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CBSE Class 12 Maths Revision Notes Chapter 4 Determinants

Determinants assign a unique scalar value to every square matrix and help study equations, areas and matrix inverses. For CBSE Class 12 Maths, this chapter covers determinants up to order three, properties, minors, cofactors, adjoint, inverse and linear equations.

Determinants are connected with square matrices. A determinant helps decide whether a system of linear equations has a unique solution. It is also used to find the area of a triangle, adjoint of a matrix and inverse of a matrix.

Use these CBSE Class 12 Maths Revision Notes Chapter 4 for the 2026–27 academic year to revise formulas, properties and applications. These Class 12 Mathematics Chapter 4 notes follow the NCERT flow and keep each concept ready for quick revision.

Key Takeaways

Square matrix: Only square matrices have determinants.
Order covered: This chapter studies determinants up to order three.
Area formula: Three points are collinear when the determinant-based area is zero.
Inverse condition: A square matrix is invertible only when its determinant is non-zero.

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Access Class 12 Maths Chapter 4 Determinants Notes in 30 Minutes

Chapter 4 becomes easier when you revise the formulas first. Then move to properties, minors, cofactors and applications.

Revision Area	What to Revise
Determinant Meaning	Scalar value linked with a square matrix
Order 1 and Order 2	Direct determinant formulas
Order 3	Expansion along row or column
Properties	Row and column operations
Area of Triangle	Determinant formula and collinearity
Minors and Cofactors	Mij and Aij
Adjoint	Transpose of cofactor matrix
Inverse	A⁻¹ = 1/
Linear Equations	Matrix method and consistency

Determinants Class 12 Notes: Meaning and Basic Concept

A determinant is a number associated with a square matrix. It is written as |A|, det(A) or Δ.

If A is a square matrix, then |A| is read as “determinant of A.” It is not read as modulus of A.

What Is a Determinant?

For every square matrix A = [aij] of order n, a unique number is associated with A. This number is called the determinant of A.

Important rule:
Only square matrices have determinants.

Matrix Type	Determinant Exists?
1 × 1 matrix	Yes
2 × 2 matrix	Yes
3 × 3 matrix	Yes
Rectangular matrix	No

Determinant of Order 1

If A = [a], then:

|A| = a

Example:
If A = [5], then |A| = 5.

Determinant of Order 2

For a 2 × 2 matrix:

A = [a b; c d]

The determinant is:

|A| = ad - bc

Example:
A = [4 2; 5 3]

|A| = 4 × 3 - 2 × 5
= 12 - 10
= 2

Determinant of Order 3

For a 3 × 3 matrix:

A = [a1 b1 c1; a2 b2 c2; a3 b3 c3]

The determinant can be expanded along the first row as:

|A| = a1(b2c3 - b3c2) - b1(a2c3 - a3c2) + c1(a2b3 - a3b2)

This is called expansion of a determinant along a row.

Expansion of Determinants in Class 12 Mathematics Chapter 4 Notes

A determinant of order 3 can be expanded along any row or any column. The final value remains the same.

Expansion Along a Row or Column

For a 3 × 3 determinant, choose the row or column with the maximum number of zeroes. This makes calculation shorter.

If:

A = [a11 a12 a13; a21 a22 a23; a31 a32 a33]

Expansion along the first row is:

|A| = a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)

Sign Pattern for Cofactor Expansion

The signs used in expansion follow this pattern:

Position	Sign
a11	+
a12	-
a13	+
a21	-
a22	+
a23	-
a31	+
a32	-
a33	+

Quick pattern:

[+ - +; - + -; + - +]

Use this pattern while finding cofactors and expanding determinants.

Properties of Determinants for Quick Revision

Properties of determinants help simplify large calculations. They are useful when a determinant has repeated rows, proportional rows or complicated expressions.

Row and Column Interchange Properties

Property	Meaning
det(A) = det(A′)	Interchanging rows and columns keeps the determinant same
Ri ↔ Rj	Interchanging two rows changes the sign
Ci ↔ Cj	Interchanging two columns changes the sign

If two rows or two columns are interchanged, the determinant becomes negative of the original determinant.

Zero, Equal and Proportional Row Properties

Situation	Determinant Value
All elements of a row are zero	0
All elements of a column are zero	0
Two rows are identical	0
Two columns are identical	0
Two rows are proportional	0
Two columns are proportional	0

Example:
If R1 = R2, then |A| = 0.

Multiplication and Addition Properties

Property	Rule
Row multiplied by k	Determinant is multiplied by k
Column multiplied by k	Determinant is multiplied by k
Every element of matrix A multiplied by k	For order n, determinant becomes kⁿ
Ri → Ri + kRj	Determinant remains unchanged
Ci → Ci + kCj	Determinant remains unchanged

If A = kB and A, B are square matrices of order n, then:

|A| = kⁿ|B|

Sum Property of Determinants

If each element of a row or column is expressed as the sum of two terms, the determinant can be written as the sum of two determinants.

This property is often used to split complex determinants into simpler ones.

Triangular Determinant Property

For an upper triangular or lower triangular determinant, the value is the product of diagonal elements.

Matrix Type	Determinant
Upper triangular matrix	Product of diagonal elements
Lower triangular matrix	Product of diagonal elements
Diagonal matrix	Product of diagonal elements

Example:
For a triangular matrix with diagonal elements 2, 3 and 5:

|A| = 2 × 3 × 5 = 30

Area of a Triangle Using Determinants

Determinants can be used to find the area of a triangle when the coordinates of its vertices are given.

Formula for Area of a Triangle

Let the vertices of a triangle be:

(x1, y1), (x2, y2), (x3, y3)

Then area of the triangle is:

Area = 1/2 | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

In determinant form:

Area = 1/2 | [x1 y1 1; x2 y2 1; x3 y3 1] |

The modulus is used because area is always positive.

Important Points for Area Questions

Situation	Rule
Area is calculated	Take positive value
Area is already given	Use both positive and negative determinant values
Points are collinear	Area = 0
Determinant value is 0	Triangle is not formed

Collinearity Condition

Three points are collinear if the area of the triangle formed by them is zero.

So:

1/2 | [x1 y1 1; x2 y2 1; x3 y3 1] | = 0

This means:

[x1 y1 1; x2 y2 1; x3 y3 1] = 0

Minors and Cofactors in Determinants

Minors and cofactors help expand determinants in a compact way. They are also used to find the adjoint and inverse of a matrix.

Minor of an Element

The minor of an element aij is the determinant obtained by deleting the ith row and jth column in which aij lies.

It is denoted by Mij.

For a determinant of order n, the minor of an element is a determinant of order n - 1.

Example:
If a23 lies in row 2 and column 3, delete row 2 and column 3 to find M23.

Cofactor of an Element

The cofactor of an element aij is denoted by Aij.

