Standard Determinants - Expressions for Standard Determinants with Examples

A determinant is defined as a quantity which is obtained by adding the products of all elements in a square matrix. To find the determinant, a particular rule is followed. In this lesson, the concept of determinants is explained in detail, along with solved examples, formulas, determinant types, and practice questions.

There are certain standard determinants whose results are given by direct formulas. The standard results of a few types of determinants are given below, which will help students solve questions more efficiently.

All Topics in Determinants

Introduction to Determinants
Minors and Cofactors
Properties of Determinants
System of Linear Equations Using Determinants
Differentiation and Integration of Determinants
Standard Determinants

Expressions for Standard Determinants

$\begin{array}{l} | \begin{array}{c} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} | = (a - b) (b - c) (c - a) \end{array}$
$\begin{array}{l} | \begin{array}{c} a & b & c \\ a^{2} & b^{2} & c^{2} \\ b c & c a & a b \end{array} | = | \begin{array}{c} 1 & 1 & 1 \\ a^{2} & b^{2} & c^{2} \\ a^{3} & b^{3} & c^{3} \end{array} | = (a - b) (b - c) (c - a) (a b + b c + c a) \end{array}$
$\begin{array}{l} | \begin{array}{c} a & b c & a b c \\ b & c a & a b c \\ c & a b & a b c \end{array} | = | \begin{array}{c} a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3} \end{array} | = a b c (a - b) (b - c) (c - a); \end{array}$
$\begin{array}{l} | \begin{array}{c} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array} | = (a - b) (b - c) (c - a) (a + b + c) \end{array}$
$\begin{array}{l} | \begin{array}{c} a & b & c \\ b & c & a \\ c & a & b \end{array} | = - a^{3} - b^{3} - c^{3} + 3 a b c \end{array}$
Determinant of order 3 × 3
$\begin{array}{l} = | \begin{array}{c} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} | = a_{1} | \begin{array}{c} b_{2} & c_{2} \\ b_{3} & c_{3} \end{array} | - b_{1} | \begin{array}{c} a_{2} & c_{2} \\ a_{3} & c_{3} \end{array} | + c_{1} | \begin{array}{c} a_{2} & b_{2} \\ b_{3} & b_{3} \end{array} | \end{array}$
In the determinant D =
$\begin{array}{l} | \begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} |, \end{array}$
minor of a₁₂ is denoted as
$\begin{array}{l} M_{12} = | \begin{array}{c} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} | \end{array}$
and so on.
Cofactor of an element
$\begin{array}{l} a_{i j} = C_{i j} = {(- 1)}^{i + j} M_{i j} \end{array}$

Evaluation of the Determinant Using the SARRUS Diagram

\begin{array}{l} A = [\begin{array}{c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}] \end{array}

is a square matrix of order 3, then the below diagram is a Sarrus Diagram, obtained by adjoining the first two columns on the right and drawing dark and dotted lines as shown.

The value of the determinant is

\begin{array}{l} (a_{11} a_{22} a_{33} + a_{12} a_{23} a_{31} + a_{13} a_{21} a_{32}) - (a_{13} a_{22} a_{31} + a_{11} a_{23} a_{32} + a_{12} a_{21} a_{33}) . \end{array}

Solving Determinants

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Solved Problems on Determinant

Illustration 1: Evaluate the determinant

\begin{array}{l} Δ = | \begin{array}{c} \sqrt{p} + \sqrt{q} & 2 \sqrt{r} & \sqrt{r} \\ \sqrt{q r} + \sqrt{2 p} & r & \sqrt{2 r} \\ q + \sqrt{p r} & \sqrt{q r} & r \end{array} |, \end{array}

where p, q and r are positive real numbers.

Solution:

Taking √r common from C₂ and C₃ of the given determinant using the scalar multiple property and then expanding it using the invariance property, we can evaluate the given problem.

