# 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 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 co-factors, 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 NCERT solutions 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 NCERT solutions for class 12 maths chapter 4 determinants. Check all **NCERT solutions ** at a single place which will help you to learn CBSE maths. Here you will get NCERT solutions for class 12 also.

The topic algebra which contains two topics matrices and determinants has 13 % weightage in the CBSE 12^{th } 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, co-factors 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 Ã— 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

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

** Question:2(i) ** Evaluate the following determinant- ** **

** Answer: **

The given two by two determinant is calculated as follows

** Question:3 ** If , then show that

** Answer: **

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

** Answer: **

Given Matrix

Calculating

So,

calculating ,

So,

** Therefore . **

** Hence proved. **

** Question:5(i) ** Evaluate the determinants.

** Answer: **

Given the determinant ;

now, calculating its determinant value,

.

** Question:5(ii) ** Evaluate the determinants.

** Answer: **

Given determinant ;

Now calculating the determinant value;

.

** Question:5(iii) ** Evaluate the determinants.

** Answer: **

Given determinant ;

Now calculating the determinant value;

** Question:5(iv) ** Evaluate the determinants.

** Answer: **

Given determinant: ,

We now calculate determinant value:

** Question:7(i) ** Find values of x, if

** Answer: **

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

** Answer: **

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

** Answer: **

Solving the L.H.S. determinant ;

and solving R.H.S determinant;

So equating both sides;

or or

** Hence answer is (B). **

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

** Question:1 ** Using the property of determinants and without expanding, prove that

** Answer: **

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. **

** Question: ** ** 2 ** Using the property of determinants and without expanding, prove that

** Answer: **

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 .

** Question:3 ** Using the property of determinants and without expanding, prove that

** Answer: **

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. **

** Question:4 ** ** ** Using the property of determinants and without expanding, prove that

** Answer: **

We have determinant:

Applying we have then;

** So, here column 3 and column 1 are proportional. **

** Therefore, . **

** Question:5 ** Using the property of determinants and without expanding, prove that

** Answer: **

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. **

** Question:6 ** Using the property of determinants and without expanding, prove that

** Answer: **

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 . **

** Question:7 ** Using the property of determinants and without expanding, prove that

** Answer: **

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:

** Answer: **

Given determinant :

,

Applying column transformation and then

We get,

Now, applying column transformation , we have:

** Hence proved. **

** Question:9 ** By using properties of determinants, show that:

** Answer: **

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:

** Answer: **

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:

** Answer: **

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:

** Answer: **

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:

** Answer: **

Given determinant

Applying we get;

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

Now, applying row transformations, and then .

we get;

** Hence proved. **

** Question:12 ** By using properties of determinants, show that:

** Answer: **

Give determinant

Applying column transformation we get;

[ ** after taking the (1+x+x ^{ 2 } ) factor common out.] **

Now, applying row transformations, and then .

we have now,

As we know

** Hence proved. **

** Question:13 ** ** ** By using properties of determinants, show that:

** Answer: **

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. **

** Question:14 ** By using properties of determinants, show that:

** Answer: **

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 ** C _{ 1, } C _{ 2, } ** and

**C**

_{ 3. }We then get;

Now, applying column transformations: and

then we have;

Now, expanding the remaining determinant:

.

** Hence proved. **

** Question:15 ** Choose the correct answer. Let A be a square matrix of order , then is equal to

** Answer: **

Assume a square matrix A of order of .

Then we have;

( ** Taking the common factors k from each row. ** )

** Therefore correct option is (C). **

** Question:16 ** Choose the correct answer.

** Answer: **

** 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 **

** Question:1(i) ** Find area of the triangle with vertices at the point given in each of the following :

** Answer: **

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

Expanding using second column

** Question:1(ii) ** Find area of the triangle with vertices at the point given in each of the following :

** Answer: **

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

** Question:1(iii) ** Find area of the triangle with vertices at the point given in each of the following :

** Answer: **

Area of the triangle by the determinant method:

Hence the area is equal to

** Question:2 ** Show that points are collinear.

** Answer: **

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. **

** Question:3(i) ** Find values of k if area of triangle is 4 sq. units and vertices are

** Answer: **

We can easily calculate the area by the formula :

or

or or

** Hence two values are possible for k. **

** Question:3(ii) ** Find values of k if area of triangle is 4 sq. units and vertices are

** Answer: **

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 . **

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

** Answer: **

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.

