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NCERT solutions for Class 12 Maths Chapter 4 Determinants
Edited By Ramraj Saini | Updated on Sep 13, 2023 09:18 PM IST | #CBSE Class 12th
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NCERT Determinants Class 12 Questions And Answers
NCERT solutions for Class 12 Maths Chapter 4 Determinants are proved here. These NCERT solutions are created by expert team at Careers360 keeping align the latest syllabus of CBSE 2023-24. In this chapter, students will be able to understand the Class 12 Maths Chapter 4 NCERT solutions. If you multiply a matrix with the coordinates of a point, it will give a new point in the space which is explained in NCERT class 12 chapter 4 maths Determinants solutions. In this sense, the matrix is a linear transformation. The determinant of the matrix is the factor by which its volume blows up. You will be familiar with these points after going through ch 4 maths class 12. Interested students can visit chapter wise NCERT solution for math.
This Story also Contains
- NCERT Determinants Class 12 Questions And Answers
- NCERT Determinants Class 12 Questions And Answers PDF Free Download
- NCERT Class 12 Maths Chapter 4 Question Answer - Important Formulae
- NCERT Class 12 Maths Chapter 4 Question Answer (Intext Questions and Exercise)
- NCERT determinants class 12 solutions: Excercise: 4.5
- NCERT solutions for class 12 Maths - Chapter wise
- NCERT solutions for class 12 - subject wise
- NCERT Solutions - Class Wise
- NCERT Books and NCERT Syllabus
The important topics of class 12 maths ch 4 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 NCERT solutions for Class 12 Maths Chapter 4 Determinants. If you are looking for determinants class 12 solutions then check all NCERT solutions at a single place which will help the students to learn CBSE maths. Here you will get NCERT solutions for class 12 also. Read further to know more about NCERT solutions for Class 12 Maths Chapter 4 PDF download.
Also read:
- Class 12 Maths Chapter 4 Determinants Notes
- NCERT Exemplar Solutions For Class 12 Maths Chapter 4 Determinants
NCERT Determinants Class 12 Questions And Answers PDF Free Download
NCERT Class 12 Maths Chapter 4 Question Answer - Important Formulae
>> Determinant of a Matrix: The determinant is the numerical value of a square matrix.
For a square matrix A of order n, the determinant is denoted by det A or |A|.
Minor and Cofactor of a Matrix:
Minor of an element aij of a determinant is a determinant obtained by deleting the ith row and jth column in which element aij lies.
The cofactor of an element aij of a determinant, denoted by Aij or Cij, is defined as Aij = (-1)(i+j) * Mij, where Mij is the minor of element aij.
Value of a Determinant (2x2 and 3x3 matrices):
For a 2x2 matrix A: |A| = a11 * a22 - a21 * a12
For a 3x3 matrix A: |A| = a11 * |A11| - a12 * |A12| + a13 * |A13|
Singular and Non-Singular Matrix:
If the determinant of a square matrix is zero, the matrix is said to be singular; otherwise, it is non-singular.
Determinant Theorems:
If A and B are non-singular matrices of the same order, then AB and BA are also non-singular matrices of the same order.
The determinant of the product of matrices is equal to the product of their respective determinants, i.e., |AB| = |A| * |B|.
Adjoint of a Matrix:
The adjoint of a square matrix A is the transpose of the matrix obtained by cofactors of each element of the determinant corresponding to A. It is denoted by adj(A).
In general, the adjoint of a matrix A = [aij]n×n is a matrix [Aji]n×n, where Aji is a cofactor of element aji.
Properties of Adjoint of a Matrix:
A(adj A) = (adj A)A = |A|In (Identity Matrix)
|adj A| = |A|(n-1)
adj(AT) = (adj A)T (Transpose of the adjoint)
Finding Area of a Triangle Using Determinants:
The area of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by
Inverse of a Square Matrix:
For a non-singular matrix A (|A| ≠ 0), the inverse A-1 is defined as A-1 = (1/|A|) * adj(A).
Properties of an Inverse Matrix:
(A-1)-1 = A
(AT)-1 = (A-1)T
(AB)-1 = B-1A-1
(ABC)-1 = C-1B-1A-1
adj(A-1) = (adj A)-1
Solving a System of Linear Equations using Inverse of a Matrix:
Given a system of equations AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
Case I: If |A| ≠ 0, the system is consistent, and X = A-1B has a unique solution.
Case II: If |A| = 0 and (adj A)B ≠ 0, the system is inconsistent and has no solution.
Case III: If |A| = 0 and (adj A)B = 0, the system may be either consistent or inconsistent, depending on whether it has infinitely many solutions or no solutions.
Free download Class 12 Determinants NCERT Solutions for CBSE Exam.
NCERT Class 12 Maths Chapter 4 Question Answer (Intext Questions and Exercise)
NCERT determinants class 12 questions and answers: Excercise- 4.1
Question:1 Evaluate the following determinant-
Answer:
The determinant is evaluated as follows
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 determinants class 12 questions and answers: 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:
Answer:
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 class 12 maths chapter 4 question answer: 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 class 12 maths chapter 4 question answer: 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 determinants class 12 solutions: 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 determinants class 12 ncert solutions: 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.
If you are interested in Determinants Class 12 NCERT Solutions exercises then these are listed below.
- Determinants Class 12 NCERT Solutions Exercise 4.1
- Determinants Class 12 NCERT Solutions Exercise 4.2
- Determinants Class 12 NCERT Solutions Exercise 4.3
- Determinants Class 12 NCERT Solutions Exercise 4.4
- Determinants Class 12 NCERT Solutions Exercise 4.5
- Determinants Class 12 NCERT Solutions Exercise 4.6
- Determinants Class 12 NCERT Solutions Miscellaneous Exercise
An insight to the NCERT solutions for Class 12 Maths Chapter 4 Determinants:
The six exercises of NCERT Class 12 Maths solutions chapter 4 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. You can also check Determinants NCERT solutions if you are facing any problems during practice.
