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# NCERT Solutions for Class 12 Maths Matrices Miscellaneous Exercise

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Matrices Miscellaneous Exercise*. This

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*NCERT Solutions** NCERT Solutions for Class 12 Maths* are not only the solutions of Maths exercise but it builds your foundation of other important subjects. Getting knowledge of depth concept of

*like Algebra, Calculus, Trigonometry, Coordinate Geometry help you to understand the concept of Physics and Physical Chemistry.*

**CBSE Class 12**^{th}Maths* CBSE Class 12^{th}* is an important school class in your life as you take some serious decision about your career. And out of all subjects, Maths is an important and core subjects. So

*is major role in your exam preparation as it has detailed chapter wise solutions for all exercise.*

**CBSE NCERT Solutions for Class 12**^{th}MathsAs we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the **CBSE NCERT Syllabus**. Thus, * NCERT Solutions* helps the students to solve the exercise questions as given in

*.*

**NCERT Books**## NCERT Solutions for Class 12 Maths Matrices Miscellaneous Exercise

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 1.**

Let , show that (aI + bA)^{n} = a^{n}I + na^{n – 1}bA, where I is the identity matrix of order 2 and n ∈ N.

**NCERT Solutions:**

To prove this, we use mathematical induction.

Given that p(n) : (aI + bA)^{n} = a^{n}I + na^{n – 1}bA.

At n = 1, we have p(1) : (aI + bA)^{1} = a^{1}I + 1a^{1 – 1}bA = al + bA

Now, at n = k, we let p(k) : (aI + bA)^{k} = a^{k}I + ka^{k – 1}bA is true, …… (1)

Thus, we need to prove that at n = k + 1, p(k + 1) : (aI + bA)^{k + 1} = a^{k + 1}I + (k + 1)a^{k + 1 – 1}bA.

LHS = (aI + bA)^{k + 1}

= (aI + bA)^{k}(aI + bA)

= (a^{k}I + ka^{k – 1}bA)(aI + bA) (From (1))

= a^{k + 1}I x I + ka^{k}bAI + a^{k}bAI + ka^{k – 1}b^{2}A^{2}

= a^{k + 1}I + ka^{k}bA + a^{k}bA + ka^{k – 1}b^{2 }x 0 (A^{2} = 0)

= a^{k + 1}I + (k + 1)a^{(k + 1) – 1}bA = RHS.

Hence, p(n) : (aI + bA)^{n} = a^{n}I + na^{n – 1}bA; it is also true at n = k + 1.

Therefore, p(n) : (aI + bA)^{n} = a^{n}I + na^{n – 1}bA is true for all values of n ∈ N.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 2.**

If

**NCERT Solutions:**

To prove this, we use mathematical induction.

Given that p(n) : .

At n = 1, we have p(1) :

Now, at n = k, we let p(k) : is true, …… (1)

Thus, we need to prove that at n = k + 1.

Therefore, by principle of mathematical induction, we get that .

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 3.**

If , then , prove that A where n is any positive integer.

**NCERT Solutions:**

To prove this, we use mathematical induction.

Given that p(n) : .

At n = 1, we have p(1) : .

Now, at n = k, we let p(k) : is true, …… (1)

Thus, we need to prove that at n = k + 1.

Therefore, by principle of mathematical induction, we get that for any positive integer n.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 4.**

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

**NCERT Solutions:**

Since A and B are symmetric matrices, A’ = A and B’ = B. To prove that (AB – BA) is skew symmetric matrix. It means, we need to prove (AB – BA)’ = -(AB – BA).

Now, LHS = (AB – BA)’ = (AB)’ – (BA)’ = B’A’ – A’B’ = BA – AB = -(AB – BA) = RHS.

Hence, (AB – BA) is skew symmetric matrix.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 5.**

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

**NCERT Solutions:**

Case I: Given that A is symmetric It means, A’ = A.

We need to prove that B’AB is symmetric. It means, we need to prove (B’AB)’ = B’AB.

LHS = (B’AB)’ = B’A'(B’)’ = B’A’B = B’AB = RHS.

Hence, B’AB is symmetric.

Case II: Given that A is skew-symmetric It means, A’ = -A.

