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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 |xsinθcosθsinθx1cosθ1x| is independent of θ.

Answer:

Calculating the determinant value of |xsinθcosθsinθx1cosθ1x|;

=x[x11x]sinΘ[sinΘ1cosΘx]+cosΘ[sinΘxcosΘ1]

=x(x21)sinΘ(xsinΘcosΘ)+cosΘ(sinΘ+xcosΘ)

=x3x+xsin2Θ+sinΘcosΘcosΘsinΘ+xcos2Θ

=x3x+x(sin2Θ+cos2Θ)

=x3x+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.

R1aR1,R2bR2, and R3cR3

we get;

=1abc|a2a3abcb2b3abcc2c3abc|

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

=|a2a31b2b31c2c31|=|1a2a31b2b31c2c3| [Applying C1C3 and C2C3]

= R.H.S.

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

Question:3 Evaluate |cosαcosβcosαsinβsinαsinβcosβ0sinαcosβsinαsinβcosα|.

Answer:

Given determinant |cosαcosβcosαsinβsinαsinβcosβ0sinαcosβsinαsinβcosα|;

=cosαcosβ|cosβ0sinαsinβcosα|cosαsinβ|sinβ0sinαcosβcosα|sinα|sinβcosβsinαcosβsinαsinβ|=cosαcosβ(cosβcosα0)cosαsinβ(cosαsinβ0)sinα(sinαsin2βsinαcos2β)

=cos2αcos2β+cos2αsin2β+sin2αsin2β+sin2αcos2β

=cos2α(cos2β+sin2β)+sin2α(sin2β+cos2β)

=cos2α(1)+sin2α(1)=1.

Question 4: If A1=[3111565522] and B=[122130021], find (AB)1.
Answer:

We know from the identity that:

(AB)1=B1A1

Then we can find easily,

Given

Given
A1=[311 1565 522],
B=[122 130 021]

Then we have to basically find the B1 matrix.

So, Given matrix B=[122130021]

|B|=1(30)2(10)2(20)=3+24=10

Hence its inverse B1 exists;

Now, as we know that

B1=1|B|adjB

So, calculating cofactors of B,

B11=(1)1+1(30)=3 B12=(1)1+2(10)=1

B13=(1)1+3(20)=2 B21=(1)2+1(24)=2

B22=(1)2+2(10)=1 B23=(1)2+3(20)=2

B31=(1)3+1(0+6)=6 B32=(1)3+2(02)=2

B33=(1)3+3(3+2)=5

adjB=[326112225]

B1=1|B|adjB=11[326112225]

Now, We have both A1 as well as B1 ;

Putting in the relation we know; (AB)1=B1A1

(AB)1=11[326112225][3111565522]

=[930+303+1212310+12315+101+6415+4630+252+1210210+10]

=[935210102]

Question 5(i): Let A=[121231115]. Verify that 100[adjA]1=adj(A1).
Answer:

Given that A=[121231115];

So, let us assume that A1=B matrix and adjA=C then;

|A|=1(151)2(101)+1(23)=14181=50

Hence its inverse exists;

A1=1|A|adjA or B=1|A|C;

so, we now calculate the value of adjA

Cofactors of A;

A11=(1)1+1(151)=14 A12=(1)1+2(101)=9

A13=(1)1+3(23)=1 A21=(1)2+1(101)=9

A22=(1)2+2(51)=4 A23=(1)2+3(12)=1

A31=(1)3+1(23)=1 A32=(1)3+2(12)=1

A33=(1)3+3(34)=1

adjA=C=[1491941111]

A1=B=1|A|adjA=15[1491941111]

Finding the inverse of C;

|C|=14(41)+9(9+1)1(9+4)=70+90+5=250

Hence its inverse exists;

C1=1|C|adjC

Now, finding the adjC;

C11=(1)1+1(41)=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(141)=15 C23=(1)2+3(149)=5

C31=(1)3+1(9+4)=5 C32=(1)3+2(149)=5

C33=(1)3+3(5681)=25

adjC=[5105101555525]

C1=1|C|adjC=125[5105101555525]=[15251525351515151]

or L.H.S.=C1=[adjA]1=[15251525351515151]

Now, finding the R.H.S.

adj(A1)=adjB

A1=B=[1459515954515151515]

Cofactors of B;

B11=(1)1+1(425125)=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(1425125)=35

B23=(1)2+3(1425925)=15

B31=(1)3+1(925+425)=15

B32=(1)3+2(1425925)=15

B33=(1)3+3(56258125)=1

R.H.S.=adjB=adj(A1)=[15251525351515151]


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

Question 5(ii):Let A=[121231115], verify that (A1)1=A.

