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NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 12 - Determinants

Edited By Komal Miglani | Updated on Apr 24, 2025 01:22 PM IST | #CBSE Class 12th

Imagine you are asked to solve different types of problems, like finding the area of a triangle using a determinant, checking if a system of equations has a solution, or finding the inverse of a matrix. How do you know which method to use? The Miscellaneous Exercise of NCERT Class 12 Maths Chapter 4 – Determinants brings together all the important topics from the chapter. It includes questions on: finding the value of determinants, using properties of determinants to simplify problems, calculating the area of a triangle using coordinates, solving systems of linear equations, finding the adjoint and inverse of a matrix. Miscellaneous Exercise of NCERT Class 12 Maths Chapter 4 helps you revise everything you have learned and tests your overall understanding. It is great for practice before exams. This article on the NCERT Solutions for miscellaneous exercise Class 12 Maths Chapter 4 - Determinant, offers clear and step-by-step solutions for the exercise problems to help the students understand the method and logic behind it. For syllabus, notes, and PDF, refer to this link: NCERT.

Class 12 Maths Chapter 4 Miscellaneous Exercise Solutions: Download PDF

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Determinants Miscellaneous Exercise

Question:1 Prove that the determinant |xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x| is independent of θ.

Answer:

Calculating the determinant value of |xsin⁡θcos⁡θ−sin⁡θ−x1cos⁡θ1x|;

=x[−x11x]−sin⁡Θ[−sin⁡Θ1cos⁡Θx]+cos⁡Θ[−sin⁡Θ−xcos⁡Θ1]

=x(−x2−1)−sin⁡Θ(−xsin⁡Θ−cos⁡Θ)+cos⁡Θ(−sin⁡Θ+xcos⁡Θ)

=−x3−x+xsin2⁡Θ+sin⁡Θcos⁡Θ−cos⁡Θsin⁡Θ+xcos2⁡Θ

=−x3−x+x(sin2⁡Θ+cos2⁡Θ)

=−x3−x+x=−x3

Clearly, the determinant is independent of Θ.

Question:2 Without expanding the determinant, prove that
|aa2bcbb2cacc2ab|=|1a2a31b2b31c2c3|

Answer:

We have the

L.H.S.=|aa2bcbb2cacc2ab|

Multiplying rows with a, b, and c respectively.

R1→aR1,R2→bR2, and R3→cR3

we get;

=1abc|a2a3abcb2b3abcc2c3abc|

=1abc.abc|a2a31b2b31c2c31| [after taking out abc from column 3].

=|a2a31b2b31c2c31|=|1a2a31b2b31c2c3| [Applying C1↔C3 and C2↔C3]

= R.H.S.

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

Question:3 Evaluate |cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α|.

Answer:

Given determinant |cos⁡αcos⁡βcos⁡αsin⁡β−sin⁡α−sin⁡βcos⁡β0sin⁡αcos⁡βsin⁡αsin⁡βcos⁡α|;

=cos⁡αcos⁡β|cos⁡β0sin⁡αsin⁡βcos⁡α|−cos⁡αsin⁡β|−sin⁡β0sin⁡αcos⁡βcos⁡α|−sin⁡α|−sin⁡βcos⁡βsin⁡αcos⁡βsin⁡αsin⁡β|=cos⁡αcos⁡β(cos⁡βcos⁡α−0)−cos⁡αsin⁡β(−cos⁡αsin⁡β−0)−sin⁡α(−sin⁡αsin2⁡β−sin⁡αcos2⁡β)

=cos2⁡αcos2⁡β+cos2⁡αsin2⁡β+sin2⁡αsin2⁡β+sin2⁡αcos2⁡β

=cos2⁡α(cos2⁡β+sin2⁡β)+sin2⁡α(sin2⁡β+cos2⁡β)

=cos2⁡α(1)+sin2⁡α(1)=1.

Question 4: If A−1=[3−11−156−55−22] and B=[12−2−1300−21], find (AB)−1.
Answer:

We know from the identity that:

(AB)−1=B−1A−1

Then we can find easily,

Given

Given
A−1=[3−11 −156−5 5−22],
B=[12−2 −130 0−21]

Then we have to basically find the B−1 matrix.

