Hello Students! In this post, you will find the complete** ****NCERT Solutions for Maths Class 12 Exercise 3.2**.

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**NCERT Solutions for Maths Class 12 Exercise 3.1**

**NCERT Solutions for Maths Class 12 Exercise 3.3**

**NCERT Solutions for Maths Class 12 Exercise 3.4**

**NCERT Solutions for Maths Class 12 Exercise 3.2**

**Maths Class 12 Ex
3.2 Question 1.**

**Let**

**Find each of the following:**

**(i) A + B (ii)
A – B (iii) 3A –
C**

**(iv) AB (v)
BA**

**Solution:**

**Maths
Class 12 Ex 3.2 Question 2.**

**Compute the following:**

**Solution:**

**Maths Class 12 Ex
3.2 Question 4.**

**If**,

**then compute (A + B) and (B – C). Also verify that A + (B – C) = (A + B) –
C.**

**Solution:**

**Solution:**

We have,

**Maths Class 12 Ex
3.2 Question 9.**

**Solution:**

We have,

Hence, x = 3 and y = 3.

**Maths
Class 12 Ex 3.2 Question 10.**

**Solve the equation for x, y, z and t, if**

**Solution:**

We have,

**Solution:**

We have,

⇒ 3x = x + 4 ⇒ x = 2

And 3y = 6 + x + y ⇒ y = 4

Also, 3w = 2w + 3 ⇒ w = 3

Again, 3z = – 1 + z + w

⇒ 2z = – 1 + 3

⇒ 2z = 2

⇒ z = 1

Hence, x = 2, y = 4, z = 1, w = 3.

**Maths Class 12 Ex
3.2 Question 15.**

**Find A² – 5A + 6I, if A =**

**Solution:**

*A*² =

*A.A*

**Maths Class 12 Ex
3.2 Question 16.**

**If A =**

**, prove that A³ – 6A² + 7A + 2I = 0**

**Solution:**

We have

*A*² =

*A.A*

**Maths Class 12 Ex
3.2 Question 17.**

**If**

**, find k so that A² = kA – 2I**

**Solution:**

We are given that,

We have to find the value of k.

Now, A² = kA – 2I

**Maths Class 12 Ex
3.2 Question 18.**

**If**

**and I is the identity matrix of order 2, show that**

**Solution:**

On the L.H.S:

**Maths Class 12 Ex
3.2 Question 19.**

**A trust fund has**₹

**30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide**₹

**30,000 among the two types of bond if the trust fund must obtain an annual total interest of**

**(a)**₹

**1800 (b)**₹

**2000**

**Solution:**

**(a)** Let ₹ *x *be
invested in the first bond. Then, the sum of money invested in the second bond
will be ₹ (30,000 − *x*).

It is given that the first bond pays 5% interest per year and the
second bond pays 7% interest per year.

Therefore, in order to obtain an annual total interest of Rs 1800,
we have:

Thus, in order to obtain an annual total interest of ₹ 1800, the
trust fund should invest ₹ 15,000 in the first bond and the remaining ₹ 15,000
in the second bond.

**(b) **Let ₹ *x *be invested in the first bond. Then, the sum of
money invested in the second bond will be ₹ (30,000 − *x*).

Therefore, in order to obtain an annual total interest of ₹ 2000,
we have:

Thus, in order to obtain an annual total interest of ₹ 2000, the trust fund should invest ₹ 5000 in the first bond and the remaining ₹ 25,000 in the second bond.

**Maths Class 12 Ex
3.2 Question 20.**

**The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling price are**₹

**80,**₹

**60 and**₹

**40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.**

**Solution:**

The bookshop has 10 dozen chemistry books, 8 dozen physics books,
and 10 dozen economics books.

The selling prices of a chemistry book, a physics book, and an
economics book are given as ₹ 80, ₹ 60 and ₹ 40, respectively.

The total amount of money that will be received from the sale of
all these books can be represented in the form of a matrix as:

Thus, the bookshop will receive ₹ 20160
from the sale of all these books.

**Assuming X, Y, Z,
W and P are the matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively. Choose the correct answer in questions 21 and 22.**

**Maths Class 12 Ex
3.2 Question 21.**

**The restrictions on n, k and p so that PY + WY will be defined are**

**(A) k = 3, p = n**

**(B) k is arbitrary, p = 2**

**(C) p is arbitrary, k = 3**

**(D) k = 2, p = 3**

**Solution:**

Given: X_{2 x n,} Y_{3 x k}, Z_{2 x p,} W_{n
x 3, }P_{p x k}

Now PY + WY = P_{p x k} × Y_{3 + k} + W_{n x 3} ×
Y_{3 x k}

Clearly, k = 3 and p = n

Hence, option (A) is correct.

**Maths Class 12 Ex
3.2 Question 22.**

**If n = p, then the order of the matrix 7X – 5Z is:**

**(A) p × 2**

**(B) 2 × n**

**(C) n × 3**

**(D) p × n**

**Solution:**

7X – 5Z = 7X

_{2 x n}– 5Z

_{2 x p}

We can add two matrices if their order is the same, i.e., n = P.

Therefore, the order of 7X – 5Z is 2 × n.

Hence, option (B) is correct.

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