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CBSE Class 12 Maths Revision Notes Chapter 4

Class 12 Mathematics Chapter 4 Notes

In Class 12, students deal with high pressure and stress, mainly in Mathematics. The marks scored in the final board examination will be required to appear for future studies in their field of interest. Mathematics is considered to be a tricky subject. A strong understanding of all basic concepts and regular practice is very important to understanding complex topics. Students are advised to use the best study material, such as the Class 12 Mathematics chapter 4 notes, to help them get a clear understanding of determinants and concepts related to it.

The concepts explained in this chapter are very useful to analyse and find the solution. In Class 12 Mathematics chapter 4 notes, students will gain knowledge of determinants up to order 3. With the help of determinants, we can find out the uniqueness of a solution. The chapter has a wide range of applications in the Engineering, Science, etc., sectors. Also, concepts based on the properties of the matrix, cofactors, inverse, adjoint, and area of a triangle are covered in the Class 12 Mathematics chapter 4 notes.

Extramarks, an online learning platform, provides professional help and gives students the best academic notes to ace their exams. With the help of the class 12 chapter 4 mathematics notes, students can prepare for both the board exams and the competitive exams. Our notes are provided with detailed information, graphical representation, and easy language for students’ overall development.

Key Topics Covered In Class 12 Mathematics Chapter 4 Notes

The main topics covered in the Class 12 Mathematics chapter 4 notes are:

Introduction
Determinant Definition
Properties
Area of a Triangle
Minors and Cofactors
Adjoint of a Matrix
The inverse of a Matrix
Applications
Solution of a system of linear equations

A brief of the key topics covered in class 12 Mathematics chapter 4 notes is as under.

INTRODUCTION:

The Class 12 Mathematics chapter 4 notes include various complex concepts related to determinants and their applications. To ace this chapter, students must learn concepts introduced in Class 11 as well as chapter 3 of the NCERT books.

RECALL:

If the system of equations is given as

a1x + b1y=c1 and a2x + b2y=c2, then the matric representation will be

	a1	b1			x			=	c1
	a2	b2			y				c2

The solution of this system is all the values of variables x and y, which satisfy the linear equations in the system.

DEFINITION OF DETERMINANT:

The determinant of any square matrix is a scalar value that is calculated from all elements of the matrix, which implies, a for every square matrix A of order n is associated with a number called the determinant of that matric A and is denoted by A (mod A), det (A) or Δ .

Consider A =

	a11	b12
	a21	b22

The scalar value for matrix A is a1b2-b1a2

A= a11b22-b12a21

	a11	b12
	a21	b22

NOTE: Only square matrices (number of rows and columns are equal) have determinants.

TYPES OF DETERMINANTS:

First Order Determinant: The determinant of a matrix, say A of order one, is known as the First-order determinant. The value of the determinant is said to be the element of matrix A.

Consider A = [2]

Therefore determinant = 2= 2

Second Order Determinant - The determinant of a matrix, say A of order two, is known as the Second-order determinant.

Consider A =

	4	2
	5	3

Therefore, A=

	4	2
	5	3

(4 x 3) - (5 x 2) = 12 - 10 = 2

Third Order Determinant - The determinant of a matrix, say A of order three, is known as the third-order determinant.

Consider A =

a1	b1	c1
a2	b2	c2
a3	b3	c3

Then A =

a1	b1	c1
a2	b2	c2
a3	b3	c3

as a1(b2c3- b3c2) - b1(a2c3- a3c2) + c1(a2b3- a3b2)

NOTE:

Consider two square matrices A and B having order n and A=kB, then |A|=kn|B|, n N

PROPERTIES OF DETERMINANTS:

Let us glance through some properties of the Determinants, which are discussed in Class 12 Mathematics Chapter 4 Notes.

Property 1- the value of the determinant remains the same even after interchanging the rows and columns. Symbolic representation Ci Ri

Consider A=

	4	2
	5	3

(4 x 3) - (5 x 2) = 12 - 10 = 2

And A=

	4	5
	2	3

(4 x 3) - (2 x 5) = 12 - 10 = 2

This implies that:

If A is a square matrix, then det (A) = det (A’), where A’ is the transpose of matrix A.

Property 2- From the matrix, if the elements of any two columns (or rows) of the determinants are interchanged, then the sign of determinants will change. Symbolic representation Ci Cj or Ri Rj

Property 3- If the matrix has any two rows or columns equal or identical, then we can say that the value of the determinant is equal to 0.

Property 4- If every element of a row or a column in a matrix is multiplied by k, then to obtain the value of the determinant, the original determinant is multiplied by k, where k is a constant.

Property 5- If the elements of any row or a column of a determinant are expressed as the sum of two or more terms, then we can say that the original determinant can also be expressed as the sum of two or more determinants.

Property 6- If for each element of a row or a column of the given determinant, the equimultiples of the analogous elements of other row or column respectively are added, then we can say that the value of the original determinant remains the same.

The symbolic representation of the operation performed is Ri Ri+ kRj or Ci Ci+ kCj.

Property 7- If every element in a row or column of the given determinant is zero, then the value of the determinant is zero.

Property 8- Consider an upper triangular matrix or low triangular matrix, i.e., all elements on one side of the diagonal of the matrix are zeroes. The value of the determinant of such a matrix can be calculated by multiplying all diagonal elements.