We get

\begin{array}{l} Δ = r | \begin{array}{c} \sqrt{p} + \sqrt{q} & 2 & 1 \\ \sqrt{q r} + \sqrt{2 p} & \sqrt{r} & \sqrt{2} \\ q + \sqrt{p r} & \sqrt{q} & \sqrt{r} \end{array} | \end{array}

\begin{array}{l} Applying C_{1} \to C_{1} - \sqrt{q} C_{2} - \sqrt{p} C_{3}, we get; \end{array}

\begin{array}{l} Δ = r | \begin{array}{c} - \sqrt{q} & 2 & 1 \\ 0 & \sqrt{r} & \sqrt{2} \\ 0 & \sqrt{q} & \sqrt{r} \end{array} | = - r \sqrt{q} (r - \sqrt{2} q) . \end{array}

Illustration 2: Let a, b, and c be positive and not equal. Show that the value of the determinant

\begin{array}{l} | \begin{array}{c} a & b & c \\ b & c & a \\ c & a & b \end{array} | is negative. \end{array}

Solution:

By applying invariance and scalar multiple properties to the given determinant, we can get the required result.

\begin{array}{l} D = | \begin{array}{c} a & b & c \\ b & c & a \\ c & a & b \end{array} |; t h e n D = | \begin{array}{c} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{array} | [C_{1} \to C_{1} + C_{2} + C_{3}] \end{array}

\begin{array}{l} = (a + b + c) | \begin{array}{c} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{array} | \end{array}

[Taking (a+b+c) common from the first column]

\begin{array}{l} = (a + b + c) | \begin{array}{c} 1 & b & c \\ 0 & c - b & a - c \\ 0 & a - b & b - c \end{array} | [R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1}] \end{array}

\begin{array}{l} = (a + b + c +) [(c - b) (b - c) (a - b) (a - c)] \\ = (a + b + c +) [b c + c a + a b - a^{2} - b^{2} - c^{2}] \end{array}

\begin{array}{l} = - (a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b) \\ = - \frac{1}{2} (a + b + c) (2 a^{2} + 2 b^{2} + 2 c^{2} - 2 b c - 2 c a - 2 a b) \end{array}

\begin{array}{l} = - \frac{1}{2} (a + b + c) [(a^{2} + b^{2} - 2 a b) + (b^{2} + c^{2} - 2 b c) + (c^{2} + a^{2} - 2 a c)] \end{array}

\begin{array}{l} = - \frac{1}{2} (a + b + c) [{(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}] \dots . (i) \end{array}

a, b, c are positive ⇒ a + b + c > 0

a, b, c are unequal ⇒ (a – b)² + (b – c)² + (c – a)² > 0 ….(ii)

From (i) and (ii), Δ <0.

Illustration 3: Show that

\begin{array}{l} Δ = | \begin{array}{c} 1 & \cos^{2} (α - β) & \cos^{2} (α - γ) \\ \cos^{2} (β - α) & 1 & \cos^{2} (β - γ) \\ \cos^{2} (γ - α) & \cos^{2} (γ - β) & 1 \end{array} | \end{array}

\begin{array}{l} = 2 \sin^{2} (β - γ) \sin^{2} (γ - α) \sin^{2} (α - β) \end{array}

Solution:

By putting

\begin{array}{l} β - γ = A, γ - α = B, α - β = C \end{array}

and then by using switching and invariance properties, we can prove the above problem.

We can write Δ as,

\begin{array}{l} Δ = | \begin{array}{c} 1 & \cos^{2} C & \cos^{2} B \\ \cos^{2} C & 1 & \cos^{2} A \\ \cos^{2} B & \cos^{2} A & 1 \end{array} | \end{array}

(Note that A + B + C = 0).

