# NCERT Solutions for Class 12 Maths Chapter 4 Determinants

NCERT Solutions for Class 12 Maths Chapter 4 Determinants: In our previous chapter, you have already learnt about matrices and properties of matrices. In this article, you will find NCERT solutions for class 12 maths chapter 4 determinants. You know that if you multiply a matrix with coordinates of a point, it will give a new point in the space. In this sense, the matrix is a linear transformation. The determinant of the matrix is the factor by which its volume blow up. For example, if the determinant is 1 which means its volume is unchanged, if the determinant is 2 means the volume is doubled after transformation. What does the physical meaning of the determinant is negative or zero mean? Why is the inverse of the matrix is not possible if its determinant is zero? You will get all these answers in the CBSE NCERT solutions for class 12 maths chapter 4 determinants article. The Important topics are determinants and their properties, finding the area of the triangle, minor and cofactors, adjoint and the inverse of the matrix, and applications of determinants like solving the system of linear equations etc are covered in this chapter. In the solutions of NCERT for class 12 maths chapter 4 determinants article, you will get detailed explanations to all these above topics.  The practice of NCERT questions is very important to get a command on this chapter otherwise you will get confused with the formulas of this chapter. You should solve every problem on your own, if you are finding difficulties, you can take help with these solutions of NCERT for class 12 maths chapter 4 determinants. Check all NCERT solutions at a single place which will help you to learn CBSE maths.

The topic algebra which contains two topics matrices and determinants has 13 % weightage in the CBSE 12th board final examination, which means 10 marks questions out of 80 marks will be asked from these two chapters matrices and determinants in the final examination. The determinant is an important part of matrices. In the solutions of NCERT class 12 maths chapter 4 determinants, you will be dealing with determinants of order up to three only. In this chapter, there are 6 exercises with 68 questions. All these questions are prepared and explained in this NCERT solutions for class 12 maths chapter 4 determinants article.

What are the Determinants?

To every square matrix of order n, we can associate a number (real or complex) called determinant of the square matrix A. Let's take a determinant (A) of order two-

If A is a then the determinant of A is written as |A|=matrix

,

The six exercises of this chapter determinants covers the properties of determinants, cofactors and applications like finding the area of triangle, solutions of linear equations in two or three variables, minors, consistency and inconsistency of system of linear equations, adjoint and inverse of a square matrix, and solution of linear equations in two or three variables using inverse of a matrix.

## Topics and sub-topics of NCERT class 12 maths chapter 4 Determinants

4.1 Introduction

4.2 Determinant

4.2.1 Determinant of a matrix of order one

4.2.2 Determinant of a matrix of order two

4.2.3 Determinant of a matrix of order 3 × 3

4.3 Properties of Determinants

4.4 Area of a Triangle

4.5 Minors and Cofactors

4.6 Adjoint and Inverse of a Matrix

4.7 Applications of Determinants and Matrices

4.7.1 Solution of a system of linear equations using the inverse of a matrix

## CBSE NCERT solutions for class 12 maths chapter-4 Determinants: Excercise- 4.1

Question:1 Evaluate the following determinant-

The determinant is evaluated as follows

Question:2(i) Evaluate the following determinant-

The given two by two determinant is calculated as follows

Question:2(ii) Evaluate the following determinant-

We have determinant

So,

Question:3 If     , then show that

Given determinant  then we have to show that ,

So,  then,

Hence we have

So, L.H.S. = |2A| = -24

then calculating R.H.S.

We have,

hence R.H.S becomes

Therefore L.H.S. =R.H.S.

Hence proved.

Question:4 If   then show that

Given Matrix

Calculating

So,

calculating ,

So,

Therefore .

Hence proved.

Question:5(i) Evaluate the determinants.

Given the determinant ;

now, calculating its determinant value,

.

Question:5(ii) Evaluate the determinants.

Given determinant ;

Now calculating the determinant value;

.

Question:5(iii) Evaluate the determinants.

Given determinant ;

Now calculating the determinant value;

Question:5(iv) Evaluate the determinants.

Given determinant: ,

We now calculate determinant value:

Question:6 If    ,  then find  .

Given the matrix  then,

Finding the determinant value of A;

Question:7(i)  Find values of x, if

Given that

First, we solve the determinant value of L.H.S. and equate it to the determinant value of R.H.S.,

and

So, we have then,

or           or

Question:7(ii) Find values of x, if

Given  ;

So, we here equate both sides after calculating each side's determinant values.

L.H.S. determinant value;

Similarly R.H.S. determinant value;

So, we have then;

or    .

Question:8 If    ,  then  is equal to

(A)         (B)        (C)      (D)

Solving the L.H.S. determinant ;

and solving R.H.S determinant;

So equating both sides;

or             or

## CBSE NCERT solutions for class 12 maths chapter -4 Determinants: Excercise - 4.2

We can split it in manner like;

So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Clearly, expanded determinants have identical columns.