So, Given matrix B=[12−2−1300−21]

|B|=1(3−0)−2(−1−0)−2(2−0)=3+2−4=1≠0

Hence its inverse B−1 exists;

Now, as we know that

B−1=1|B|adjB

So, calculating cofactors of B,

B11=(−1)1+1(3−0)=3 B12=(−1)1+2(−1−0)=1

B13=(−1)1+3(2−0)=2 B21=(−1)2+1(2−4)=2

B22=(−1)2+2(1−0)=1 B23=(−1)2+3(−2−0)=2

B31=(−1)3+1(0+6)=6 B32=(−1)3+2(0−2)=2

B33=(−1)3+3(3+2)=5

adjB=[326112225]

B−1=1|B|adjB=11[326112225]

Now, We have both A−1 as well as B−1 ;

Putting in the relation we know; (AB)−1=B−1A−1

(AB)−1=11[326112225][3−11−156−55−22]

=[9−30+30−3+12−123−10+123−15+10−1+6−41−5+46−30+25−2+12−102−10+10]

=[9−35−210102]

Question 5(i): Let A=[121231115]. Verify that 100[adjA]−1=adj(A−1).
Answer:

Given that A=[121231115];

So, let us assume that A−1=B matrix and adjA=C then;

|A|=1(15−1)−2(10−1)+1(2−3)=14−18−1=−5≠0

Hence its inverse exists;

A−1=1|A|adjA or B=1|A|C;

so, we now calculate the value of adjA

Cofactors of A;

A11=(−1)1+1(15−1)=14 A12=(−1)1+2(10−1)=−9

A13=(−1)1+3(2−3)=−1 A21=(−1)2+1(10−1)=−9

A22=(−1)2+2(5−1)=4 A23=(−1)2+3(1−2)=1

A31=(−1)3+1(2−3)=−1 A32=(−1)3+2(1−2)=1

A33=(−1)3+3(3−4)=−1

⇒adjA=C=[14−9−1−941−11−1]

A−1=B=1|A|adjA=1−5[14−9−1−941−11−1]

Finding the inverse of C;

|C|=14(−4−1)+9(9+1)−1(−9+4)=−70+90+5=25≠0

Hence its inverse exists;

C−1=1|C|adjC

Now, finding the adjC;

C11=(−1)1+1(−4−1)=−5 C12=(−1)1+2(9+1)=−10

C13=(−1)1+3(−9+4)=−5 C21=(−1)2+1(9+1)=−10

C22=(−1)2+2(−14−1)=−15 C23=(−1)2+3(14−9)=−5

C31=(−1)3+1(−9+4)=−5 C32=(−1)3+2(14−9)=−5

C33=(−1)3+3(56−81)=−25

adjC=[−5−10−5−10−15−5−5−5−25]

C−1=1|C|adjC=125[−5−10−5−10−15−5−5−5−25]=[−15−25−15−25−35−15−15−15−1]

or L.H.S.=C−1=[adjA]−1=[−15−25−15−25−35−15−15−15−1]

Now, finding the R.H.S.

adj(A−1)=adjB

A−1=B=[−145951595−45−1515−1515]

Cofactors of B;

B11=(−1)1+1(−425−125)=−15

B12=(−1)1+2(925+125)=−25

B13=(−1)1+3(−925+425)=−15

B21=(−1)2+1(925+125)=−25

B22=(−1)2+2(−1425−125)=−35

B23=(−1)2+3(1425−925)=−15

B31=(−1)3+1(−925+425)=−15

B32=(−1)3+2(1425−925)=−15

B33=(−1)3+3(5625−8125)=−1

R.H.S.=adjB=adj(A−1)=[−15−25−15−25−35−15−15−15−1]


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

Question 5(ii):Let A=[121231115], verify that (A−1)−1=A.