Using C₂ → C₂ – C₁, C₁ → C₃ – C₁, we get

\begin{array}{l} Δ = | \begin{array}{c} 1 & - \sin^{2} C & - \sin^{2} B \\ \cos^{2} C & \sin^{2} C & \cos^{2} A - \cos^{2} C \\ \cos^{2} B & \cos^{2} A - \cos^{2} B & \sin^{2} B \end{array} | | \begin{array}{c} 1 & - \sin^{2} C & - \sin^{2} B \\ \cos^{2} C & \sin^{2} C & \sin B \sin (C - A) \\ \cos^{2} B & \sin C \sin (B - A) & \sin^{2} B \end{array} | \end{array}

\begin{array}{l} = {(- 1)}^{2} | \begin{array}{c} 1 & \sin^{2} C & \sin^{2} B \\ \cos^{2} C & - \sin^{2} C & \sin B \sin (- A) \\ \cos^{2} B & \sin C \sin (B - A) & - \sin^{2} B \end{array} | \end{array}

Since,

\begin{array}{l} [\cos^{2} A - \cos^{2} B = \sin (A + B) \sin (B - A), A + B = - C, C + A = - B]; = \sin C \sin B [Δ_{1}] \end{array}

Where

\begin{array}{l} Δ_{1} = | \begin{array}{c} 1 & \sin^{2} C & \sin B \\ \cos^{2} C & - \sin^{2} C & \sin (C - A) \\ \cos^{2} B & \sin (B - A) & - \sin B \end{array} | U s i n g R_{2} \to R_{2} - R_{1} a n d R_{3} \to R_{3} - R_{1} \end{array}

Using R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we get

\begin{array}{l} Δ_{1} = | \begin{array}{c} 1 & \sin C & \sin B \\ - \sin^{2} C & - 2 \sin^{2} C & \sin (C - A) - s i n B \\ - \sin^{2} B & \sin (B - A) - s i n C & - 2 \sin^{2} B \end{array} | \end{array}

But sin (C – A) – sin B = sin (C – A) + sin (C + A) = 2 sin C cos A and sin (B – A) – sin C = 2 sin B cos A

Therefore,

\begin{array}{l} Δ_{1} = \sin C \sin B Δ_{2} w h e r e Δ_{2} = | \begin{array}{c} 1 & \sin C & \sin B \\ \sin C & 2 & - 2 \cos A \\ \sin B & - 2 \cos A & 2 \end{array} | \end{array}

Applying R₂ → R₂ – sin C R₁ and R₃ → R₃ – sin B R₁, we get

\begin{array}{l} Δ_{2} = | \begin{array}{c} 1 & \sin C & \sin B \\ 0 & 2 - \sin^{2} C & - 2 \cos A - \sin B \sin C \\ 0 & - 2 \cos A - \sin B \sin C & 2 - \sin^{2} B \end{array} | \\ = (2 - \sin^{2} B) (2 - \sin C) - {(2 \cos A + \sin B \sin C)}^{2} \end{array}

\begin{array}{l} = 4 - 2 \sin^{2} B - 2 \sin^{2} C + \sin^{2} B \sin^{2} C - [4 \cos^{2} A + 4 \cos A \sin B \sin C + \sin^{2} B \sin^{2} C] \end{array}

\begin{array}{l} = 4 \sin^{2} A - 2 \sin^{2} B - 2 \sin^{2} C - 4 \cos A \sin B \sin C \end{array}

\begin{array}{l} = 2 \sin^{2} A - 2 [\sin^{2} B + \sin^{2} C - \sin^{2} A + 2 \cos A \sin B \sin C] \end{array}

But A + B + C = 0 implies;

\begin{array}{l} \sin^{2} B + \sin^{2} C - \sin^{2} A = - 2 \cos A \sin B \sin C \end{array}

\begin{array}{l} Δ_{2} = 2 \sin^{2} A; \end{array}

Hence,

\begin{array}{l} D = \sin C \sin B Δ_{1} = \sin^{2} C \sin^{2} B Δ_{2} \end{array}

\begin{array}{l} = 2 \sin^{2} A \sin^{2} B \sin^{2} C = 2 \sin^{2} (α - β) \sin^{2} (β - γ) \sin^{2} (γ - α) . \end{array}

Illustration 4: Prove that the following determinant vanishes if any two of x, y, z are equal

\begin{array}{l} Δ = | \begin{array}{c} \sin x & \sin y & \sin z \\ \cos x & \cos y & \cos z \\ \cos^{3} x & \cos^{3} y & \cos^{3} z \end{array} | \end{array}

Solution:

Taking cos x, cos y, and cos z common from the first, second and third columns using scalar multiple and then using the invariance property, we can prove the given statement.

