## NCERT Solutions for Class 12 Maths Exercise Miscellaneous NCERT Solutions for Class 12 Maths Chapter 3 – Matrices – FREE PDF Download

NCERT Class 12 Maths Ch 3 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 12 Maths Chapter 3 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. Class 12 Maths Chapter 3 – Matrices solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Matrices Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

# NCERT Solutions for Class 12 Maths Chapter 3 – Matrices

1. Let A = show that where I is the identity matrix of order 2 and N.

**Ans.**Using Mathematical Induction, we see the result is true for for

Given: is true, i.e.

To prove:

Proof: L.H.S. = = =

=

=

= = R.H.S.

Thus, is true, therefore, is true.

### 2. If A = , prove that A’’ = N.

**Ans.**Given: A = ……….(i)

Let

=

is true for

Now ……….(ii)

Multiplying eq. (ii) by eq. (i),

=

Therefore, is true for all natural numbers by P.M.I.

### 3. If A = then prove that A’’ = where is any positive integer.

**Ans.**Given: A’’ =

which is true for

Now, ……….(i)

Again ……….(ii)

[From eq. (i)]

=

Therefore, the result is true for

Hence, by the principal of mathematical induction, the result is true for all positive integers

### 4. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

**Ans.**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.

5. Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

**Ans.**(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.

### 6. Find the values of if the matrix A = satisfies the equation A’A = I.

**Ans.**Given: A =

A’ =

Now A’A = I

Equating corresponding entries, we have

And

And

, ,

### 7. For what value of ?

**Ans.**Given:

Equating corresponding entries, we have

### 8. If A = show that A^{2} – 5A + 7I = 0.

**Ans.**Given: A =

A^{2} – 5A + 7I =

= =

= = =

= = 0 = R.H.S. Proved.

### 9. Find if

**Ans.**Given:

Equating corresponding entries, we have

### 10. A manufacturer produces three products, 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 |

(a) If unit sales prices of and are ` 2.50, ` 1.50 and ` 1.00 respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.

**Ans.**According to question, the matrix A =

(a) Let B be the column matrix representing sale price of each unit of products

Then B =

Now Revenue = Sale price c Number of items sold

= =

Therefore, the revenue collected by sale of all items in Market I = ` 46,000 and the revenue collected by sale of all items in Market II = ` 53,000.

(b) Let C be the column matrix representing cost price of each unit of products

Then C =

Total cost = AC =

= =

The profit collected in two markets is given in matrix form as

Profit matrix = Revenue matrix – Cost matrix

Therefore, the gross profit in both the markets = ` 15000 + ` 17000 = ` 32,000.

11. Find the matrix X so that X

**Ans.**Given: X ……….(i)

Putting X = in eq. (i),

Equating corresponding entries, we have

…..(ii) …..(iii)

…..(iv) …..(v)

…..(vi) …..(vi)

Solving eq. (ii) and (iii), we have and

Solving eq. (v) and (vi), we have and

Putting these values in X = , X =

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.

**Ans.**Given: AB = BA …..(i)

Let ……(ii)

For becomes AB = BA

is true for

For

Multiplying both sides by B,

[From eq. (i)]

is also true.

Therefore, is true for all N by P.M.I.

13. If A = is such that A^{2} = I, then:

(A)

(B)

(C)

(D)

**Ans.**Given: A = and A

^{2}= I

Equating corresponding entries, we have

Therefore, option (C) is correct.

### 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

**Ans.**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.

### 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

**Ans.**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.