Formula:

Aij = (-1)i+j Mij

Here, Mij is the minor of aij.

i + j	Cofactor Sign
Even	Positive
Odd	Negative

Expansion Using Cofactors

A determinant can be expanded as the sum of products of elements of any row or column with their corresponding cofactors.

Expansion along first row:

Δ = a11A11 + a12A12 + a13A13

Expansion along first column:

Δ = a11A11 + a21A21 + a31A31

Important Cofactor Rule

If elements of a row are multiplied with cofactors of another row, the sum is zero.

Example:

a11A21 + a12A22 + a13A23 = 0

This happens because the determinant then has two identical rows.

Adjoint and Inverse of a Matrix

Adjoint and inverse are important applications of determinants. These ideas connect Chapter 4 with matrices from Chapter 3.

Adjoint of a Matrix

The adjoint of a square matrix A is the transpose of its cofactor matrix.

It is denoted by adj A.

Steps to find adj A:

Find the cofactor of each element of A.
Form the cofactor matrix.
Take the transpose of the cofactor matrix.

Adjoint of a 2 × 2 Matrix

If:

A = [a b; c d]

Then:

adj A = [d -b; -c a]

This shortcut is useful for 2 × 2 inverse questions.

Singular and Non-Singular Matrices

Matrix Type	Condition	Meaning
Singular Matrix		A
Non-Singular Matrix		A

A square matrix is invertible if and only if it is non-singular.

Theorem on Adjoint

For any square matrix A of order n:

A(adj A) = (adj A)A = |A|I

Here, I is the identity matrix of the same order.

Inverse Formula Using Determinants

If A is a non-singular square matrix, then:

A⁻¹ = 1/|A| adj A

This formula is valid only when |A| ≠ 0.

If |A| = 0, A⁻¹ does not exist.

Applications of Determinants in Linear Equations

Determinants are used to solve systems of linear equations. They also help check whether a system is consistent or inconsistent.

Matrix Form AX = B

A system of linear equations can be written in matrix form as:

AX = B

Here:

Symbol	Meaning
A	Coefficient matrix
X	Variable matrix
B	Constant matrix

For two equations:

a1x + b1y = c1
a2x + b2y = c2

Matrix form:

[a1 b1; a2 b2] [x; y] = [c1; c2]

Case 1: Unique Solution

If A is non-singular, then |A| ≠ 0.

So, A⁻¹ exists.

AX = B
A⁻¹AX = A⁻¹B
IX = A⁻¹B
X = A⁻¹B

This gives a unique solution.

This is called the matrix method.

Case 2: No Solution

If A is singular, then |A| = 0.

Now calculate (adj A)B.

If (adj A)B is a non-zero matrix, the system is inconsistent.

This means no solution exists.

Case 3: Infinite Solutions

If A is singular and (adj A)B is a zero matrix, the system may have infinitely many solutions.

In this case, the equations are dependent or require further checking.

Solution Cases Summary

Condition	Result
	A
	A
	A

Quick Revision Tables for Determinants

These tables cover the main formulas in Determinants Class 12 Notes.

Determinant Formula Table

Concept	Formula
Order 1 determinant
Order 2 determinant
Order 3 expansion	a11(a22a33 - a23a32) - a12(a21a33 - a23a31) + a13(a21a32 - a22a31)
Area of triangle	1/2
Cofactor	Aij = (-1)i+j Mij
Inverse	A⁻¹ = 1/

Properties of Determinants Table

Property	Result
Rows changed into columns	Determinant remains same
Two rows interchanged	Sign changes
Two columns interchanged	Sign changes
Two rows identical	Determinant is zero
Two columns identical	Determinant is zero
One row multiplied by k	Determinant multiplied by k
Ri → Ri + kRj	Determinant unchanged
Triangular determinant	Product of diagonal elements

Matrix Inverse Table

Term	Meaning
adj A	Transpose of cofactor matrix
	A
Singular matrix	Matrix with determinant zero
Non-singular matrix	Matrix with non-zero determinant
Invertible matrix	Matrix whose inverse exists

Important Terms in Determinants

Term	Meaning
Determinant	Scalar value associated with a square matrix
Square Matrix	Matrix with equal number of rows and columns
Minor	Determinant after deleting a row and column
Cofactor	Signed minor of an element
Adjoint	Transpose of the cofactor matrix
Singular Matrix	Square matrix with determinant zero
Non-Singular Matrix	Square matrix with non-zero determinant
Matrix Method	Method of solving AX = B using inverse
Consistent System	System with at least one solution
Inconsistent System	System with no solution

Common Mistakes in Class 12 Mathematics Chapter 4 Notes

Mistake	Correct Point
Finding determinant of a rectangular matrix	Only square matrices have determinants
Reading	A
Forgetting sign change after row interchange	Interchanging two rows changes sign
Ignoring modulus in area	Area must be positive
Using inverse when	A
Confusing minor and cofactor	Cofactor = signed minor
Expanding along a difficult row	Choose row or column with more zeroes

Useful Links for Class 12 Maths

Section	Useful Links
Syllabus	CBSE Class 12 Maths Syllabus
Revision Notes	CBSE Class 12 Maths Revision Notes
Maths Notes	CBSE Class 12 Maths Revision Notes Chapter 1
NCERT Solutions	NCERT Solutions Class 12 Maths
Sample Papers	CBSE Sample Papers for Class 12 Maths
Important Questions	Important Questions Class 12 Maths
NCERT Books	NCERT Books for Class 12 Maths
Previous Year Papers	CBSE Maths Question Paper Class 12

Q.1 Find the area of a triangle whose vertices are given by (x₁, y₁), (x₂, y₂) and (x₃, y₃).

Ans

$Area of Δ = \frac{1}{2} |\begin{array}{l} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}|$

Q.2

$Evaluate |\begin{array}{l} 1 2 4 \\ - 1 3 0 \\ 4 1 0 \end{array}|$

Ans

$\begin{array}{l} |\begin{array}{l} 1 2 4 \\ - 1 3 0 \\ 4 1 0 \end{array}| = 1 |\begin{array}{l} 30 \\ 10 \end{array}| - 2 |\begin{array}{l} - 1 0 \\ 4 0 \end{array}| + 4 |\begin{array}{l} - 1 3 \\ 4 1 \end{array}| \\ = 1 (0) - 2 (0) + 4 (- 1 - 12) \\ = 4 (- 13) = - 52 \end{array}$

Q.3

$\begin{array}{l} Show that |\begin{array}{l} a b c \\ a + 2 x b +2 y c +2 z \\ x y z \end{array}| = 0 by \\ using properties of determinants . \end{array}$

Ans

$\begin{array}{l} |\begin{array}{l} a b c \\ a + 2 x b +2 y c +2 z \\ x y z \end{array}| = |\begin{array}{l} a b c \\ a b c \\ x y z \end{array}| + |\begin{array}{l} a b c \\ 2 x 2 y 2 z \\ x y z \end{array}| \\ (\begin{array}{l} All elements of R_{2} are expressed as sum of \\ two terms . Therefore, the determinant can \\ be expressed as sum of two determinant . \end{array}) \\ = 0 + 2 |\begin{array}{l} a b c \\ a b c \\ x y z \end{array}| (2 is taken out common from R_{2} .) \\ (\begin{array}{l} If any two rows or columns of a determinant are \\ identical then the value of determinant is zero . \end{array}) \\ = 0 + 0 = 0 \end{array}$

Q.4 If P, Q, R are three non-null square matrices of the same order, write the condition on P such that PQ = PR ⇒ Q = R.