** Answer: **

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 . **

** Question:5 ** If the area of triangle is 35 sq units with vertices and . Then k is

** Answer: **

Area of triangle is given by:

or

or

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

** Therefore option (D) is correct. **

** NCERT solutions for class 12 maths chapter 4 Determinants-Excercise: 4.4 **

** Question:1(i) ** Write Minors and Cofactors of the elements of following determinants:

** Answer: **

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:

** Question:1(ii) ** Write Minors and Cofactors of the elements of following determinants:

** Answer: **

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:

** Question:2(i) ** Write Minors and Cofactors of the elements of following determinants:

** Answer: **

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 . **

** Question:2(ii) ** Write Minors and Cofactors of the elements of following determinants:

** Answer: **

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 . **

** Question:3 ** Using Cofactors of elements of second row, evaluate .

** Answer: **

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,

** Question:4 ** Using Cofactors of elements of third column, evaluate

** Answer: **

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 . **

** Question:5 ** If and is Cofactors of , then the value of is given by

** Answer: **

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

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

** Question:1 ** Find adjoint of each of the matrices.

** Answer: **

Given matrix:

Then we have,

Hence we get:

** Question:2 ** Find adjoint of each of the matrices

** Answer: **

Given the matrix:

Then we have,

Hence we get:

** Question:3 ** Verify .

** Answer: **

Given the matrix:

Let

Calculating the cofactors;

Hence,

Now,

aslo,

Now, ** calculating |A|; **

** So, **

** Hence we get **

** **

** Question:4 ** Verify .

** Answer: **

Given matrix:

Let

Calculating the cofactors;

Hence,

Now,

also,

Now, ** calculating |A|; **

** So, **

** Hence we get, **

** ** .

** Question:5 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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:

** Question:6 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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:

** Question:7 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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:8 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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:9 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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:10 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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:11 ** Find the inverse of each of the matrices (if it exists).

** Answer: **

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 .

** Answer: **

We have and .

then calculating;

** Finding the inverse of AB. **

Calculating the cofactors fo AB:

** **

** **

Then we have adj(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

** Answer: **

Given then we have to show the relation

So, calculating each term;

therefore ;

Hence .

[ ** Post multiplying by , also ] **

** Question:14 ** For the matrix , find the numbers and such that ** . **

** Answer: **

Given then we have the relation

So, calculating each term;

therefore ;

So, we have equations;

and

We get .

** Question:15 ** For the matrix Show that Hence, find .

** Answer: **

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 .

** Answer: **

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 : **

** Question:17 ** Let A be a nonsingular square matrix of order . Then is equal to

** Answer: **

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. **

** Question:18 ** If A is an invertible matrix of order 2, then det is equal to

** Answer: **

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. **

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

** Question:1 ** Examine the consistency of the system of equations.

** Answer: **

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. **

** Question:2 ** Examine the consistency of the system of equations

** Answer: **

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. **

** Question:3 ** Examine the consistency of the system of equations.

** Answer: **

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. **

** Question:4 ** Examine the consistency of the system of equations.

** Answer: **

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. **

** Question:5 ** Examine the consistency of the system of equations.

** Answer: **

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. **

** Question:6 ** Examine the consistency of the system of equations.

** Answer: **

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 non- singular matrix therefore there exist .

** Hence, the given system of equations is consistent. **

** Question:7 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

, and

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

So, the solutions can be found by

Hence the solutions of the given system of equations;

** x = 2 and y =-3 ** .

** Question:8 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

, and

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:9 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

, and

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:10 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

, and

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:11 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

, and

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

Now, we will find the cofactors;

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:12 ** ** ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

,

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

Now, we will find the cofactors;

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:13 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

,

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

Now, we will find the cofactors;

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:14 ** Solve system of linear equations, using matrix method.

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

,

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

Now, we will find the cofactors;

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Question:15 ** If , find . Using solve the system of equations

** Answer: **

The given system of equations

can be written in the matrix form of AX =B, where

,

we have,

.