We need to prove that B’AB is skew-symmetric. It means, we need to prove (B’AB)’ = -B’AB.

LHS = (B’AB)’ = B’A'(B’)’ = B’A’B = B'(-A)B = -B’AB = RHS.

Hence, B’AB is skew-symmetric.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 6.**

Find the values of x, y, z if the matrix satisfy the equation A′A = I.

**NCERT Solutions:**

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 7.**

For what values of x : ?

**NCERT Solutions:**

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 8.**

If , show that A^{2} – 5A + 7I = 0.

**NCERT Solutions:**

Given that .

Now,

Put these values in A^{2} – 5A + 7I, we get

LHS = = RHS

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 9.**

Find x, if .

**NCERT Solutions:**

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 10.**

A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below.

(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

**NCERT Solutions:**

Let quantity matrix is .

And, unit selling price matrix is .

Now, total selling price matrix is

Thus, total revenue in Market 1 is Rs. 46000 and in Market 2 is Rs. 53000.

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

**NCERT Solutions:**

Let quantity matrix is .

Now, cost price matrix is

Thus,

Total cost price, CP = 31000 + 36000 = Rs. 67000.

Total selling price, SP = 46000 + 53000 = Rs. 99000.

Profit = SP – CP = 99000 – 67000 = Rs. 32000.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 11.**

A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below.

**NCERT Solutions:**

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 12.**

If A and B are square matrices of the same order such that AB = BA, then prove by induction that AB^{n} = B^{n}A. Further, prove that (AB)^{n} = A^{n}B^{n} for all n ∈ N.

**NCERT Solutions:**

To prove this, we use mathematical induction.

Given that p(n) : AB^{n} = B^{n}A.

At n = 1, we have p(1) : AB = BA.

Now, at n = k, we let p(k) : AB^{k} = B^{k}A is true, …… (1)

Thus, we need to prove that at n = k + 1, p(k + 1) : AB^{k + 1} = B^{k + 1}A.

LHS = AB^{k + 1 }= AB^{k}B = (AB)B^{k} = (BA)B^{k} = B^{k+1}A = RHS

Hence, p(n) : AB^{n} = B^{n}A; it is also true at n = k + 1.

Therefore, p(n) : AB^{n} = B^{n}A is true for all values of n ∈ N.

Now, Again given that p(n) : (AB)^{n} = A^{n}B^{n}.

At n = 1, we have p(1) : AB = BA.

Now, at n = k, we let p(k) : (AB)^{k} = A^{k}B^{k} is true, …… (1)

Thus, we need to prove that at n = k + 1, p(k + 1) : (AB)^{k+1} = A^{k+1}B^{k+1}.

LHS = (AB)^{k + 1 }= (AB)^{k}(AB) = A^{k}(B^{k}B)A = A^{k}(B^{k+1})A = A^{k}(AB^{k+1}) = A^{k+1}B^{k+1} = RHS.

Hence, p(n) : (AB)^{n} = A^{n}B^{n}; it is also true at n = k + 1.

Therefore, p(n) : (AB)^{n} = A^{n}B^{n} is true for all values of n ∈ N.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 13.**

If is such that A² = I, then

(A) 1 + α² + βγ = 0

(B) 1 – α² + βγ = 0

(C) 1 – α² – βγ = 0

(D) 1 + α² – βγ = 0

**NCERT Solutions:**

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 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

**NCERT Solutions:**

**B**.

Since A is symmetric and skew-symmetric both, we have A’ = A and A’ = -A.

Then, A = -A ⟹ 2A = 0 ⟹ A = 0 ⟹ A is a zero matrix.

**NCERT Solutions for Class 12 Maths Matrices Miscellaneous ****Exercise****: Ques No 15.**

If A is square matrix such that A^{2} = A, then (I + A)³ – 7A is equal to

(A) A

(B) I – A

(C) I

(D) 3A

**NCERT Solutions:**

**C**.

Since A^{2} = A,

(I + A)³ – 7A

= I^{3} + A^{3} + 3IA(I + A) – 7A

= I + (A2)A + 3A(I + A) – 7A

= I + (A)A + 3A + 3A^{2} – 7A

= I + A2 + 3A2 – 4A

= I + 4A2 – 4A

= I + 4A – 4A = I