Ans

P must be invertible or |P| ≠ 0.

Q.5

$If A = |\begin{array}{l} 12 \\ 42 \end{array}|, then show that |2 A| = 4 |A|$

Ans

$\begin{array}{l} A = |\begin{array}{l} 12 \\ 42 \end{array}| \\ \Rightarrow 2 A = A = |\begin{array}{l} 24 \\ 84 \end{array}| \\ ∴ |A| = |\begin{array}{l} 12 \\ 42 \end{array}| = 2 - 8 = - 6 \\ |2 A| = |\begin{array}{l} 24 \\ 84 \end{array}| = 8 - 32 = - 24 \\ \Rightarrow |2 A| = -24 = 4 \times - 6 = 4 |A| \end{array}$

Q.6

$\begin{array}{l} Examine the consistency of the system of equations : \\ 5 x - y +4 z = 5 \\ 2 x +3 y +5 z = 2 \\ 5 x -2 y +6 z = -1 \end{array}$

Ans

$\begin{array}{l} The system of equations can be written as AX = B \\ where A = |\begin{array}{l} 5 -1 4 \\ 2 3 5 \\ 5 -2 6 \end{array}|, X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 5 \\ 2 \\ - 1 \end{array}] \\ Here, |A| = 5 (18 + 10) + (12 - 25) + 4 (- 4 - 15) \\ = 5 (28) + (- 13) + 4 (- 19) \\ = 140 - 13 - 76 \\ = 51 \neq 0 \\ Since |A| \neq 0, the system of equations is consistent . \end{array}$

Q.7

$\begin{array}{l} If A = |\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}| and B |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| then find AB . \\ Using AB, solve the folowing system of equations : \\ x - y +2 z =1 \\ 2 y - 3 z =1 \\ 3 x - 2 y +4 z = 2 \end{array}$

Ans

$\begin{array}{l} Here, AB = |\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}| |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| \\ = |\begin{array}{l} - 2 - 9 +1 2 0 - 2 + 2 1+3 - 4 \\ 0 +18 - 18 0+4 - 3 0 - 6 + 6 \\ - 6 - 18 + 24 0 - 4 +4 3 +6 - 8 \end{array}| \\ = |\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}| \\ Then AB = I \\ Thus A^{- 1} = B \\ {|\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}|}^{- 1} = |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| \\ Now given equations can be written as A_{1} X = B_{1} \\ where, A_{1} = |\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}| X = [\begin{array}{l} x \\ y \\ z \end{array}] and B_{1} = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}] \\ Now A_{1} = A \\ ∴ {A_{1}}^{- 1} = A^{- 1} \\ \Rightarrow {A_{1}}^{- 1} = |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| \\ as X = {A_{1}}^{- 1} B_{1} \\ ∴ X = |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}] \\ X = |\begin{array}{l} - 2 + 0 + 1 \\ 9 +2 -6 \\ 6 +1 -4 \end{array}| [\begin{array}{l} 0 \\ 5 \\ 3 \end{array}] \\ Hence, x = 0, y = 5 and z = 3 \end{array}$

Q.8 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it we get 11. By adding first and third numbers we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Ans

$\begin{array}{l} Let the first, second and third numbers be x, y and z . \\ ∴ x + y + z = 6 \\ y + 3 z = 11 \\ x + z = 2 y or x -2 y + z = 0 \\ This system of equations can be written as AX = B, where \\ A = [\begin{array}{l} 1 1 1 \\ 0 1 3 \\ 1 -2 1 \end{array}], X = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 6 \\ 11 \\ 0 \end{array}] \\ Here, A is non singular matrix and so its inverse exists . \\ Now we find adjA \\ c_{11} = 7 c_{12} = 3 c_{13} = - 1 \\ c_{21} = - 3 c_{22} = 0 c_{23} = 3 \\ c_{31} = 2 c_{32} = - 3 c_{33} = 1 \\ Hence, adj A = A = [\begin{array}{l} 7 -3 2 \\ 3 0 -3 \\ - 1 3 1 \end{array}] \\ Thus A^{- 1} = \frac{adjA}{|A|} = \frac{1}{9} [\begin{array}{l} 7 -3 2 \\ 3 0 -3 \\ - 1 3 1 \end{array}] \\ Since X = A^{- 1} B \\ Therefore X = \frac{1}{9} [\begin{array}{l} 7 -3 2 \\ 3 0 -3 \\ - 1 3 1 \end{array}] [\begin{array}{l} 6 \\ 11 \\ 0 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{9} [\begin{array}{l} 42 -33 0 \\ 18 +0 -0 \\ - 6 + 3 +0 \end{array}] = \frac{1}{9} [\begin{array}{l} 9 \\ 18 \\ 3 \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] \\ Thus x =1, y = 2 and z = 3 \end{array}$

Q.9

$\begin{array}{l} Solve the following system of equations by matrix method . \\ |\begin{array}{l} 3 x -2 y +3 z = 8 \\ 2 x + y - z = 1 \\ 4 x -3 y +2 z =4 \end{array}| \end{array}$

Ans

$\begin{array}{l} The system of equations can be written as AX = B, where \\ A = [\begin{array}{l} 3 -2 3 \\ 2 1 -1 \\ 4 -3 2 \end{array}] x = [\begin{array}{l} x \\ y \\ z \end{array}] and B = [\begin{array}{l} 8 \\ 1 \\ 4 \end{array}] \\ Here, |A| = 3 (2 - 3) + 2 (4 + 4) + 3 (- 6 - 4) \\ = - 17 \neq 0 \\ Hence, A is nonsingular matrix and so its inverse exists . \\ Now cofactors of A are \\ c_{11} = - 1 c_{12} = - 8 c_{13} = - 10 \\ c_{12} = - 1 c_{22} = - 8 c_{23} = - 1 \\ c_{31} = - 1 c_{32} = - 8 c_{33} = - 7 \\ adjoint of A = C^{T} = [\begin{array}{l} - 1 - 5 - 1 \\ - 8 - 6 9 \\ - 10 1 7 \end{array}] \\ ∴ A^{- 1} = \frac{adjA}{|A|} = \frac{- 1}{7} = [\begin{array}{l} - 1 - 5 - 1 \\ - 8 - 6 9 \\ - 10 1 7 \end{array}] \\ So X = A^{- 1} B \frac{- 1}{17} = [\begin{array}{l} - 1 - 5 - 1 \\ - 8 - 6 9 \\ - 10 1 7 \end{array}] [\begin{array}{l} 8 \\ 1 \\ 4 \end{array}] \\ [\begin{array}{l} x \\ y \\ z \end{array}] = \frac{- 1}{17} [\begin{array}{l} - 17 \\ - 34 \\ - 51 \end{array}] = [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] \\ Hence, x = 1, y = 2 and z = 3. \end{array}$

