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.

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

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

1. $\left| \begin{matrix} 1 & a & {{a}^{2}} \ 1 & b & {{b}^{2}} \ 1 & c & {{c}^{2}} \ \end{matrix} \right|=(a-b)(b-c)(c-a)$

2. (\left| \begin{matrix} a & b & c \ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \ bc & ca & ab \ \end{matrix} \right|=\left| \begin{matrix} 1 & 1 & 1 \ {{a}^{2}} & {{b}^{2}} & {{c}^{2}} \ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( ab+bc+ca \right))

3. $$\left| \begin{matrix} a & bc & abc \\ b & ca & abc \\ c & ab & abc \\ \end{matrix} \right|=\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=abc\left( a-b \right)\left( b-c \right)\left( c-a \right);$$

4. $$\left| \begin{matrix} 1 & 1 & 1 \ a & b & c \ {{a}^{3}} & {{b}^{3}} & {{c}^{3}} \ \end{matrix} \right|=\left( a-b \right)\left( b-c \right)\left( c-a \right)\left( a+b+c \right)$$

5. $\left| \begin{matrix} a & b & c \ b & c & a \ c & a & b \ \end{matrix} \right|=-a^3-b^3-c^3+3abc$

The determinant of order 3 × 3 is:

$$\begin{array}{l}=\left| \begin{matrix} {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \ {{a}_{3}} & {{b}_{3}} & {{c}_{3}} \ \end{matrix} \right|={{a}_{1}}\left| \begin{matrix} {{b}_{2}} & {{c}_{2}} \ {{b}_{3}} & {{c}_{3}} \ \end{matrix} \right|-{{b}_{1}}\left| \begin{matrix} {{a}_{2}} & {{c}_{2}} \ {{a}_{3}} & {{c}_{3}} \ \end{matrix} \right|+{{c}_{1}}\left| \begin{matrix} {{a}_{2}} & {{b}_{2}} \ {{b}_{3}} & {{b}_{3}} \ \end{matrix} \right|\end{array}$$

7. In the determinant D = (\begin{array}{l}\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \ \end{matrix} \right|,\end{array} ), the minor of ${a}_{12}$ is denoted as (\begin{array}{l}{{M}_{12}}=\left| \begin{matrix} {{a}_{21}} & {{a}_{23}} \ {{a}_{31}} & {{a}_{33}} \ \end{matrix} \right|\end{array} ) and so on.

Cofactor of an element: $\begin{array}{l}{C_{i,j}} = (-1)^{i+j}M_{i,j}\end{array}$

SARRUS Diagram for Evaluating the Determinant

If $\begin{bmatrix} a_{11} & a_{12} & a_{13} \ a_{21} & a_{22} & a_{23} \ a_{31} & a_{32} & a_{33} \end{bmatrix}$

A Sarrus Diagram of order 3 is obtained by adjoining the first two columns on the right and draw dark and dotted lines as shown in the below diagram, which is a matrix.

The value of the determinant is $$\left( {{a}_{11}}{{a}_{22}}{{a}_{33}}+{{a}_{12}}{{a}_{23}}{{a}_{31}}+{{a}_{13}}{{a}_{21}}{{a}_{32}} \right)-\left( {{a}_{13}}{{a}_{22}}{{a}_{31}}+{{a}_{11}}{{a}_{23}}{{a}_{32}}+{{a}_{12}}{{a}_{21}}{{a}_{33}} \right).$$

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Solved Problems on Determinant

Illustration 1: Evaluate the determinant $\left| \begin{matrix} \sqrt{p}+\sqrt{q} & 2\sqrt{r} & \sqrt{r} \ \sqrt{qr}+\sqrt{2p} & r & \sqrt{2r} \ q+\sqrt{pr} & \sqrt{qr} & r \ \end{matrix} \right|$

Where $p$, $q$, and $r$ are positive real numbers.