Hence the sum is zero.

Given determinant

Applying the rows addition    then we have;

So, we have two rows  and  identical hence we can say that the value of determinant = 0

Therefore .

Given determinant

So, we can split it in two addition determinants:

[ Here two columns are identical ]

and     [ Here two columns are identical ]

Therefore we have the value of determinant = 0.

We have determinant:

Applying  we have then;

So, here column 3 and column 1 are proportional.

Therefore, .

Given determinant :

Splitting the third row; we get,

.

Then we have,

On Applying row transformation    and then  ;

we get,

Applying Rows exchange transformation    and   , we have:

also

On applying rows transformation,  and then

and then

Then applying rows exchange transformation;

and then . we have then;

So, we now calculate the sum =

Hence proved.

We have given determinant

Applying transformation,  we have then,

We can make the first row identical to the third row so,

Taking another row transformation:  we have,

So, determinant has two rows  identical.

Hence .

Given determinant :

As we can easily take out the common factors a,b,c from rows  respectively.

So, get then:

Now, taking common factors a,b,c from the columns  respectively.

Now, applying rows transformations    and then  we have;

Expanding to get R.H.S.

Question:8(i) By using properties of determinants, show that:

We have the determinant

Applying the row transformations   and then   we have:

Now, applying  we have:

or

Hence proved.

Question:8(ii) By using properties of determinants, show that:

Given determinant :

,

Applying column transformation  and then

We get,

Now, applying column transformation , we have:

Hence proved.

We have the determinant:

Applying the row transformations  and then  , we have;

Now, applying ; we have

Now, expanding the remaining determinant;

Hence proved.

Question:10(i) By using properties of determinants, show that:

Given determinant:

Applying row transformation:   then we have;

Taking a common factor: 5x+4

Now, applying column transformations   and

Question:10(ii) By using properties of determinants, show that:

Given determinant:

Applying row transformation   we get;

[taking common (3y + k) factor]

Now, applying column transformation   and

Hence proved.

Question:11(i) By using properties of determinants, show that:

Given determinant:

We apply row transformation:  we have;

Taking common factor (a+b+c) out.

Now, applying column tranformation     and then

We have;

Hence Proved.

Question:11(ii) By using properties of determinants, show that:

Given determinant

Applying    we get;

Taking 2(x+y+z) factor out, we get;

Now, applying row transformations,   and then .

we get;

Hence proved.

Give determinant

Applying column transformation  we get;

[after taking the (1+x+x) factor common out.]

Now, applying row transformations,     and then .

we have now,

As we know

Hence proved.

We have determinant:

Applying row transformations,     and    then we have;

taking common factor out of the determinant;

Now expanding the remaining determinant we get;

Hence proved.

Given determinant:

Let

Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C1,C2,and C3.

We then get;

Now, applying column transformations:   and

then we have;

Now, expanding the remaining determinant:

.

Hence proved.

(A)           (B)         (C)         (D)

Assume a square matrix A of order of .

Then we have;

(Taking the common factors k from each row.)

Therefore correct option is (C).

Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these

The answer is (C) Determinant is a number associated to a square matrix.

As we know that To every square matrix of order n, we can associate a number (real or complex) called determinant of the square matrix A, where element of A.

## NCERT solutions for class 12 maths chapter 4 Determinants: Excercise-4.3

We can find the area of the triangle with vertices  by the following determinant relation:

Expanding using second column

We can find the area of the triangle with given coordinates by the following method:

Area of the triangle by the determinant method:

Hence the area is equal to

Question:2 Show that points    are collinear.

If the area formed by the points is equal to zero then we can say that the points are collinear.

So, we have an area of a triangle given by,

calculating the area:

Hence the area of the triangle formed by the points is equal to zero.

Therefore given points  are collinear.

We can easily calculate the area by the formula :

or

or      or

Hence two values are possible for k.

The area of the triangle is given by the formula:

Now, calculating the area:

or

Therefore we have two possible values of 'k' i.e.,    or  .

As we know the line joining  ,  and let say a point on line  will be collinear.

Therefore area formed by them will be equal to zero.

So, we have:

or

Hence, we have the equation of line .

Question:4(ii)  Find equation of line joining  and  using determinants.

We can find the equation of the line by considering any arbitrary point  on line.

So, we have three points which are collinear and therefore area surrounded by them will be equal to zero.

Calculating the determinant:

Hence we have the line equation:

or  .

Area of triangle is given by:

or

or

Hence the possible values of k are 12 and -2.

Therefore option (D) is correct.

Solutions of NCERT for class 12 maths chapter 4 Determinants-Excercise: 4.4

GIven determinant:

Minor of element  is .

Therefore we have

= minor of element  = 3

= minor of element  = 0

= minor of element  = -4

= minor of element  = 2

and finding cofactors of  is   = .