Q.10

$\begin{array}{l} If a, b, care in A . P then without expanding show that \\ |\begin{array}{l} x +2 x +3 x +2 a \\ x +3 x +4 x +2 b \\ x +4 x +5 x +2 c \end{array}| = 0 \end{array}$

Ans

$\begin{array}{l} Here, Δ = |\begin{array}{l} x +2 x +3 x +2 a \\ x +3 x +4 x +2 b \\ x +4 x +5 x +2 c \end{array}| \\ = |\begin{array}{l} x +2 x +3 x +2 a \\ x +3 x +4 x +2 b \\ 1 1 2 (c - b) \end{array}| (by applying R_{3} \to R_{3} - R_{2}) \\ = |\begin{array}{l} x +2 x +3 x +2 a \\ 1 1 2 (b - a) \\ 1 1 2 (c - b) \end{array}| (by applying R_{2} \to R_{2} - R_{1}) \\ Now because a, b, c are in A . P ., we have \\ b - a = c - b \\ Hence in Δ R_{2} = R_{3} \\ If any two rows or columns of a determinant are identical \\ then the value of determinant is zero . \\ \Rightarrow Δ = 0 \end{array}$

Q.11

$\begin{array}{l} Without expanding, prove that \\ |\begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ a + b p + q x + y \end{array}| = 2 |\begin{array}{l} a q x \\ b q y \\ c r z \end{array}| \end{array}$

Ans

$\begin{array}{l} Here, LHS = |\begin{array}{l} b + c q + r y + z \\ c + a r + p z + x \\ a + b p + q x + y \end{array}| \\ By applying R_{1} \to R_{1} - (R_{2} + R_{3}), we get \\ LHS = |\begin{array}{l} - 2 a - 2 p - 2 x \\ c + a r + p z + x \\ a + b p + q x + y \end{array}| \\ = 2 |\begin{array}{l} - a - p - x \\ c + a r + p z + x \\ a + b p + q x + y \end{array}| [Taking 2 common from R_{1}] \\ = 2 |\begin{array}{l} - a - p - x \\ c r z \\ b q y \end{array}| [By applying R_{2} \to R_{2} + R_{1} and R_{3} \to R_{3} + R_{1}] \\ = 2 |\begin{array}{l} a p x \\ c r z \\ b q y \end{array}| [Taking - 1 common from R_{1}] \\ = 2 |\begin{array}{l} a p x \\ b q y \\ c r z \end{array}| [By applying R_{2} \leftrightarrow R_{3}] \\ = RHS \end{array}$

Q.12

$\begin{array}{l} Using properties of determinants, show that \\ |\begin{array}{l} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{array}| = abc (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = abc + bc + ca + ab . \end{array}$

Ans

$\begin{array}{l} Let Δ = |\begin{array}{l} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{array}| \\ By applying c_{1} \to c_{2} and c_{3} \to c_{3} - c_{2}, we get \\ |\begin{array}{l} a 1 0 \\ - b 1 + b - b \\ 0 1 c \end{array}| \\ Expanding along R_{1}, we obtain \\ Δ = a |\begin{array}{l} 1 + b - b \\ 1 c \end{array}| - 1 |\begin{array}{l} - b - b \\ 0 c \end{array}| + 0 \\ = a (c + bc + b) - (- bc) \\ = ac + abc + ab + bc \\ = abc + bc + ca + ab \\ = abc [a + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}] \end{array}$

Q.13

$\begin{array}{l} If x, y, z are different and Let Δ = |\begin{array}{l} x x^{2} 1 + x^{3} \\ y y^{2} 1 + y^{3} \\ z z^{2} 1 + z^{3} \end{array}| = 0, \\ then show that 1+ xyz is a factor of Δ . \end{array}$

Ans

$\begin{array}{l} Let Δ = |\begin{array}{l} x x^{2} 1 + x^{3} \\ y y^{2} 1 + y^{3} \\ z z^{2} 1 + z^{3} \end{array}| = 0, \\ All elements of c_{3} are expressed as sum of two terms . \\ Therefore, the determinant can be expressed as sum of \\ two determinants . \\ = |\begin{array}{l} x x^{2} 1 \\ y y^{2} 1 \\ z z^{2} 1 \end{array}| + |\begin{array}{l} x x^{2} x^{3} \\ y y^{2} y^{3} \\ z z^{2} z^{3} \end{array}| \\ Using c_{3} \leftrightarrow c_{2} and then c_{1} \leftrightarrow c_{2} in first determinant and \\ taking out x, y and z common from R_{1}, R_{2} and R_{3} respectively \\ in second determinant, we get \\ = {(- 1)}^{2} |\begin{array}{l} 1 x x^{2} \\ 1 y y^{2} \\ 1 z z^{2} \end{array}| + xyz |\begin{array}{l} 1 x x^{2} \\ 1 y y^{2} \\ 1 z z^{2} \end{array}| \\ Δ = |\begin{array}{l} 1 x x^{2} \\ 1 y y^{2} \\ 1 z z^{2} \end{array}| (1 + xyz) \\ Hence, (1 + xyz) is factor of Δ . \end{array}$

Q.14

$\begin{array}{l} Using properties of determinants, prove that \\ |\begin{array}{l} a a + b a + b + c \\ 2 a 3 a + 2 b 4 a + 3 b + 2 c \\ 3 a 6 a + 3 b 10 a + 6 b + 3 c \end{array}| = a^{3} \end{array}$

Ans

$\begin{array}{l} By applying R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - 3 R_{1} \\ Δ = |\begin{array}{l} a a + b a + b + c \\ 0 a 2 a + b \\ 0 3 a 7 a + 3 b \end{array}| \\ Now applying R_{3} \to R_{3} - R_{2}, we get \\ Δ = |\begin{array}{l} a a + b a + b + c \\ 0 a 2 a + b \\ 0 0 a \end{array}| \\ Expanding along c_{1} we obtain \\ Δ = a |\begin{array}{l} a 2 a + b \\ 0 a \end{array}| + 0 + 0 \\ = a (a^{2} - 0) = a (a^{2}) = a^{3} \end{array}$

