NCERT Solutions Class 12

NCERT SolutionsMathematics
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 1)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 2)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 4)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 5)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 1)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 2)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 1)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 4)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 1)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 2)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 3)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 4)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 6)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 2)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 3)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 4)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 6)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 8)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 9)

NCERT SolutionsChemistry
 Aldehydes, Ketones and Carboxylic Acids : NCERT Solutions – Class 12 Chemistry
 Alcohols, Phenols and Ethers : NCERT Solutions – Class 12 Chemistry
 Amines : NCERT Solutions – Class 12 Chemistry
 Biomolecules : NCERT Solutions – Class 12 Chemistry
 Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
 Chemistry in Everyday Life : NCERT Solutions – Class 12 Chemistry
 Coordination Compounds : NCERT Solutions – Class 12 Chemistry
 Electrochemistry : NCERT Solutions – Class 12 Chemistry
 General Principles and Processes of Isolation of Elements : NCERT Solutions – Class 12 Chemistry
 Haloalkanes and Haloarenes : NCERT Solutions – Class 12 Chemistry
 Polymers : NCERT Solutions – Class 12 Chemistry
 Surface Chemistry : NCERT Solutions – Class 12 Chemistry
 The dand fBlock Elements : NCERT Solutions – Class 12 Chemistry
 The pBlock Elements : NCERT Solutions – Class 12 Chemistry
 The Solid State : NCERT Solutions – Class 12 Chemistry
 Solutions : NCERT Solutions – Class 12 Chemistry

NCERT SolutionsBiology

NCERT SolutionsPhysics
 Electrostatic Potential And Capacitance : NCERT Solutions – Class 12 Physics
 Electric Charges And Fields : NCERT Solutions – Class 12 Physics
 Semiconductor Electronics: Materials, Devices And Simple Circuits : NCERT Solutions – Class 12 Physics
 Ray Optics And Optical Instruments : NCERT Solutions – Class 12 Physics
 Nuclei : NCERT Solutions – Class 12 Physics
 Moving Charges And Magnetism : NCERT Solutions – Class 12 Physics
 Magnetism And Matter : NCERT Solutions – Class 12 Physics
 Electromagnetic Induction : NCERT Solutions – Class 12 Physics
 Dual Nature Of Radiation And Matter : NCERT Solutions – Class 12 Physics
 Current Electricity : NCERT Solutions – Class 12 Physics
 Communication Systems : NCERT Solutions – Class 12 Physics
 Atoms : NCERT Solutions – Class 12 Physics
 Alternating Current : NCERT Solutions – Class 12 Physics
Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
Exercise 3.2
1. Let A = B = C = . Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
Ans. (i) A + B = =
(ii) A – B = =
(iii) 3A – C = =
(iv) AB = =
(v) BA = =
2. Compute the following:
(i)
(ii)
(iii)
(iv)
Ans. (i) =
(ii)
=
(iii) =
(iv) =
3. Compute the indicated products:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ans. (i) =
(ii) =
(iii) =
(iv)
=
=
(v)
=
(vi)
=
4. If A = B = and C = then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Ans. A + B = = =
B – C = = =
Now, A + (B – C) = (A + B) – C
=
=
=
L.H.S. = R.H.S. Proved.
5. If A = and B = then compute 3A – 5B.
Ans. 3A – 5B =
=
=
6. Simplify:
Ans. Given:
=
=
7. Find X and Y, if:
(i) X + Y = and X – Y =
(ii) 2X + 3Y = and 3X + 2Y =
Ans. (i) Given: X + Y = …..(i)
and X – Y = …..(ii)
Adding eq. (i) and (ii), we get
2X =
X =
Subtracting eq. (i) and (ii), we get
2Y =
Y =
(ii) Given: 2X + 3Y = …..(i)
and 3X + 2Y = …..(ii)
Multiplying eq. (i) by 2, 4X + 6Y = ……….(iii)
Multiplying eq. (ii) by 3, 9X + 6Y = ………(iv)
Eq. (iv) – Eq. (iii) = 5X = =
X =
Now, From eq. (i), 3Y = 2X =
3Y = =
Y =
8. Fin X if Y = and 2X + Y =
Ans. 2X + Y =
2X = – Y
2X =
2X =
X = =
9. Find and if
Ans. Given:
Equating corresponding entries, we have
and
and
and
and
10. Solve the equation for and if
Ans. Given:
Equating corresponding entries, we have
And
And
And
, , ,
11. If find the values of and
Ans. Given:
Equating corresponding entries, we have
……….(i) and ……….(ii)
Adding eq. (i) and (ii), we have
Putting in eq. (ii),
12. Given: find the values of and
Ans. Given:
Equating corresponding entries, we have
And
And ……….(i)
And
Putting in eq. (i),
, , ,
13. If show that
Ans. Given: ……….(i)
Changing to in eq. (i),
L.H.S. =
=
=
=
= R.H.S. [changing to in eq. (i)]
14. Show that:
(i)
(ii)
Ans. (i) L.H.S. = = =
R.H.S. = = =
L.H.S. R.H.S.
(ii) L.H.S. =
=
=
R.H.S. =
=
=
L.H.S. R.H.S.
15. Find A^{2} – 5A + 6I if A = .
Ans. A^{2} – 5A + 6I =
=
= =
=
16. If A = prove that A^{3} – 6A^{2} + 7A + 2I = 0.
Ans. L.H.S. = A^{3} – 6A^{2} + 7A + 2I
=
=
=
= =
= =
= = 0 (Zero matrix)
= R.H.S. Proved.
17. If A = and I = find so that
Ans. Given: A = and I =
Equating corresponding entries, we have
And and
18. If A = and I is the identity matrix of order 2, show that
Ans. L.H.S. = I + A =
Now, I – A =
R.H.S. = =
=
=
=
= = =
L.H.S. = R.H.S. Proved.
19. A trust fund has ` 30,000 that must be invested in two different types of bond. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ` 30,000 in two types of bonds, if the trust fund must obtain an annual interest of (a) ` 1800, (b) ` 2000.
Ans. Let the investment in first bond be ` , then the investment in the second bond = `
Interest paid by first bond = 5% = per rupee and interest paid by second bond = 5% = per rupee.
Matrix of investment is A =
Matrix of annual interest per rupee B =
Matrix of total annual interest is AB = =
= =
Total annual interest = `
(a) According to question,
Therefore, Investment in first bond = ` 15,000
And Investment in second bond = ` (30000 – 15000) = ` 15,000
(b) According to question,
Therefore, Investment in first bond = ` 5,000
And Investment in second bond = ` (30000 – 15000) = ` 25,000
20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ` 80, ` 60 and ` 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Ans. Let the number of books as a 1 x 3 matrix B =
Let the selling prices of each book as a 3 x 1 matrix S =
Total amount received by selling all books = BS =
= =
Therefore, Total amount received by selling all the books = ` 20160
21. The restriction on and so that PY + WY will be define are:
(A)
(B) is arbitrary,
(C) is arbitrary,
(D)
Ans. Given:
Now,
On comparing, and
Therefore, option (A) is correct.
22. If then order of matrix 7X – 5Z is:
(A)
(B)
(C)
(D)
Ans. Here (given), the order of matrices X and Z are equal.
7X – 5Z is well defined and the order of 7X – 5Z is same as the order of X and Z.
The order of 7X – 5Z is either equal to or
But it is given that
Therefore, the option (B) is correct.