Given:

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Solution:

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Taking the common factor $\sqrt{r}$ 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 $$\Delta =r\left| \begin{matrix} \sqrt{p}+\sqrt{q} & 2 & 1 \ \sqrt{qr}+\sqrt{2p} & \sqrt{r} & \sqrt{2} \ q+\sqrt{pr} & \sqrt{q} & \sqrt{r} \ \end{matrix} \right|$$

(\begin{array}{l}\text{Subtracting }\sqrt{q}{{C}_{2}}\text{ and }\sqrt{p}{{C}_{3}}\text{ from }{{C}_{1}},\ \text{we get};\end{array} )

(\Delta=r\left| \begin{matrix} -\sqrt{q} & 2 & 1 \ 0 & \sqrt{r} & \sqrt{2} \ 0 & \sqrt{q} & \sqrt{r} \ \end{matrix} \right| = -r\sqrt{q}\left(r - \sqrt{2q}\right))

Illustration 2: Let $a, b, c$ be positive and not equal. Show that the value of the determinant $\left| \begin{matrix} a & b & c \ b & c & a \ c & a & b \ \end{matrix} \right|$ is negative.

Given:

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Solution:

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By utilizing the properties of invariance and scalar multiple to the given determinant, we can obtain the desired result.

(\begin{array}{l}D=\left| \begin{matrix} a+b+c & b & c \ a+b+c & c & a \ a+b+c & a & b \ \end{matrix} \right|;;;\left[ {{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}} \right]\end{array} )

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

(\begin{array}{l}=\left( a+b+c \right)\left| \begin{matrix} 1 & b & c \ 0 & 0 & a-2c \ 0 & a-b & b-c \ \end{matrix} \right|;; \left[ {{R}_{2}}\to {{R}_{2}}-{{R}_{1}},,,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \right]\end{array} )

(\begin{array}{l}=\left( a+b+c \right)\left[ bc+ca+ab-a^2-b^2-c^2 \right]\end{array})

(-\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab \right)=-\frac{1}{2}\left( a+b+c \right)\left( 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2bc-2ca-2ab \right))

(-\frac{1}{2}\left( a+b+c \right)\left[ {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2\left( ab+bc+ac \right) \right])

(\begin{array}{l}=-\frac{1}{2}\left( a+b+c \right)\left[ {{\left( a-b \right)}^{2}}+{{\left( b-c \right)}^{2}}+{{\left( c-a \right)}^{2}} \right] \\ \hspace{2cm} \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \

a + b + c > 0

(a - b)² + (b - c)² + (c - a)² > 0 ….(ii)

From (i) and (ii), Δ < 0.

Illustration 3: Show that (\Delta = \left| \begin{matrix} 1 & \cos^2(\alpha - \beta) & \cos^2(\alpha - \gamma) \ \cos^2(\beta - \alpha) & 1 & \cos^2(\beta - \gamma) \ \cos^2(\gamma - \alpha) & \cos^2(\gamma - \beta) & 1 \ \end{matrix}\right| )

(\begin{array}{l}2\sin^2(\beta - \gamma)\sin^2(\gamma - \alpha)\sin^2(\alpha - \beta)\end{array})

Given:

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Solution:

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By using the switching and invariance properties of (\begin{array}{l}\beta -\gamma =A,\gamma -\alpha =B,\alpha -\beta =C\end{array} ), we can prove the problem.

We can write $\Delta$ as, $$\Delta = \left| \begin{matrix} 1 & \cos^2 C & \cos^2 B \ \cos^2 C & 1 & \cos^2 A \ \cos^2 B & \cos^2 A & 1 \ \end{matrix} \right|$$

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

Using C2 - C1 → C2, and C1 - C1 → C3, we get;

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

(\begin{vmatrix} -1 & {{\sin }^{2}}C & {{\sin }^{2}}B \ {{\cos }^{2}}C & -{{\sin }^{2}}C & \sin B\sin \left( -A \right) \ {{\cos }^{2}}B & \sin C\sin \left( B-A \right) & -{{\sin }^{2}}B \ \end{vmatrix})