Q.15

$If A = [\begin{array}{l} 25 \\ 36 \end{array}], verify that A (adjA) = |A| I .$

Ans

$\begin{array}{l} We have |A| = 4 - 6 = - 2 \\ |A| I = -2 [\begin{array}{l} 1 0 \\ 0 1 \end{array}] \\ = [\begin{array}{l} - 2 0 \\ 0 - 2 \end{array}] . \dots.. (1) \\ Cofactors of A \\ c_{11} = 4 c_{12} = - 2 \\ c_{21} = - 3 c_{22} = 1 \\ \Rightarrow adjA = c^{T} \\ = [\begin{array}{l} 4 - 2 \\ - 3 1 \end{array}] \\ Now A (adjA) = [\begin{array}{l} 1 2 \\ 3 4 \end{array}] [\begin{array}{l} 4 - 2 \\ - 3 1 \end{array}] \\ = [\begin{array}{l} 1 \times 4 + 2 \times (- 3) 1 \times (- 2) + 2 \times 1 \\ 3 \times 4 + 4 \times (- 3) 3 \times (- 2) + 4 \times 1 \end{array}] \\ = [\begin{array}{l} 4 - 6 - 2 + 2 \\ 12 - 12 - 6 + 4 \end{array}] \\ = [\begin{array}{l} - 2 2 \\ 0 - 2 \end{array}] . \dots (2) \\ by (1) and (2) \\ A (adjA) = |A| I \end{array}$

Q.16

Ans

$\begin{array}{l} The area of triangle is given by \\ Δ = \frac{1}{2} |\begin{array}{l} 3 8 1 \\ - 4 2 1 \\ 5 1 1 \end{array}| \\ = \frac{1}{2} [3 (2 - 1) - 8 (- 4 - 5) + 1 (- 4 - 10)] \\ = \frac{1}{2} (3 + 72 - 14) \\ = \frac{61}{2} Sq . units \end{array}$

Q.17

$Without expanding, evaluate Δ = |\begin{array}{l} 1 a bc \\ 2 b ca \\ 5 c ab \end{array}| .$

Ans

$\begin{array}{l} Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, We get \\ Δ = |\begin{array}{l} 1 a bc \\ 0 b - c c (a - b) \\ 0 c - a b (a - c) \end{array}| \\ Taking factors (b - a) and (c - a) common from R_{2} and R_{3}, \\ respectively we get \\ Δ = (b - a) (c - a) |\begin{array}{l} 1 a bc \\ 0 1 - c \\ 0 1 - b \end{array}| \\ = (b - a) (c - a) (- b + c) \\ = (a - b) (b - c) (c - a) \end{array}$

Q.18

$If A = |\begin{array}{l} 1 1 - 2 \\ 2 1 - 3 \\ 5 4 - 9 \end{array}|, find |A| .$

Ans

$\begin{array}{l} Expanding with R_{1} \\ |A| = 1 |\begin{array}{l} 1 - 3 \\ 4 - 9 \end{array}| - 1 |\begin{array}{l} 2 - 3 \\ 5 - 9 \end{array}| - 2 |\begin{array}{l} 2 1 \\ 5 4 \end{array}| \\ = (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5) \\ = 3 - (- 3) - 2 (3) \\ = 3 + 3 - 6 \\ = 0 \end{array}$

Q.19

$Find values of x for which |\begin{array}{l} 3 x \\ x 1 \end{array}| = |\begin{array}{l} 3 2 \\ 4 1 \end{array}| .$

Ans

$\begin{array}{l} We have, |\begin{array}{l} 3 x \\ x 1 \end{array}| = |\begin{array}{l} 3 2 \\ 4 1 \end{array}| . \\ \Rightarrow 3 - x^{2} = 3 - 8 \\ \Rightarrow x^{2} = 8 \\ \Rightarrow x = \pm 2 \sqrt{2} \end{array}$

Q.20

$Find values of x for which |\begin{array}{l} 1 x \\ x 1 \end{array}| = |\begin{array}{l} 4 1 \\ 4 1 \end{array}|$

Ans

$\begin{array}{l} We have, |\begin{array}{l} 1 x \\ x 1 \end{array}| = |\begin{array}{l} 4 1 \\ 4 1 \end{array}| \\ \Rightarrow 1 - x^{2} = 4 - 4 \\ \Rightarrow 1 - x^{2} = 0 \\ \Rightarrow x^{2} = 1 \\ \Rightarrow x = \pm 1 \end{array}$

Q.21

$Evaluate: |\begin{array}{l} x x + 1 \\ x -1 x \end{array}| .$

Ans

$|\begin{array}{l} x x + 1 \\ x -1 x \end{array}| = x (x) - (x - 1) (x - 1) = x^{2} - x^{2} + 1 = 1$

Q.22 The value of determinant will _______ if its rows and columns are interchanged.

Ans

remain unchanged

Q.23 The value of determinant will be _____, if its two rows or two columns are interchanged with each other.

Ans

negative

Q.24 If A is an invertivle square matrix, then write the matrix adj (A^T) – (adj A)^T.

Ans

If A is an invertible square matrix, then adj (A^T) = (adj A)^T

∴ adj (A^T) – (adj A)^T is a null matrix.

Q.25

$Without expanding, prove that |\begin{array}{l} x y z \\ y + z z + x x + y \\ 1 1 1 \end{array}| = 0$

Ans

$\begin{array}{l} By applying R_{2} \to R_{2} + R_{1}, we get \\ |\begin{array}{l} x y z \\ y + z z + x x + y \\ 1 1 1 \end{array}| = |\begin{array}{l} x y z \\ x + y x + y x + y + z \\ 1 1 1 \end{array}| \\ = |\begin{array}{l} x y z \\ 1 1 1 \\ 1 1 1 \end{array}| (Taking x + y + z common from R_{2} .) \\ = (x + y + z) \times 0 \\ = 0 \end{array}$

Q.26 Let A be a 3 × 3 square matrix such that A(adj A) = 2I, where I is the identity matrix. Write the value of |adj A|.

Ans

∵ A (adj (A)) = |A|I

∴ A (adj A) = 2I ⇒ |A| = 2

Now, |adj A| = |A|^n-1 ⇒ |adj A| = 2^3-1 = 4

Q.27 If A is a square matrix of order 3 such that adj (2A) = kadj (A), then write the value of k.