What are the Determinants?
To every square matrix ![A=\left [ a_{ij} \right ]](https://cache.careers360.mobi/media/articles/uploads/froala_editor/images/2022/4/14/1649937226771.png)
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 × 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
Also read,
NCERT exemplar solutions class 12 maths chapter 4
Topics of NCERT Class 12 Maths Chapter Determinants
The main topics covered in chapter 4 maths class 12 are:
Determinants
Ch 4 maths class 12 includes concepts of calculation of determinants with respect to their order one, two, three. Also class 12 NCERT topics discuss concepts related to the expansion of the matrix to calculate the determinant. there are good quality questions in Determinants class 12 solutions.
Properties of determinants
This ch 4 maths class 12 comprehensively and elaborately discussed the properties of determinants, which are vastly used. To get a good hold of these concepts you can refer to NCERT solutions for class 12 maths chapter 4.
Area of triangle
This ch 4 maths class 12 also includes concepts of the area of a triangle in which vertices are given. You can refer to class 12 NCERT solutions for questions about these concepts.
- Minors and Cofactors
Maths class 12 chapter 4 discussed the minors and cofactors. To get command of these concepts you can go through the NCERT solution for class 12 maths chapter 4.
Adjoint and Inverse of a matrix
concepts related to adjoints and inverse of the matrix are detailed in maths class 12 chapter 4. And it also concerns conditions for the existence of the inverse of a matrix. Determinants class 12 solutions include quality questions to understand the concepts.
- Applications of determinants and matrix
ch 4 maths class 12 deliberately discussed the applications of determinants and matrices. it also includes the terms consistent system inconsistent system. concepts related to the solution of a system of linear equations using the inverse of a matrix. For questions on these concepts, you can browse NCERT solutions for class 12 chapter 4.
Topics mentioned in class 12 NCERT are very important and students are suggested to go through all the concepts discussed in the topics. Questions related to all the above topics are covered in the NCERT solutions for class 12 maths chapter 4
- NCERT Exemplar Class 12 Chemistry Solutions
- NCERT Exemplar Class 12 Mathematics Solutions
- NCERT Exemplar Class 12 Biology Solutions
- NCERT Exemplar Class 12 Physics Solutions
NCERT solutions for class 12 Maths - Chapter wise
NCERT solutions for class 12 - subject wise
- NCERT solutions for class 12 mathematics
- NCERT solutions class 12 chemistry
- NCERT solutions for class 12 physics
- NCERT solutions for class 12 biology
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Download EBookNCERT Solutions - Class Wise
- NCERT solutions for class 12
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Benefits of NCERT solutions for Class 12 Maths Chapter 4 Determinants:
NCERT Class 12 Maths solutions chapter 4 will assist the students in the exam preparation in a strategic way.
Class 12 Maths Chapter 4 NCERT solutions are prepared by the experts, therefore, students can rely upon the same without any second thought .
NCERT solutions for Class 12 Maths Chapter 4 provides the detailed solution for all the questions. This will help the students in analysing and understanding the questions in a better way.
NCERT Books and NCERT Syllabus
Frequently Asked Questions (FAQs)
1. What are the key themes discussed in chapter 4 maths class 12 ncert solutions?
NCERT Solutions for Class 12 Maths Chapter 4 primarily focuses on the topic of determinants. This chapter covers the following key themes:
Definition of determinants
Properties of determinants
Area of a parallelogram and a triangle
The inverse of a matrix
Adjoint and inverse of a matrix
Solutions of linear equations using matrices
Determinant as scaling factor
2. What is the weightage of the chapter determinants for CBSE board exam?
The topic algebra which contains two topics matrices and determinants which has 13 % weightage in the maths CBSE 12th board final examination. students can prioritise their subjects according to respective weightage and study accordingly.
3. How are the NCERT solutions helpful in the board exam?
Only knowing the answer does 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 know how best to write answers in the board exam in order to get good marks in the board exam.
4. Which is the best book for CBSE class 12 Maths ?
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.
5. What applications of determinants are discussed in ncert solutions class 12 maths chapter 4?
According to NCERT Solutions for Class 12 Maths Chapter 4, determinants play a crucial role in algebra and have multiple practical applications. The concept of determinants is valuable in solving systems of linear equations. With determinants, students can explore concepts such as changes in area, volume, and variables through integrals. Additionally, determinants can be used to determine the values of square matrices. Interested students can study determinants class 12 ncert pdf both online and offline.
6. Where can I find the complete solutions of NCERT class 12 Maths ?
Here you will get the detailed NCERT solutions for class 12 maths by clicking on the link. also you can find these in official web page of careers360.
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12 min ⋆ Feb 11, 2025 10:02 AM ISTQuestions related to CBSE Class 12th
Have a question related to CBSE Class 12th ?
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Manasvini Gupta 30 January,2025
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
Jefferson 25 September,2024
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
-
Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
-
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
-
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
-
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
-
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.
Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.
I hope this information helps you.
Kanishka kaushikii 17 September,2024
Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
Type C: For candidates scoring between 40% and 50%.
Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
Yuvan 30 August,2024
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
Ashvadha 12 July,2024
Popular CBSE Class 12th Questions
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
In the reaction,
| Option 1)
|
Option 2)
|
| Option 3)
|
Option 4)
|
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
| Option 1)
0.02 |
Option 2)
3.125 × 10-2 |
| Option 3)
1.25 × 10-2 |
Option 4)
2.5 × 10-2 |
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