So, A is non-singular, Therefore, its inverse exists.

as we know

Now, we will find the cofactors;

So, the solutions can be found by

Hence the solutions of the given system of equations;

** Answer: **

So, let us assume the cost of onion, wheat, and rice be ** x ** , ** y ** and ** z ** respectively.

Then we have the equations for the given situation :

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix .

We have;

Now, we will find the cofactors of A;

Now we have ** adjA; **

s

So, the solutions can be found by

Hence the solutions of the given system of equations;

Therefore, we have the cost of onions is ** Rs. 5 ** per Kg, the cost of wheat is ** Rs. 8 ** per Kg, and the cost of rice is ** Rs. ** ** 8 ** per kg.

## ** NCERT solutions for class 12 maths chapter 4 Determinants: Miscellaneous exercise **

** Question:1 ** Prove that the determinant is independent of .

** Answer: **

Calculating the determinant value of ;

** Clearly, the determinant is independent of . **

** Question:2 ** Without expanding the determinant, prove that

** Answer: **

We have the

Multiplying rows with a, b, and c respectively.

we get;

= ** R.H.S. **

** Hence proved. L.H.S. =R.H.S. **

** Question:4 ** If and are real numbers, and

** Answer: **

We have given

Applying the row transformations; we have;

Taking out common factor 2(a+b+c) from the first row;

Now, applying the column transformations;

we have;

and given that the determinant is equal to zero. i.e., ;

So, either or .

we can write as;

are non-negative.

Hence .

we get then

Therefore, if given = 0 then either or .

** Question:5 ** Solve the equation

** Answer: **

Given determinant

Applying the row transformation; we have;

Taking common factor (3x+a) out from first row.

Now applying the column transformations; and .

we get;

as ,

or or

** Question:6 ** Prove that .

** Answer: **

Given matrix

Taking common factors a,b and c from the column respectively.

we have;

Applying , we have;

Then applying , we get;

Applying , we have;

Now, applying column transformation; , we have

So we can now expand the remaining determinant along we have;

** Hence proved. **

** Question:7 ** If and , find .

** Answer: **

We know from the identity that;

.

Then we can find easily,

Given and

Then we have to basically find the matrix.

So, Given matrix

Hence its inverse exists;

Now, as we know that

So, calculating cofactors of B,

Now, We have both as well as ;

Putting in the relation we know;

** Question:8(i) ** Let . Verify that,

** Answer: **

Given that ;

So, let us assume that matrix and then;

Hence its inverse exists;

or ;

so, we now calculate the value of

Cofactors of A;

Finding the inverse of C;

Hence its inverse exists;

Now, finding the ;

or

Now, finding the ** R.H.S. **

Cofactors of B;

** Hence L.H.S. = R.H.S. proved. **

** Question:8(ii) ** Let , Verify that

** Answer: **

Given that ;

So, let us assume that

Hence its inverse exists;

or ;

so, we now calculate the value of

Cofactors of A;

Finding the inverse of B ;

Hence its inverse exists;

Now, finding the ;

** Hence proved L.H.S. =R.H.S. ** ** . **

** Question:9 ** Evaluate

** Answer: **

We have determinant

Applying row transformations; , we have then;

Taking out the common factor ** 2(x+y) ** from the row first.

Now, applying the column transformation; and we have ;

Expanding the remaining determinant;

.

** Question:10 ** Evaluate

** Answer: **

We have determinant

Applying row transformations; and then we have then;

Taking out the common factor ** -y ** from the row first.

Expanding the remaining determinant;

** Question:11 ** Using properties of determinants, prove that

** Answer: **

Given determinant

Applying Row transformations; and , then we have;

Expanding the remaining determinant;

** hence the given result is proved. **

** Question:12 ** Using properties of determinants, prove that

** Answer: **

Given the determinant

Applying the row transformations; and then we have;

Applying row transformation we have then;

Now we can expand the remaining determinant to get the result;

** hence the given result is proved. **

** Question:13 ** Using properties of determinants, prove that

** Answer: **

Given determinant

Applying the column transformation, we have then;

Taking common factor ** (a+b+c) ** out from the column first;

Applying and , we have then;

Now we can expand the remaining determinant along we have;

** Hence proved. **

** Question:14 ** Using properties of determinants, prove that

** Answer: **

Given determinant

Applying the row transformation; and we have then;

Now, applying another row transformation we have;

We can expand the remaining determinant along , we have;

** Hence the result is proved. **

** Question:15 ** Using properties of determinants, prove that

** Answer: **

Given determinant

Multiplying the first column by and the second column by , and expanding the third column, we get

Applying column transformation, we have then;

Here we can see that two columns are identical.