In simple words, the determinant of an upper (lower) triangular matrix is equal to the product of all the elements in the diagonal.

Visit the Extramarks platform to access the class 12 mathematics chapter 4 notes. Gain in-depth information and understand each property in detail.

AREA OF A TRIANGLE:

Let the vertices of a triangle be (x1, y1),(x2, y2) and (x3, y3). Then the area of the triangle is defined as

A=12[x1(y2−y3) + x3(y3−y1) + x3(y1−y3)].

Representation of area in the form of determinant = = 12 x determinant

Determinant is given as

x1	y1	1
x2	y2	1
x3	y3	1

Remember:

The area is a positive quantity. Therefore determinant will be positive
In case the area is already given, then use both the positive and negative values of the determinant.
The area of a triangle with three collinear points is equal to zero.

The class 12 mathematics notes chapter 4 includes several problems based on the area of the triangle for students to practice and master this concept.

Minors of the Determinant:

Representation: Mij

Minor of any element is calculated by deleting the ith row and jth column in which the element lies. For a21 we delete the 2nd row and first column.

Minor of any element of the given determinant of order n, where n ≥ 2 is a determinant of order n – 1.

Cofactor of the Determinant:

Representation: Aij

Cofactor of an element aij is calculated by multiplying the minor of the element with (-1)i+j

Cofactor of element aij (Aij) = (-1)i+jMij

Adjoint of a Matrix:

The adjoint of a matrix is the matrix obtained by the transpose of a cofactor matrix.

Singular Matrices

A singular matrix is defined as the square matrix whose determinant is zero.

Non-Singular Matrices

A Non-singular matrix is defined as the square matrix whose determinant is a non-zero value.

THEOREMS:

Theorem 1 - Let A be any square matrix of order n, then A (adj A) = (adj A) A =|A| I, where I is the identity matrix of the same order.

Theorem 2 - Let A and B be the non-singular matrices of order n, then their product, i.e., AB and BA, will also be non-singular matrices of the same order n.

Theorem 3 - Let A and B be two square matrices of order n. The determinant of the product of matrix AB is given as the product of the respective determinants. It is written as AB=|A||B|.

Theorem 4 - A square matrix is invertible (inverse exists) iff the matrix is non-singular. So, for any matrix A, the inverse of the matrix is A-1= 1A(adj A).

Refer to the class 12 chapter 4 mathematics notes to access unlimited problems for extra practice.

APPLICATIONS:

Matrices and determinants are used to solve systems of linear equations with respect to two or three variables. They are also used to check the consistency of the given system of linear equations.
A Consistent system is said to be a system of equations whose solution can be found or exists.
An Inconsistent system is said to be a system of equations whose solution cannot be found or does not exist.
The value of the determinant is a number that is capable of determining the uniqueness of the solution of the given system of linear equations.

SOLUTION TO THE SYSTEM OF LINEAR EQUATIONS:

Consider a system,

a1x + b1y=c1 and a2x + b2y=c2, then the matric representation will be

	a1	b1			x			=	c1
	a2	b2			y				c2

It is represented as AX = B ….. (1)

Case 1: A is a non-singular matrix

In this case, we premultiply A-1to both sides of equation 1

Therefore, we get (A-1) AX = (A-1).B

Using associativity, (A-1A) X = (A-1).B

I X = (A-1).B

We get, X = (A-1).B

The value of X provides a unique solution. This method is commonly known as the matrix method.

Case 2: A is a singular matrix

In this case, firstly, we calculate (adj A) B

Depending on the value of (adj A) B, we get our result.

(adj A)B is a non-zero matrix: Solution does not exist. The system of equations is said to be inconsistent with no solution.

(adj A)B is a zero matrix: Solution exists. The system of equations is said to be either consistent with many solutions or inconsistent.

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Chapter 4 Mathematics Class 12 Notes: Exercises & Answer Solutions

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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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CBSE Class 12 Maths Revision Notes Chapter 4

Class 12 Mathematics Chapter 4 Notes

Key Topics Covered In Class 12 Mathematics Chapter 4 Notes

INTRODUCTION:

DEFINITION OF DETERMINANT:

TYPES OF DETERMINANTS:

PROPERTIES OF DETERMINANTS:

THEOREMS:

Chapter 4 Mathematics Class 12 Notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features Of Class 12 Mathematics Chapter 4 Notes

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FAQs (Frequently Asked Questions)

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CBSE Class 12 Maths Revision Notes Chapter 4

Class 12 Mathematics Chapter 4 Notes

Key Topics Covered In Class 12 Mathematics Chapter 4 Notes

INTRODUCTION:

DEFINITION OF DETERMINANT:

TYPES OF DETERMINANTS:

PROPERTIES OF DETERMINANTS:

THEOREMS:

Chapter 4 Mathematics Class 12 Notes: Exercises & Answer Solutions

NCERT Exemplar Class 12 Mathematics

Key Features Of Class 12 Mathematics Chapter 4 Notes

Please register to view this section

FAQs (Frequently Asked Questions)

1. Should I practice all the questions included in the Class 12 Mathematics chapter 4 notes?

2. How do I prepare for the Class 12 Mathematics board exams?

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