Here,

\begin{array}{l} Δ = \cos x \cos y \cos z | \begin{array}{c} \tan x & \tan y & \tan z \\ 1 & 1 & 1 \\ \cos^{2} x & \cos^{2} y & \cos^{2} z \end{array} | \end{array}

\begin{array}{l} = \cos x \cos y \cos z | \begin{array}{c} \tan x & \tan y - \tan x & \tan z - \tan y \\ 1 & 0 & 0 \\ \cos^{2} x & \cos^{2} y - \cos^{2} x & \cos^{2} z - \cos^{2} y \end{array} | (C_{3} \to C_{3} - C_{2}, C_{2} \to C_{2} - C_{1}) \end{array}

Expanding along R₂,

\begin{array}{l} Δ = - \cos x \cos y \cos z | \begin{array}{c} \tan y - \tan x & \tan z - \tan y \\ \cos^{2} y - \cos^{2} x & \cos^{2} z - \cos^{2} y \end{array} | \end{array}

\begin{array}{l} = - \cos x \cos y \cos z | \begin{array}{c} \frac{\sin (y - x)}{\cos x \cos y} & \frac{\sin (z - y)}{\cos y . \cos z} \\ \sin^{2} x - \sin^{2} y & \sin^{2} y - \sin^{2} z \end{array} | \\ = | \begin{array}{c} \cos z . \sin (x - y) & \cos x . \sin (y - z) \\ \sin (x + y) . \sin (x - y) & \sin (y + z) . \sin (y - z) \end{array} | \dots . (i) \end{array}

\begin{array}{l} = \sin (x - y) \sin (y - z) | \begin{array}{c} \cos z & \cos x \\ \sin (x + y) & \sin (y + z) \end{array} | \\ = \sin (x - y) \sin (y - z) [\sin (y + z) \cos z - \sin (x + y) \cos x)] \end{array}

\begin{array}{l} = \frac{1}{2} \sin (x - y) \sin (y - z) [{\sin (y + 2 z) + \sin y} - {\sin (y + 2 x) + \sin y}] \end{array}

\begin{array}{l} = \frac{1}{2} \sin (x - y) \sin (y - z) [\sin (y + 2 x)] \\ = \frac{1}{2} \sin (x - y) \sin (y - z) 2 \cos (x + y + z) \sin (z - x) \end{array}

\begin{array}{l} = \sin (x - y) \sin (y - z) \sin (z - x) \cos (x + y + z) \end{array}

Clearly, Δ is zero when any two of x, y, z are equal or

\begin{array}{l} x + y + z = \frac{π}{2} . \end{array}

Hence proved.

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Frequently Asked Questions

When 2 rows or columns are interchanged, what happens to a determinant?

When 2 rows or columns are interchanged, the determinant changes its sign.

What is the value of the determinant, if all the elements of a row or column are zero?

If all the elements of a row or column are zero, then the determinant is equal to zero.

What are determinants used for?

Determinants are used to give formulas for the area or volume of certain geometric figures and also to find the inverse of a matrix.

Are determinants always positive?

No, determinants can be positive, negative or zero.

Standard Determinant

Expressions for Standard Determinants

Evaluation of the Determinant Using the SARRUS Diagram

Solved Problems on Determinant

Video Lessons

Matrices and Determinants – Important Topics

Important Questions for JEE Matrices and Determinants

Matrices and Determinants – Top 10 Most Important and Expected JEE Main Questions

Frequently Asked Questions

When 2 rows or columns are interchanged, what happens to a determinant?

What is the value of the determinant, if all the elements of a row or column are zero?

What are determinants used for?

Are determinants always positive?

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