
Step 1: A is singular, so its determinant is zero:
\[ \begin{vmatrix} -1 & x \\ -3 & 6 \end{vmatrix} = 0 \]
Step 2: Expanding the determinant:
\[ (-1) \times 6 - x \times (-3) = 0 \]
Step 3: Working out the products:
\[ -6 + 3x = 0 \]
Step 4: Adding 6 to both sides:
\[ 3x = 6 \]
Step 5: Dividing both sides by 3:
\[ x = 2 \]
Check: with \( x = 2 \), the determinant is \( (-1)(6) - (2)(-3) = -6 + 6 = 0 \), so A really is singular.
Worked Example: Determinant of a Product
If \( P = \begin{bmatrix} 5 & -3 \\ 2 & -4 \end{bmatrix} \) and \( Q = \begin{bmatrix} 4 & 2 \\ -3 & 5 \end{bmatrix} \), verify that \( |PQ| = |P||Q| \). We work out each side separately and compare.
Step 1: Expanding the determinant of P:
\[ |P| = 5 \times (-4) - (-3) \times 2 \]
Step 2: Simplifying:
\[ |P| = -20 + 6 = -14 \]
Step 3: Expanding the determinant of Q:
\[ |Q| = 4 \times 5 - 2 \times (-3) \]
Step 4: Simplifying:
\[ |Q| = 20 + 6 = 26 \]
Step 5: Multiplying the two determinants gives the right side, which we call result (i):
\[ |P||Q| = -14 \times 26 = -364 \]
Step 6: For the left side, first write the product of the matrices:
\[ PQ = \begin{bmatrix} 5 & -3 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ -3 & 5 \end{bmatrix} \]
Step 7: Multiplying each row of P by each column of Q:
\[ PQ = \begin{bmatrix} 20 + 9 & 10 - 15 \\ 8 + 12 & 4 - 20 \end{bmatrix} \]
Step 8: Simplifying each element:
\[ PQ = \begin{bmatrix} 29 & -5 \\ 20 & -16 \end{bmatrix} \]
Step 9: Expanding the determinant of PQ:
\[ |PQ| = 29 \times (-16) - (-5) \times 20 \]
Step 10: Simplifying gives the left side, which we call result (ii):
\[ |PQ| = -464 + 100 = -364 \]
Results (i) and (ii) are both \( -364 \), so \( |PQ| = |P||Q| \) is verified.
The determinant of a product of two square matrices of the same order equals the product of their determinants: \( |PQ| = |P||Q| \). Multiplying the two determinants is much quicker than multiplying the matrices first.
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