Ans

Since, adj (kA) = k^{n – 1}adj(A), where n is the order
∴ adj (2A) = 2^{3 – 1}adj A
⇒ adj (2A) = 4 adj A
⇒ k = 4

Q.28

$\begin{array}{l} Without expanding evaluate the determinant \\ |\begin{array}{l} 41 1 5 \\ 79 7 9 \\ 29 5 3 \end{array}| \end{array}$

Ans

$\begin{array}{l} Let Δ = |\begin{array}{l} 41 1 5 \\ 79 7 9 \\ 29 5 3 \end{array}| . \\ Then, applying c_{1} \to c_{2} + (- 8) c_{3}, we get \\ Δ = |\begin{array}{l} 1 1 5 \\ 7 7 9 \\ 5 5 3 \end{array}| = 0 [∵ c_{1} and c_{2} are identical] \end{array}$

Q.29

$If A = [\begin{array}{l} 1 2 \\ 3 4 \end{array}], verify A (adjA) = |A| I .$

Ans

$\begin{array}{l} We have |A| I = - 2 [\begin{array}{l} 1 0 \\ 0 1 \end{array}] \\ = [\begin{array}{l} - 2 0 \\ 0 - 2 \end{array}] . \dots.. (1) \\ Cofactors of A \\ c_{11} = 4 c_{12} = - 3 \\ c_{21} = - 2 c_{22} = 1 \\ \Rightarrow adj A = C^{T} \\ = [\begin{array}{l} 4 - 2 \\ - 3 1 \end{array}] \\ Now A (adjA) = [\begin{array}{l} 1 2 \\ 3 4 \end{array}] [\begin{array}{l} 4 - 2 \\ - 3 1 \end{array}] \\ = [\begin{array}{l} 1 \times 4 + 2 \times (- 3) 1 \times (- 2) + 2 \times 1 \\ 3 \times 4 + 4 \times (- 3) 3 \times (- 2) + 4 \times 1 \end{array}] \\ = [\begin{array}{l} 4 - 6 - 2 + 2 \\ 12 - 12 - 6 + 4 \end{array}] \\ = [\begin{array}{l} - 2 0 \\ 0 - 2 \end{array}] . \dots\dots (2) \\ by (1) and (2) \\ A (adjA) = |A| I \end{array}$

Q.30

$\begin{array}{l} Using properties of determinants, prove that \\ |\begin{array}{l} b^{2 +} c^{2} ab ac \\ ba c^{2} + a^{2} bc \\ ca cb a^{2} + b^{2} \end{array}| = 4 a^{2} b^{2} c^{2} . \end{array}$

Ans

$\begin{array}{l} Let Δ |\begin{array}{l} b^{2 +} c^{2} ab ac \\ ba c^{2} + a^{2} bc \\ ca cb a^{2} + b^{2} \end{array}| \\ Multiplying R_{1}, R_{2} and R_{3} by a, b and c respectively, we get \\ Δ = \frac{1}{abc} |\begin{array}{l} a (b^{2 +} c^{2}) a^{2} b a^{2} c \\ b^{2} a b (c^{2} + a^{2}) b^{2} c \\ c^{2} a c^{2} b c (a^{2} + b^{2}) \end{array}| \\ \Rightarrow Δ = \frac{abc}{abc} |\begin{array}{l} b^{2 +} c^{2} a^{2} a^{2} \\ b^{2} c^{2} + a^{2} b^{2} \\ c^{2} c^{2} a^{2} + b^{2} \end{array}| [\begin{array}{l} Taking a, b anc c common from \\ c_{1}, c_{2} and c_{3} respectively . \end{array}] \\ \Rightarrow Δ = |\begin{array}{l} 2 (b^{2 +} c^{2}) 2 (a^{2} + c^{2}) 2 (a^{2} + b^{2}) \\ b^{2} c^{2} + a^{2} b^{2} \\ c^{2} c^{2} a^{2} + b^{2} \end{array}| [Applying R_{1}, R_{2} and R_{3}] \\ \Rightarrow Δ = 2 |\begin{array}{l} (b^{2 +} c^{2}) (a^{2} + c^{2}) (a^{2} + b^{2}) \\ b^{2} c^{2} + a^{2} b^{2} \\ c^{2} c^{2} a^{2} + b^{2} \end{array}| [Taking 2 common from R_{1}] \\ \Rightarrow Δ = 2 |\begin{array}{l} (b^{2 +} c^{2}) (a^{2} + c^{2}) (a^{2} + b^{2}) \\ - c^{2} 0 - a^{2} \\ - b^{2} - a^{2} 0 \end{array}| [Applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}] \\ \Rightarrow Δ = 2 |\begin{array}{l} 0 c^{2} b^{2} \\ - c^{2} 0 - a^{2} \\ - b^{2} - a^{2} 0 \end{array}| [Applying R_{1} \to R_{1} + R_{2} + R_{3}] \\ \Rightarrow Δ = 2 \{- c^{2} |\begin{array}{l} - c^{2} - a^{2} \\ - b^{2} 0 \end{array}| + b^{2} |\begin{array}{l} - c^{2} 0 \\ - b^{2} - a^{2} \end{array}|\} \\ \Rightarrow Δ = 2 \{a^{2} b^{2} c^{2} + a^{2} b^{2} c^{2}\} = 4 a^{2} b^{2} c^{2} \end{array}$

Q.31

$\begin{array}{l} If f (x) = {ax}^{2} + bx + c isaquadratic function such that f (1) = 8, \\ f (2) = 11 and f (- 3) = 6, find f (x) by using determinants . \\ Also, find f (0) . \end{array}$

Ans

$\begin{array}{l} We have f (x) = {ax}^{2} + bx + c \\ ∴ f (1) = 8 \Rightarrow a + b + c = 8 \\ f (2) = 11 \Rightarrow 4 a + 2 b + c = 11 \\ and f (- 3) = 6 \Rightarrow 9 a - 3 b + c = 6 \\ Thus, we obtain the following system of equations \\ a + b + c = 8 \\ 4 a +2 b + c = 11 \\ 9 a -3 b + c = 6 \\ Here, \\ D = |\begin{array}{l} 1 1 1 \\ 4 2 1 \\ 9 -3 1 \end{array}| = 1 (2 + 3) - (4 - 9) + 1 (12 - 18) \\ = 5 + 5 - 30 = - 20 \\ D_{1} = |\begin{array}{l} 8 1 1 \\ 11 2 1 \\ 6 -3 1 \end{array}| = 8 (2 + 3) - 1 (11 - 6) + 1 (33 - 12) \\ = 40 - 5 - 45 = - 10 \\ D_{2} = |\begin{array}{l} 1 8 1 \\ 4 11 1 \\ 9 6 1 \end{array}| = 1 (11 - 6) - 8 (4 - 9) + 1 (24 - 99) \\ = 5 + 40 - 75 = - 30 \\ D_{3} = |\begin{array}{l} 1 1 8 \\ 4 2 11 \\ 9 -3 6 \end{array}| = 1 (12 + 33) - 1 (24 - 99) + 8 (- 12 - 18) \\ = 45 + 75 - 240 = - 120 \\ a = \frac{D_{1}}{D} = \frac{- 10}{- 20} = \frac{1}{2} \\ b = \frac{D_{2}}{D} = \frac{- 30}{- 20} = \frac{3}{2} \\ c = \frac{D_{3}}{D} = \frac{- 120}{- 20} = 6 \\ Hence, f (x) = \frac{1}{2} x^{2} + \frac{3}{2} x + 6 \\ \Rightarrow f (0) = 6 . \end{array}$