Answer:

Given that A=[121231115];

So, let us assume that A1=B

|A|=1(151)2(101)+1(23)=14181=50

Hence its inverse exists;

A1=1|A|adjA or B=1|A|C;

so, we now calculate the value of adjA

Cofactors of A;

A11=(1)1+1(151)=14 A12=(1)1+2(101)=9

A13=(1)1+3(23)=1 A21=(1)2+1(101)=9

A22=(1)2+2(51)=4 A23=(1)2+3(12)=1

A31=(1)3+1(23)=1 A32=(1)3+2(12)=1

A33=(1)3+3(34)=1

adjA=C=[1491941111]

A1=B=1|A|adjA=15[1491941111]=[1459515954515151515]

Finding the inverse of B ;

|B|=145(425125)95(925+125)+15(925+425)

=70125901255125=25125=150

Hence its inverse exists;

B1=1|B|adjB

Now, finding the adjB;

A1=B=1|A|adjA=15[1491941111]=[1459515954515151515]

B11=(1)1+1(425125)=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(1425125)=35 B23=(1)2+3(1425925)=15

B31=(1)3+1(925+425)=15

B32=(1)3+2(1425925)=15

B33=(1)3+3(56258125)=2525=1

adjB=[15251525351515151]

B1=1|B|adjB=51[15251525351515151]=[121231115]

L.H.S.=B1=(A1)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; R1R1+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; C1C1C2 and C2C2C1 we have ;

=2(x+y)|001xyxyxyy|

Expanding the remaining determinant;

=2(x+y)(x(xy)y2)=2(x+y)[x2+xyy2]

=2(x+y)[x2xy+y2]=2(x3+y3).

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

Answer:

We have determinant =|1xy1x+yy1xx+y|

Applying row transformations; R1R1R2 and R2R2R3 then we have then;

=|0y00yx1xx+y|

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

=y|0100yx1xx+y|

Expanding the remaining determinant;

y[1(xo)]=xy


Question:8 Solve the system of equations

2x+3y+10z=4

4x6y+5z=1

6x+9y20z=2

Answer:

We have a system of equations;

2x+3y+10z=4

4x6y+5z=1

6x+9y20z=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

4a6b+5c=1

6a+9b20c=2

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

A=[23104656920],X=[abc] and B=[412].

Now, Finding the determinant value of A;

|A|=2(12045)3(8030)+10(36+36)

=150+330+720

=12000

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 A1=1|A|adjA

=11200[75150751101003072024]

Now we will find the solutions by relation X=A1B.

[abc]=11200[75150751101003072024][412]

=11200[300+150+150440100+60288+048]

=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) [x1000y1000z1]

(B) xyz[x1000y1000z1]

(C) 1xyz[x000y000z]

(D) 1xyz[100010001]


Answer:

Given Matrix A=[x000y000z],

|A|=x(yz0)=xyz

As we know,

A1=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 A1=1|A|adjA=1xyz[yz000xz000xy]

A1=[1x0001y0001z]

Therefore, the correct answer is (A)

Question:10 Choose the correct answer.

Let A=|1sinθ1sinθ1sinθ1sinθ1|, where 0θ2π. Then


(A)Det(A)=0 nbsp; (B) Det(A)(2,)

(C) Det(A)(2,4) (D)Det(A)[2,4]

Answer:

Given determinant A=|1sinθ1sinθ1sinθ1sinθ1|

|A|=1(1+sin2Θ)sinΘ(sinΘ+sinΘ)+1(sin2Θ+1)

=1+sin2Θ+sin2Θ+1

=2+2sin2Θ=2(1+sin2Θ)

Now, given the range of Θ from 0Θ2π

0sinΘ1

0sin2Θ1

11+sin2Θ2

22(1+sin2Θ)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=A1B to find the solution of a system of equations expressed in matrix form.

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

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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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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