Answer:

Given that A=[121231115];

So, let us assume that A−1=B

|A|=1(15−1)−2(10−1)+1(2−3)=14−18−1=−5≠0

Hence its inverse exists;

A−1=1|A|adjA or B=1|A|C;

so, we now calculate the value of adjA

Cofactors of A;

A11=(−1)1+1(15−1)=14 A12=(−1)1+2(10−1)=−9

A13=(−1)1+3(2−3)=−1 A21=(−1)2+1(10−1)=−9

A22=(−1)2+2(5−1)=4 A23=(−1)2+3(1−2)=1

A31=(−1)3+1(2−3)=−1 A32=(−1)3+2(1−2)=1

A33=(−1)3+3(3−4)=−1

⇒adjA=C=[14−9−1−941−11−1]

A−1=B=1|A|adjA=1−5[14−9−1−941−11−1]=[−145951595−45−1515−1515]

Finding the inverse of B ;

|B|=−145(−425−125)−95(925+125)+15(−925+425)

=70125−90125−5125=−25125=−15≠0

Hence its inverse exists;

B−1=1|B|adjB

Now, finding the adjB;

A−1=B=1|A|adjA=1−5[14−9−1−941−11−1]=[−145951595−45−1515−1515]

B11=(−1)1+1(−425−125)=−15 B12=(−1)1+2(925+125)=−25

B13=(−1)1+3(−925+425)=−15 B21=(−1)2+1(925+125)=−25

B22=(−1)2+2(−1425−125)=−35 B23=(−1)2+3(1425−925)=−15

B31=(−1)3+1(−925+425)=−15

B32=(−1)3+2(1425−925)=−15

B33=(−1)3+3(5625−8125)=−2525=−1

adjB=[−15−25−15−25−35−15−15−15−1]

B−1=1|B|adjB=−51[−15−25−15−25−35−15−15−15−1]=[121231115]

L.H.S.=B−1=(A−1)−1=[121231115]

R.H.S.=A=[121231115]

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

Question:6 Evaluate |xyx+yyx+yxx+yxy|

Answer:

We have determinant △=|xyx+yyx+yxx+yxy|

Applying row transformations; R1→R1+R2+R3 , we have then;

△=|2(x+y)2(x+y)2(x+y)yx+yxx+yxy|

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

=2(x+y)|111yx+yxx+yxy|

Now, applying the column transformation; C1→C1−C2 and C2→C2−C1 we have ;

=2(x+y)|001−xyxyx−yy|

Expanding the remaining determinant;

=2(x+y)(−x(x−y)−y2)=2(x+y)[−x2+xy−y2]

=−2(x+y)[x2−xy+y2]=−2(x3+y3).

Question:7 Evaluate |1xy1x+yy1xx+y|

Answer:

We have determinant △=|1xy1x+yy1xx+y|

Applying row transformations; R1→R1−R2 and R2→R2−R3 then we have then;

△=|0−y00y−x1xx+y|

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

△=−y|0100y−x1xx+y|

Expanding the remaining determinant;

−y[1(−x−o)]=xy


Question:8 Solve the system of equations

2x+3y+10z=4

4x−6y+5z=1

6x+9y−20z=2

Answer:

We have a system of equations;

2x+3y+10z=4

4x−6y+5z=1

6x+9y−20z=2

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

Let us take, 1x=a, 1y=b and 1z=c

Then we have the equations;

2a+3b+10c=4

4a−6b+5c=1

6a+9b−20c=2

We can write it in the matrix form as AX=B , where

A=[23104−6569−20],X=[abc] and B=[412].

Now, Finding the determinant value of A;

|A|=2(120−45)−3(−80−30)+10(36+36)

=150+330+720

=1200≠0

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

Finding cofactors of A;

A11=75 , A12=110, A13=72

A21=150, A22=−100, A23=0

A31=75, A31=30, A33=−24

∴ as we know A−1=1|A|adjA

=11200[7515075110−10030720−24]

Now we will find the solutions by relation X=A−1B.

⇒[abc]=11200[7515075110−10030720−24][412]

=11200[300+150+150440−100+60288+0−48]

=11200[600400240]=[121315]

Therefore we have the solutions a=12, b=13, and c=15.

Or in terms of x, y, and z;

x=2, y=3, and z=5


Question 9: Choose the correct answer.