Q.32

$\begin{array}{l} If m \in N and m \geq 2, without expanding, prove that \\ |\begin{array}{l} 1 1 1 \\ m_{c_{1}} m {+1}_{c_{1}} m {+2}_{c_{1}} \\ m_{c_{2}} m {+1}_{c_{2}} m {+2}_{c_{2}} \end{array}| = 1 \end{array}$

Ans

$\begin{array}{l} Let Δ = |\begin{array}{l} 1 1 1 \\ m_{c_{1}} m {+1}_{c_{1}} m {+2}_{c_{1}} \\ m_{c_{2}} m {+1}_{c_{2}} m {+2}_{c_{2}} \end{array}| . Then \\ Δ = |\begin{array}{l} 1 1 1 \\ m_{c_{1}} m {+1}_{c_{1}} m {+1}_{c_{0}} + m + 1_{c_{1}} \\ m_{c_{2}} m {+1}_{c_{2}} m {+1}_{c_{1}} + m {+1}_{c_{2}} \end{array}| [∵ n_{c_{t}} + n_{c_{t - 1}} = n + 1_{c_{r}}] \\ Δ = |\begin{array}{l} 1 1 1 \\ m_{c_{1}} m {+1}_{c_{1}} m {+1}_{c_{0}} \\ m_{c_{2}} m {+1}_{c_{2}} m {+1}_{c_{1}} \end{array}| [Applying c_{3} \to c_{3} - c_{2}] \\ \Rightarrow Δ = |\begin{array}{l} 1 1 1 \\ m_{c_{1}} m {+1}_{c_{1}} + m_{c_{0}} + m + 1_{c_{0}} \\ m_{c_{2}} m_{c_{1}} + m_{c_{2}} m {+1}_{c_{1}} \end{array}| [∵ n_{c_{r}} + n_{c_{r - 1}} = n + 1_{c_{r}}] \\ \Rightarrow Δ = |\begin{array}{l} 1 0 0 \\ m_{c_{1}} m {+1}_{c_{0}} m + 1_{c_{0}} \\ m_{c_{2}} m_{c_{1}} m {+1}_{c_{1}} \end{array}| [Applying c_{2} \to c_{2} - c_{1}] \\ \Rightarrow Δ = |\begin{array}{l} m_{c_{0}} m {+1}_{c_{0}} \\ m_{c_{1}} m {+1}_{c_{1}} \end{array}| [Expanding along R_{1}] \\ \Rightarrow Δ = |\begin{array}{l} 1 1 \\ m m +1 \end{array}| = (m + 1 - m) = 1 \end{array}$

Q.33

$\begin{array}{l} Solve the following system of equations by cramer ‘ s rule : \\ \frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4, \frac{4}{x} + \frac{6}{y} + \frac{5}{z} = 1, and \frac{6}{x} + \frac{9}{y} + \frac{20}{z} = 2. \end{array}$

Ans

$\begin{array}{l} Let \frac{1}{x} = μ, \frac{1}{y} = ν and \frac{1}{z} = ω . Then, the above system of equation can be \\ written as \\ 2 μ + 3 ν + 10 ω = 4 \\ 4 μ - 6 ν + 5 ω = 1 \\ 6 μ + 9 ν - 20 ω = 2 \\ Here, D = |\begin{array}{l} 2 3 10 \\ 4 -6 5 \\ 6 9 -20 \end{array}| \\ = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) \\ \Rightarrow D = 150 + 330 + 720 = 1200 \\ D_{1} = |\begin{array}{l} 4 3 10 \\ 1 -6 5 \\ 2 9 -20 \end{array}| \\ = 4 (120 - 45) - 3 (- 20 - 10) + 10 (9 + 12) \\ \Rightarrow D_{1} = 300 + 90 + 210 = 600 \\ D_{2} = |\begin{array}{l} 2 4 10 \\ 4 1 5 \\ 6 2 -20 \end{array}| \\ = 2 (- 20 - 10) - 4 (- 80 - 30) + 10 (8 + 6) \\ \Rightarrow D_{2} = - 60 + 440 + 20 = 400 \\ D_{3} = |\begin{array}{l} 2 3 4 \\ 4 -6 1 \\ 6 9 2 \end{array}| \\ = 2 (- 12 - 9) - 3 (8 - 6) + 4 (36 + 36) \\ \Rightarrow D_{3} = - 42 - 6 + 288 = 240 \\ ∴ μ = \frac{D_{1}}{D} = \frac{600}{1200} = \frac{1}{2} \\ \Rightarrow \frac{1}{x} = \frac{1}{2} \\ \Rightarrow x = 2 \\ ν = \frac{D_{2}}{D} = \frac{400}{1200} = \frac{1}{3} \\ \Rightarrow \frac{1}{y} = \frac{1}{3} \\ \Rightarrow y = 3, \\ ω = \frac{D_{3}}{D} = \frac{240}{1200} = \frac{1}{5} \\ \Rightarrow \frac{1}{z} = \frac{1}{5} \\ Hence, x = 2, y = 3 and z = 5. \end{array}$

Q.34

$\begin{array}{l} A triangle has its theree sides equal to a, b and c, If the coordinates \\ of its vertices are A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}), show that \\ {|\begin{array}{l} x_{1} y_{1} 2 \\ x_{2} y_{2} 2 \\ x_{3} y_{3} 2 \end{array}|}^{2} = (a + b + c)) (b + c - a) (a + b - c) \\ by using determinants . \end{array}$

