

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.1
NCERT Solutions Class 12 Maths Chapter 3 PDF Download Link
Solve The Following Questions NCERT Solution for Class 12 Maths Chapter 3 Miscellaneous Exercise Matrices:
Question 1. Let A =
show that
where I is the identity matrix of order 2 and ∈ N. Solution : Given:
NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.2
Question 2. If A =
, prove that A n =
,n ∈ N. Solution : Given: A =
To show :P(n): A n
,n ∈ N We shall prove the result by using the principle of mathematical induction. For n = 1, we have:
NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3
Question 3. If A =
then prove that A n =
where n is any positive integer. Solution : Given: A n =
To show :P(n): A n
,n ∈ N We shall prove the result by using the principle of mathematical induction. For n = 1, we have:
NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4
Question 4. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix. Solution : A and B are symmetric matrices. A’ = A and B’ = B ……….(i) Now, (AB – BA)’ = (AB)’ – (BA)’ ∴ (AB – BA)’ = B’A’ – A’B’ [Reversal law] ∴ (AB – BA)’ = BA – AB [Using eq. (i)] ∴ (AB – BA)’ = – (AB – BA) Therefore, (AB – BA) is a skew symmetric. Question 5. Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric. Solution : (B’AB)’ = [B’(AB]’ = (AB)’ (B’)’ [∴ (CD)’ = D’C’] ∴ (B’AB)’ = B’A’B ……….(i) Case I: A is a symmetric matrix, then A’ = A From eq. (i) (B’AB)’ = B’AB B’AB is a symmetric matrix. Case II: A is a skew symmetric matrix. A’ = – A Putting A’ = – A in eq. (i), (B’AB)’ = B’(– A)B = – B’AB B’AB is a skew symmetric matrix. Question 6. Find the values of x,y,z if the matrix
satisfies the equation A’A = I.
Question 7. For what value of x
= O? Solution :
| NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.1 | NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 |
| NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise | |
show that A 2 – 5A + 7I = 0. Solution : Given: A =
Question 9. Find x if
Solution :
Question 10. A manufacturer produces three products, x,y,z which he sells in two markets. Annual sales are indicated below:
| Market | Products | |
| I. 10,000 | 2,000 | 18,000 |
| II. 6,000 | 20,000 | 8,000 |
= 10000 x 2.50 + 2000 x 1.50 + 18000 x 1.00 = 25000 + 3000 + 18000 = 46000 The total revenue in market II can be represented in the form of a matrix as:
= 6000 x 2.50 + 20000 x 1.50 + 8000 x 1.00 = 15000 + 30000 + 8000 = 53000 Therefore, the total revenue in market I isRs 46000 and the same in market II isRs 53000. (b) The unit cost prices of x, y, and z are respectively given as Rs 2.00, Rs 1.00, and 50 paise. Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:
= 10000 x 2.00 + 2000 x 1.00 + 18000 x 0.50 = 20000 + 2000 + 9000 = 31000 Since the total revenue in market I isRs 46000, the gross profit in this marketis (Rs 46000 − Rs 31000) Rs 15000.
= 6000 x 2.00 + 20000 x 1.00 + 8000 x 0.50 = 12000 + 20000 + 4000 = 36000 The total cost prices of all the products in market II can be represented in the form of a matrix as: Since the total revenue in market II isRs 53000, the gross profit in this market is (Rs 53000 − Rs 36000) Rs 17000. Question 11. Find the matrix X so that X
Solution : Given: X
……….(i) The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of the equation is a 2 × 3 matrix. Therefore, X has to be a 2 × 2 matrix.
Question 12. If A and B are square matrices of the same order such that AB = BA, then prove by induction that AB’’ = B’’A. Further prove that (AB)’’ = A’’B’’ for all n ∈ N. Solution : Given: AB = BA …..(i)
Now, we prove that the result is true for n = k + 1.
Therefore, the result is true for n = k + 1. Thus, by the principle of mathematical induction, we have (AB)’’ = A’’B’’, for all natural numbers. Question 13. If A =
is such that A 2 = I, then:
Solution : Given:
and A 2 = I
Equating corresponding entries, we haveTherefore, option (C) is correct. Question 14. If the matrix A is both symmetric and skew symmetric, then: (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these Solution : Since, A is symmetric, therefore, A’ = A ……..(i) And A is skew-symmetric, therefore, A’ = – A A = – A [From eq. (i)] A + A = 0 ⇒2A = 0 ⇒A = 0 Therefore, A is zero matrix. Therefore, option (B) is correct. Question 15. If A is a square matrix such that A 2 = A, then (I + A) 3 – 7A is equal to: (A) A (B) I – A (C) I (D) 3A Solution : Given: A 2 = A …..(i) Multiplying both sides by A, A 3 = A 2 = A [From eq. (i)] ……(ii) Also given (I + A) 3 – 7A = I 3 + A 3 + 3I 2 A + 3IA 2 – 7A Putting A 2 = A [from eq. (i)] and A 3 = A [from eq. (ii)], = I + A + 3IA + 3IA – 7A = I + A + 3A + 3A – 7A [ IA = A] = I + 7A – 7A = I Therefore, option (C) is correct.