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

A=[x000y000z]

is

(A) [x−1000y−1000z−1]

(B) xyz[x−1000y−1000z−1]

(C) 1xyz[x000y000z]

(D) 1xyz[100010001]


Answer:

Given Matrix A=[x000y000z],

|A|=x(yz−0)=xyz

As we know,

A−1=1|A|adjA

So, we will find the adjA,

Determining its cofactor first,

A11=yz A12=0 A13=0

A21=0 A22=xz A23=0

A31=0 A32=0 A33=xy

Hence A−1=1|A|adjA=1xyz[yz000xz000xy]

A−1=[1x0001y0001z]

Therefore, the correct answer is (A)

Question:10 Choose the correct answer.

Let A=|1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1|, where 0≤θ≤2π. Then


(A)Det(A)=0 nbsp; (B) Det(A)∈(2,∞)

(C) Det(A)∈(2,4) (D)Det(A)∈[2,4]

Answer:

Given determinant A=|1sin⁡θ1−sin⁡θ1sin⁡θ−1−sin⁡θ1|

|A|=1(1+sin2⁡Θ)−sin⁡Θ(−sin⁡Θ+sin⁡Θ)+1(sin2⁡Θ+1)

=1+sin2⁡Θ+sin2⁡Θ+1

=2+2sin2⁡Θ=2(1+sin2⁡Θ)

Now, given the range of Θ from 0≤Θ≤2π

⇒0≤sin⁡Θ≤1

⇒0≤sin2⁡Θ≤1

⇒1≤1+sin2⁡Θ≤2

⇒2≤2(1+sin2⁡Θ)≤4

Therefore the |A| ϵ [2,4].

Hence, the correct answer is D.


Also read,

Topics Covered in Chapter 4, Determinants: Miscellaneous Exercise

Here are the main topics covered in NCERT Class 12 Chapter 4, Determinants: Miscellaneous Exercise.

1. Evaluation of Determinants: Practice of expanding 3×3 determinants along any row or column using minors and cofactors.

2. Properties of Determinants: Apply key properties like row/column interchange, scalar multiplication, and linear combinations to simplify determinants.

3. Area of a Triangle Using Determinants: Use coordinate geometry and the determinant formula to find the area of a triangle formed by three points.

4. Consistency and Inconsistency of Systems: Determine whether systems of linear equations have unique solutions, no solutions, or infinitely many solutions based on determinant values.

5. Adjoint and Inverse of a Matrix: Calculate the adjoint and use it to find the inverse of a square matrix, then verify it through matrix multiplication.

6. Solving Linear Equations Using Matrix Inverse: Apply the formula X=A−1B to find the solution of a system of equations expressed in matrix form.

Also, read,

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NCERT Solutions of Class 12 Subject Wise

Given below are some useful links for subject-wise NCERT solutions of class 12.

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Subject-Wise NCERT Exemplar Solutions

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Frequently Asked Questions (FAQs)

1. What is the value of determinant, If any two rows of a determinant are identical ?

The value of the determinant is zero if any two rows of a determinant are identical.

2. What is the value of determinant if any two rows of a determinant are proportional ?

The value of the determinant is zero If any two rows of a determinant are proportional.

3. If |A| = 3 and the first row matrix A is multiplied by 3 then what is the value of |A|?

If first row of matrix A is multiplied by 3 then the new value of |A| = 3x3 = 9.

4. Can you find the inverse of a singular matrix ?

No, singular matrices are non-invertible.

5. If transpose of matrix A and matrix A are equal then such matrix is called ?

If the transpose of matrix A and matrix A are equal then such matrix is called symmetric matrix.

6. What is the definition of diagonal matrix ?

The matrix has all the non-diagonal elements zero is called a diagonal matrix.

7. Do all the identity matrices are diagonal matrices ?

Yes,  all the identity matrices are diagonal matrices.

8. Do CBSE provides solutions for miscellaneous exercise ?

No, CBSE doesn't provide NCERT solutions for miscellaneous exercises.

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Manasvini Gupta 30 January,2025
quartely question paper for mathematics cbse

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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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Ashvadha 12 July,2024
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Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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