Ans

$\begin{array}{l} Let Δ be the area of triangle ABC . Then, \\ Δ = \frac{1}{2} |\begin{array}{l} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}| \\ \Rightarrow 2 Δ = |\begin{array}{l} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}| \\ \Rightarrow 4 Δ =2 |\begin{array}{l} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}| (Multiplying both sides by 2.) \\ \Rightarrow 4 Δ = |\begin{array}{l} x_{1} y_{1} 2 \\ x_{2} y_{2} 2 \\ x_{3} y_{3} 2 \end{array}| (All the elements of c_{3} are multiplied by 2.) \\ \Rightarrow 16 Δ^{2} = {|\begin{array}{l} x_{1} y_{1} 2 \\ x_{2} y_{2} 2 \\ x_{3} y_{3} 2 \end{array}|}^{2} (On squaring both sides .) . \dots. (i) \\ We also know that the area of triangle ABC is given by \\ Δ = \sqrt{s (s - a) (s - b) (s - c)}, where s = \frac{1}{2} (a + b + c) \\ But, s = \frac{1}{2} (a + b + c) \\ \Rightarrow s - a = \frac{1}{2} (a + b + c) - c \\ = \frac{1}{2} (b + c - a) \\ Similarly, s - c = \frac{1}{2} (a + b - c) . \\ ∴ Δ^{2} = \frac{1}{2} (a + b + c) . \frac{1}{2} (b + c - a) . \frac{1}{2} (c + a - b) . \frac{1}{2} (a + b - c) . \\ \Rightarrow 16 Δ^{2} = (a + b + c) (b + c - a) (c + a - b) (a + b - c) . \dots. (ii) \\ From (i) and (ii) we get \\ |\begin{array}{l} x_{1} y_{1} 2 \\ x_{2} y_{2} 2 \\ x_{3} y_{3} 2 \end{array}| = (a + b + c) (b + c - a) (c + a - b) (a + b - c) \end{array}$

Q.35

$\begin{array}{l} Using cramer ‘ s rule, find values of a and b for which \\ the system of equations \\ 2 x + ay + 6 z = 8 \\ a + 2 y + bz = 5 \\ x + y + 3 z = 4 \\ has \\ (i) a unique solution \\ (ii) infinitely many soutions \\ (iii) no solutions . \end{array}$

Ans

$\begin{array}{l} We have, \\ D = |\begin{array}{l} 2 6 6 \\ 12 b \\ 1 1 3 \end{array}| \\ \Rightarrow D = 2 (6 - b) - a (3 - b) + 6 (1 - 2) \\ \Rightarrow D = 12- 2 b - 3 b + ab - 6 \\ \Rightarrow D = 6 - 3 a - 2 b + ab = (b - 3) (a - 2) \\ D = |\begin{array}{l} 8 a 6 \\ 52 b \\ 4 1 3 \end{array}| \\ D_{1} = 8 (6 - b) - a (15 - 4 b) + 6 (5 - 8) \\ \Rightarrow D_{1} = 48 - 8 b - 15 a + 4 ab - 18 \\ \Rightarrow D_{1} = 30 - 15 a - 8 b + 4 ab = (a - 2) (4 b - 15) \\ D_{2} |\begin{array}{l} 2 8 6 \\ 15 b \\ 1 4 3 \end{array}| \\ D_{2} = 2 (15 - 4 b) - 8 (3 - b) + 6 (4 - 5) \\ \Rightarrow D_{2} = 30 - 8 b - 24 + 8 b - 6 = 0 \\ and \\ D_{3} |\begin{array}{l} 2 a 8 \\ 1 2 5 \\ 1 1 4 \end{array}| \\ D_{3} = 2 (8 - 5) - a (4 - 50 + 8) (1 - 2) \\ \Rightarrow D_{3} = 6 + a - 8 = a - 2 \\ (i) For unique solution, we must have \\ D \neq 0 \Rightarrow (a - 2) (b - 3) \neq 0 \\ \Rightarrow a \neq 2, or b \neq 3 . \\ (ii) for infinitely many solutions, we must have \\ D = D_{1} = D_{2} = D_{3} = 0 \\ \Rightarrow (a - 2) (b - 3) = 0, (a - 2) (4 b - 15) = 0 and a - 2 = 0 . \\ \Rightarrow a = 2 \\ Putting a = 2 in hte given system of equations, we get \\ 2 x + 2 y + 6 z = 8 \\ x + 2 y + 6 z = 5 \\ x + y + 3 z = 4 \\ x + y + z = 4 \\ x + 2 y + bz = 5 \\ Putting z = k, we get \\ x + y = 4 - k \\ x + 2 y = 5 - bk \\ Solving these two equations, we get \\ x = 3 - 3 k + bk, y = 1 - bk + k \\ Thus, the given system has infinitely many solutions given by \\ x = 3 - 2 k + bk, y = 1 - bk + k, z = k, where k \in R . \\ Hence, the system has infinitely many solutions for a = 2. \\ (iii) For no solution, we must have \\ D = 0 and at least one of D_{1}, D_{2} and is non zero . \\ Clearly, for b = 3, we have \\ D = 0 and D_{3} \neq 0 . \\ Hence, the system has no solution for b = 3. \end{array}$

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FAQs (Frequently Asked Questions)

Only square matrices have determinants because determinants are defined for matrices with the same number of rows and columns. Rectangular matrices do not have a determinant value.

Choose the row or column with the maximum number of zeroes. This reduces multiplication and makes the determinant easier to evaluate.

The determinant may give a positive or negative value depending on the order of points. Since area cannot be negative, the absolute value is used.

A minor is obtained by deleting a row and column. A cofactor is the signed minor. The sign is decided by (-1)i+j.

A system AX = B has a unique solution when the coefficient matrix A is non-singular. This means |A| ≠ 0, so A⁻¹ exists and X = A⁻¹B.

CBSE Class 12 Maths Revision Notes Chapter 4 Determinants

Key Takeaways

Access Class 12 Maths Chapter 4 Determinants Notes in 30 Minutes

Determinants Class 12 Notes: Meaning and Basic Concept

What Is a Determinant?

Determinant of Order 1

Determinant of Order 2

Determinant of Order 3

Expansion of Determinants in Class 12 Mathematics Chapter 4 Notes

Expansion Along a Row or Column

Sign Pattern for Cofactor Expansion

Properties of Determinants for Quick Revision

Row and Column Interchange Properties

Zero, Equal and Proportional Row Properties

Multiplication and Addition Properties

Sum Property of Determinants

Triangular Determinant Property

Area of a Triangle Using Determinants

Formula for Area of a Triangle

Important Points for Area Questions

Collinearity Condition

Minors and Cofactors in Determinants

Minor of an Element

Cofactor of an Element

Expansion Using Cofactors

Important Cofactor Rule

Adjoint and Inverse of a Matrix

Adjoint of a Matrix

Adjoint of a 2 × 2 Matrix

Singular and Non-Singular Matrices

Theorem on Adjoint

Inverse Formula Using Determinants

Applications of Determinants in Linear Equations

Matrix Form AX = B

Case 1: Unique Solution

Case 2: No Solution

Case 3: Infinite Solutions

Solution Cases Summary

Quick Revision Tables for Determinants

Determinant Formula Table

Properties of Determinants Table

Matrix Inverse Table

Important Terms in Determinants

Common Mistakes in Class 12 Mathematics Chapter 4 Notes

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FAQs (Frequently Asked Questions)

1. Why do only square matrices have determinants?

2. How do I choose the best row or column for expanding a determinant?

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4. What is the main difference between a minor and a cofactor?

5. When does a system of linear equations have a unique solution?

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