The determinant value is equal to zero.

** Hence proved. **

** Question:16 ** Solve the system of equations

** Answer: **

We have a system of equations;

So, we will convert the given system of equations in a simple form to solve the problem by the matrix method;

Let us take, ,

Then we have the equations;

We can write it in the matrix form as , where

Now, Finding the determinant value of A;

Hence we can say that A is non-singular its invers exists;

Finding cofactors of A;

, ,

, ,

, ,

as we know

Now we will find the solutions by relation .

Therefore we have the solutions

Or in terms of ** x, y, and z; **

** Question:17 ** Choose the correct answer.

** ** If are in A.P, then the determinant

is

** Answer: **

Given determinant and given that a, b, c are in A.P.

That means , ** 2b =a+c **

Applying the row transformations, and then we have;

Now, applying another row transformation, , we have

Clearly we have the determinant value equal to zero;

** Hence the option (A) is correct. **

** Question:18 ** Choose the correct answer.

** ** If x, y, z are nonzero real numbers, then the inverse of matrix is

** Answer: **

Given Matrix ,

As we know,

So, we will find the ,

Determining its cofactor first,

Hence

** Therefore the correct answer is (A) **

** Question:19 ** Choose the correct answer.

** Answer: **

Given determinant

Now, given the range of from

Therefore the .

** Hence the correct answer is D. **

** NCERT solutions for class 12 maths - Chapter wise **

** NCERT solutions for class 12 subject wise **

** NCERT Solutions class wise **

**NCERT solutions for class 12****NCERT solutions for class 11****NCERT solutions for class 10****NCERT solutions for class 9**

Have you thought about the two questions in the first paragraph? Well, here are the answers.

i) What does it really mean if determinant is negative or zero?

Answer: As you have learnt in the NCERT solutions for class 12 maths chapter 4 determinants article, that determinant is the factor by which volume is blow up. The determinant of the matrix is zero means it has flattened the volume to zero by p rojecting everything on a plane.

The determinant of a matrix is negative means it turns everything inside out. Reflection has determinant -1 as the volume is unchanged.

ii) W hy is the inverse of the matrix is not possible if its determinant is zero?

Answer: Imagine you have a box (3*3 matrix), after the transformation you will get all points single line as the determinant of the matrix is zero. So reverse transformation is not possible as everything collapsed on a single line. The inverse of the matrix is not possible if its determinant is zero. You will see such questions in the solutions of NCERT class 12 maths chapter 4 determinant article, where an inverse of matrix doesn't exist.

** Happy learning!!! **

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 12 Maths Chapter 4 Determinants

**Question: **What are the important topics in chapter determinants ?

**Answer: **

Determinant of a matrix of order upto three, properties of determinants, area of a triangle, minors and co-factors, adjoint and inverse of a matrix, applications of determinants and matrices, and solution of a system of linear equations using the inverse of a matrix are the important topics from this chapter.

**Question: **What is the weightage of the chapter determinants for CBSE board exam ?

**Answer: **

The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12^{th }board final examination.

**Question: **How the NCERT solutions are helpful in the board exam ?

**Answer: **

Only knowing the answer is not guarantee to score good marks in the exam. One should know how to answer in order to get good marks. NCERT solutions are provided by the experts who knows how best to write answer in the board exam in order to get good marks in the board exam.

**Question: **Which is the best book for CBSE class 12 maths ?

**Answer: **

NCERT textbook is the best book for CBSE class 12 maths. Most of the questions in CBSE class 12 board exam are directly asked from NCERT textbook. So you don't need to buy any supplementary books for CBSE class 12 maths.

**Question: **Does CBSE provides the solutions of NCERT class 12 maths ?

**Answer: **

No, CBSE doesn’t provided NCERT solutions for any class or subject.

**Question: **Where can I find the complete solutions of NCERT class 12 maths ?

**Answer: **

A Here you will get the detailed NCERT solutions for class 12 maths by clicking on the link.

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