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

Where (\begin{array}{l}{{\Delta }_{1}}=\left| \begin{matrix} 1 & {{\sin }^{2}}C & \sin B \ 0 & -2{{\sin }^{2}}C & \sin (C-A)-\sin B \ 0 & \sin (B-A)-\sin B & -\sin B \ \end{matrix} \right|\end{array} ) Using ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}; and ;{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}$

Using R2 - R1 and R3 - R1, we get;

(\begin{vmatrix} 1 & \sin C & \sin B \ -{{\sin }^{2}}C & -2{{\sin }^{2}}C & \sin (C-A)-\sin B \ -{{\sin }^{2}}B & \sin (B-A)-\sin C & -2{{\sin }^{2}}B \ \end{vmatrix})

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, $$\Delta_1 = \sin C \sin B \Delta_2 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\Delta_2 = \begin{vmatrix} 1 & \sin C & \sin B \ \sin C & 2 & -2\cos A \ \sin B & -2\cos A & 2 \ \end{vmatrix}$$

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

$$\Delta_{2} = \begin{vmatrix} 1 & \sin C & \sin B \ 0 & 2-{{\sin }^{2}}C & -2\cos A-\sin B\sin C \ 0 & -2\cos A-\sin B\sin C & 2-{{\sin }^{2}}B \ \end{vmatrix} = \left( 2-{{\sin }^{2}}B \right)\left( 2-\sin C \right)-{{\left( 2\cos A+\sin B\sin C \right)}^{2}}$$

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

But A + B + C = 0 implies; (\begin{array}{l}{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\cos A\sin B\sin C\end{array} )

Hence, (\begin{array}{l}D=\sin C\sin B{{\Delta }_{1}}=\sin C\sin B\cdot 2{{\sin }^{2}}A\end{array} )

$\begin{array}{l}2\sin^2 A; \sin^2 B; \sin^2 C = 2\sin^2(\alpha - \beta); \sin^2(\beta - \gamma); \sin^2(\gamma - \alpha).\end{array}$

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

$$\Delta = \left| \begin{matrix} \sin x & \sin y & \sin z \ \cos x & \cos y & \cos z \ {{\cos }^{3}}x & {{\cos }^{3}}y & {{\cos }^{3}}z \ \end{matrix} \right|$$

Given:

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Solution:

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Taking the common terms $\cos x$, $\cos y$, and $\cos z$ from the first, second, and third column respectively using scalar multiplication and then using the invariance property, we can prove the given statement.

Here, $\Delta = \left| \begin{matrix} \cos x \cos y \cos z & \tan x & \tan y & \tan z \ 1 & 1 & 1 & 1 \ {{\cos }^{2}}x & {{\cos }^{2}}y & {{\cos }^{2}}z & 1 \ \end{matrix} \right|$

(\begin{array}{l}=\cos x\cos y\cos z\left| \begin{matrix} \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{matrix} \right|\left( {{C}_{3}}\to {{C}_{3}}-{{C}_{2}},{{C}_{2}}\to {{C}_{2}}-{{C}_{1}} \right)\end{array} )

(\Delta =-\cos x\cos y\cos z\left| \begin{matrix} \tan y-\tan x & \tan z-\tan y \ {{\cos }^{2}}y-{{\cos }^{2}}x & {{\cos }^{2}}z-{{\cos }^{2}}y \ \end{matrix} \right|) when expanded along R2

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

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

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

(\frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)\sin \left( y+2x \right) = \frac{1}{2}\sin \left( x-y \right)\sin \left( y-z \right)2\cos \left( x+y+z \right)\sin \left( z-x \right))

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

Clearly, $\Delta = 0$ when any two of $x$, $y$, and $z$ are equal or when $x + y + z = \frac{\pi}{2}$.

Therefore, it is proven.

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Frequently Asked Questions

When 2 rows or columns of a determinant are interchanged, the value of the determinant is multiplied by -1.

When 2 rows or columns are interchanged, the sign of the determinant is reversed.

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

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

What are Determinants Used For?

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

No, determinants are not always positive.

No, determinants can be positive, negative, or zero.