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. In this chapter, there are 6 exercises & a miscellaneous exercise.
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, 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 subtopics 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 chapter4 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).
CBSE 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: Excercise4.3
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.
Solutions of NCERT for class 12 maths chapter 4 DeterminantsExcercise: 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.
CBSE 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) = 40874089 = 2
Therefore we have inverse:
.....................................(1)
Now, calculating inverses of A and B.
A = 1514 = 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;
Postmultiplying 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;
Postmultiplying 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.
CBSE 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 nonzero.
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 nonzero 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 nonsingular, 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 nonsingular, 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 nonsingular, 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 nonsingular, 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 nonsingular, 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 nonsingular, 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 nonsingular, 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 nonsingular, 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 nonsingular, 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.
Solutions of NCERT 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 nonnegative.
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 nonsingular 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 chapterwise
chapter 1 
Solutions of NCERT for class 12 maths chapter 1 Relations and Functions 
chapter 2 
NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions 
chapter 3  
chapter 4 
Solutions of NCERT for class 12 maths chapter 4 Determinants 
chapter 5 
NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability 
chapter 6 
CBSE NCERT solutions for class 12 maths chapter 6 Application of Derivatives 
chapter 7 

chapter 8 
NCERT solutions for class 12 maths chapter 8 Application of Integrals 
chapter 9 
CBSE NCERT solutions for class 12 maths chapter 9 Differential Equations 
chapter 10 
Solutions of NCERT for class 12 maths chapter 10 Vector Algebra 
chapter 11 
NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry 
chapter 12 
CBSE NCERT solutions for class 12 maths chapter 12 Linear Programming 
chapter 13 
Solutions of NCERT for class 12 maths chapter 13 Probability 
NCERT solutions for class 12
NCERT solutions
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!!!