# Plus Two Math's Solution Ex 3.2 Chapter3 Matrices

Discover how to improve your maths grades with our high-quality, easy-to-understand maths textbook solutions. Our solutions are packed with information to ensure you get the best out of your studying Here is the solution for Exercise 3.2 Chapter 3 Matrices of NCERT plus two maths. Here we have given a detailed explanation of each and every exercise so that students can understand the concepts easily without any difficulty. The solution to each and every question is provided here so you can solve them by yourself if you don’t get the answer here.

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 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Math's Textbook Solution Chapter Chapter 3 Exercise Ex 3.2 Chapter Name Matrices Category Plus Two Kerala

## Kerala Syllabus Plus Two Math's Textbook Solution Chapter  3 Matrices Exercises 3.2

### Chapter 3 : Matrices Solution

Kerala plus two maths NCERT textbooks, we provide complete solutions for the exercise and answers provided at the end of each chapter. We also cover the entire syllabus given by the Board of secondary education, Kerala state.

### Chapter 3: Matrices Exercise   3.2

If andthen compute.

Simplify

If, find values of x and y.

Comparing the corresponding elements of these two matrices, we get:

2x − y = 10 and 3x + y = 5

Adding these two equations, we have:

5x = 15

⇒ x = 3

Now, 3x + y = 5

⇒ y = 5 − 3x

⇒ y = 5 − 9 = −4

x = 3 and y = −4

Let

Find each of the following

(i)  (ii)  (iii)

(iv)  (v)

(i)

(ii)

(iii)

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as:

Compute the following:

(i)  (ii)

(iii)

(iv)

In the addition of matrices , we add similar row and column matrix with  another matrix . For example in first part a will be added with a , b with b , -b with b and a with a .

Compute the indicated products

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

If, and, then compute and. Also, verify that

Find and Y, if

(i) and

(ii) and

(i)

Adding equations (1) and (2), we get:

(ii)

Multiplying equation (3) with (2), we get:

Multiplying equation (4) with (3), we get:

From (5) and (6), we have:

Now,

Comparing the corresponding elements of these two matrices, we have:

x = 3 and y = 3

Solve the equation for xyz and t if

Comparing the corresponding elements of these two matrices, we get:

Given, find the values of xyz and w.

Comparing the corresponding elements of these two matrices, we get:

If, show that.

(i)

(ii)

(i)

(ii)

Find if

We have A2 = A × A

If, prove that

$A^{3}-6A^{2}+7A+2I=egin{bmatrix}&space;1&space;&0&space;&2&space;\&space;0&space;&2&space;&1&space;\&space;2&space;&0&space;&3&space;\end{bmatrix}^{3}-6egin{bmatrix}&space;1&space;&0&space;&2&space;\&space;0&space;&2&space;&1&space;\&space;2&space;&0&space;&3&space;\end{bmatrix}^{2}+7egin{bmatrix}&space;1&space;&0&space;&2&space;\&space;0&space;&2&space;&1&space;\&space;2&space;&0&space;&3&space;\end{bmatrix}+2egin{bmatrix}&space;1&space;&0&space;&0&space;\&space;0&space;&1&space;&0&space;\&space;0&space;&0&space;&1&space;\end{bmatrix}&space;=egin{bmatrix}&space;21&space;&0&space;&34&space;\&space;12&space;&8&space;&23&space;\&space;34&space;&0&space;&55&space;\end{bmatrix}-6egin{bmatrix}&space;5&space;&0&space;&8&space;\&space;2&space;&4&space;&5&space;\&space;8&space;&0&space;&13&space;\end{bmatrix}+7egin{bmatrix}&space;1&space;&0&space;&2&space;\&space;0&space;&2&space;&1&space;\&space;2&space;&0&space;&3&space;\end{bmatrix}+2egin{bmatrix}&space;1&space;&0&space;&0&space;\&space;0&space;&1&space;&0&space;\&space;0&space;&0&space;&1&space;\end{bmatrix}&space;=egin{bmatrix}&space;21&space;&0&space;&34&space;\&space;12&space;&8&space;&23&space;\&space;34&space;&0&space;&55&space;\end{bmatrix}-egin{bmatrix}&space;30&space;&0&space;&48&space;\&space;12&space;&24&space;&30&space;\&space;48&space;&0&space;&78&space;\end{bmatrix}+egin{bmatrix}&space;7&space;&0&space;&14&space;\&space;0&space;&14&space;&7&space;\&space;14&space;&0&space;&21&space;\end{bmatrix}+egin{bmatrix}&space;2&space;&0&space;&0&space;\&space;0&space;&2&space;&0&space;\&space;0&space;&0&space;&2&space;\end{bmatrix}&space;=egin{bmatrix}&space;0&space;&0&space;&0&space;\&space;0&space;&0&space;&0&space;\&space;0&space;&0&space;&0&space;\end{bmatrix}&space;=O$

If and, find k so that

$A^{2}=kA-2I$

$\Rightarrow&space;egin{bmatrix}&space;3&space;&-2&space;\&space;4&space;&-2&space;\end{bmatrix}^{2}=kegin{bmatrix}&space;3&space;&-2&space;\&space;4&space;&-2&space;\end{bmatrix}-2egin{bmatrix}&space;1&space;&0&space;\&space;0&space;&1&space;\end{bmatrix}$

$\Rightarrow&space;egin{bmatrix}&space;1&space;&-2&space;\&space;4&space;&-4&space;\end{bmatrix}=egin{bmatrix}&space;3k-2&space;&-2k&space;\&space;4k&space;&-2k-2&space;\end{bmatrix}$

$\Rightarrow&space;3k-2=1$

$\Rightarrow&space;k=1$

Ifand I is the identity matrix of order 2, show that

A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1,800 (b) Rs 2,000

(a) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − 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 Rs 1800, the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond.

(b) Let Rs x be invested in the first bond. Then, the sum of money invested in the second bond will be Rs (30000 − x).

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

Thus, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

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 respectively given as Rs 80, Rs 60, and Rs 40.

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 Rs 20160 from the sale of all these books.

Assume XYZW and P are matrices of order, and respectively. The restriction on nk and p so that 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

Matrices P and Y are of the orders p × k and 3 × k respectively.

Therefore, matrix PY will be defined if k = 3. Consequently, PY will be of the order p × k.

Matrices W and Y are of the orders × 3 and 3 × k respectively.

Since the number of columns in W is equal to the number of rows in Y, matrix WY is well-defined and is of the order × k.

Matrices PY and WY can be added only when their orders are the same.

However, PY is of the order p × k and WY is of the order n × k. Therefore, we must have p = n.

Thus, = 3 and p = n are the restrictions on nk, and p so that will be defined.

Assume XYZW and P are matrices of order, and respectively. If n = p, then the order of the matrix is

(A) p × 2 (B) 2 × n (C) n × 3 (D) p × n

Matrix X is of the order 2 × n.

Therefore, matrix 7X is also of the same order.

Matrix Z is of the order 2 × p, i.e., 2 × n [Since n = p]

Therefore, matrix 5Z is also of the same order.

Now, both the matrices 7X and 5Z are of the order 2 × n.

Thus, matrix 7X − 5Z is well-defined and is of the order 2 × n.

#### Chapter 3: Matrices Exercise 3.2 Textbook Solution- Preview

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