Therefore we have:

GIven determinant:

Minor of element  is .

Therefore we have

= minor of element  = d

= minor of element  = b

= minor of element  = c

= minor of element  = a

and finding cofactors of  is   = .

Therefore we have:

Given determinant :

Finding Minors: by the definition,

minor of                  minor of

minor of                  minor of

minor of                  minor of

minor of                  minor of

minor of

Finding the cofactors:

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of .

Given determinant :

Finding Minors: by the definition,

minor of        minor of

minor of              minor of

minor of              minor of

minor of

minor of

minor of

Finding the cofactors:

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of

cofactor of .

Given determinant :

First finding Minors of the second rows by the definition,

minor of

minor of

minor of

Finding the Cofactors of the second row:

Cofactor of

Cofactor of

Cofactor of

Therefore we can calculate   by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

Given determinant :

First finding Minors of the third column by the definition,

minor of

minor of

minor of

Finding the Cofactors of the second row:

Cofactor of

Cofactor of

Cofactor of

Therefore we can calculate   by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

Thus, we have value of .

(A)

(B)

(C)

(D)

Answer is (D)  by the definition itself,  is equal to the product of the elements of the row/column with their corresponding cofactors.

## CBSE NCERT solutions for class 12 maths chapter 4 Determinants- Excercise: 4.5

Question:1 Find adjoint of each of the matrices.

Given matrix:

Then we have,

Hence we get:

Question:2  Find adjoint of each of the matrices

Given the matrix:

Then we have,

Hence we get:

Question:3 Verify .

Given the matrix:

Let

Calculating the cofactors;

Hence,

Now,

aslo,

Now, calculating |A|;

So,

Hence we get

Question:4 Verify .

Given matrix:

Let

Calculating the cofactors;

Hence,

Now,

also,

Now, calculating |A|;

So,

Hence we get,

.

Given matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

|A| = (6+8) = 14

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Given the matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

|A| = (-2+15) = 13

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Given the matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Given the matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Given the matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Given the matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Given the matrix :

To find the inverse we have to first find adjA then as we know the relation:

So, calculating |A| :

Now, calculating the cofactors terms and then adjA.

So, we have

Therefore inverse of A will be:

Question:12 Let      and  .  Verify that  .

We have  and .

then calculating;

Finding the inverse of AB.

Calculating the cofactors fo AB:

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse:

.....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1   and |B| = 54- 56 = -2

and

therefore we have

and

Now calculating .

........................(2)

From (1) and (2) we get

Hence proved.

Question:13 If   ? , show that  . Hence find

Given  then we have to show the relation

So, calculating each term;

therefore  ;

Hence .

[Post multiplying by , also ]

Given  then we have the relation

So, calculating each term;

therefore  ;

So, we have equations;

and

We get .

Given matrix: ;

To show:

Finding each term:

So now we have,

Now finding the inverse of A;

Post-multiplying by  as,

...................(1)

Now,

From equation (1) we get;

Question:16 If    , verify that . Hence find .

Given matrix: ;

To show:

Finding each term:

So now we have,

Now finding the inverse of A;

Post-multiplying by  as,

...................(1)

Now,

From equation (1) we get;

Hence inverse of A is :

(A)       (B)       (C)       (D)

We know the identity

Hence we can determine the value of .

Taking both sides determinant value we get,

or

or taking R.H.S.,

or, we have then

Therefore

Hence the correct answer is B.

(A)        (B)         (C)         (D)

Given that the matrix is invertible hence  exists and

Let us assume a matrix of the order of 2;

.

Then .

and

Now,

Taking determinant both sides;

Therefore we get;

Hence the correct answer is B.

CBSE NCERT solutions for class 12 chapter 4 Determinants: Excercise- 4.6

We have given the system of equations:18967

The given system of equations can be written in the form of the matrix;

where ,    and .

So, we want to check for the consistency of the equations;

Here A is non -singular therefore there exists .

Hence, the given system of equations is consistent.

We have given the system of equations:

The given system of equations can be written in the form of matrix;

where ,    and .

So, we want to check for the consistency of the equations;

Here A is non -singular therefore there exists .

Hence, the given system of equations is consistent.

We have given the system of equations:

The given system of equations can be written in the form of the matrix;

where ,    and .

So, we want to check for the consistency of the equations;

Here A is singular matrix therefore now we will check whether the  is zero or non-zero.

So,

As,  , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

We have given the system of equations:

The given system of equations can be written in the form of the matrix;

where ,    and .

So, we want to check for the consistency of the equations;

[If zero then it won't satisfy the third equation]

Here A is non- singular matrix therefore there exist .

Hence, the given system of equations is consistent.

We have given the system of equations:

The given system of equations can be written in the form of matrix;

where ,    and .

So, we want to check for the consistency of the equations;

Therefore matrix A is a singular matrix.

So, we will then check

As,  is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.