10th Maths - Book Back Answers - Chapter 3 Exercise 3.17 - English Medium Guides

  

 

SSLC / 10th - Maths - Book Back Answers - Chapter 3 Exercise 3.17 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 3 Exercise 3.17: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 3 Exercise 3.17 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 3 Exercise 3.17 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 3 Algebra Ex 3.17

1. In the

write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄.
Answer Key:
(i) The number of elements is 16

(ii) The order of the matrix is 4 × 4

(iii) Elements corresponds to

2. If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer Key:

The possible orders of the matrix having 18 elements are

The possible orders of the matrix having 6 elements are

3. Construct a 3 × 3 matrix whose elements are given by
(i) a_ij = |i – 2j|

Answer Key:

a_ij = |i – 2j|
The general 3 × 3 matrices is

a₁₁ = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a₁₂ = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a₁₃ = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a₂₁ = |2 – 2(1)| = |2 – 2| = 0 = 0
a₂₂ = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a₂₃ = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a₃₁ = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a₃₂ = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a₃₃ = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix

 

(ii) a_ij = $\frac{(i+j{)}^3}{3}$
Answer Key:
a₁₁ = $\frac{(1+1{)}^3}{3}$ = 233 = $\frac{8}{3}$
a₁₂ = $\frac{(1+2{)}^3}{3}$ = $\frac{27}{3}$ = 9
a₁₃ = $\frac{(1+3{)}^3}{3}$ = $\frac{64}{3}$ = $\frac{64}{3}$ 
a₂₁ = $\frac{(2+1{)}^3}{3}$ = $\frac{27}{3}$ = 9
a₂₂ = $\frac{(2+2{)}^3}{3}$ = $\frac{64}{3}$ = $\frac{64}{3}$
a₂₃ = $\frac{(2+3{)}^3}{3}$ = $\frac{125}{3}$ = $\frac{125}{3}$
a₃₁ = $\frac{(3+1{)}^3}{3}$ = $\frac{64}{3}$ = $\frac{64}{3}$
a₃₂ = $\frac{(3+2{)}^3}{3}$ = $\frac{125}{3}$ = $\frac{125}{3}$
a₃₃ = $\frac{(3+3{)}^3}{3}$ = $\frac{216}{3}$= 72
The required matrix

 

4. If A=  then find the tranpose of A.
Answer Key:

A=

transpose of A = (A^T)

5. If A= then find the tranpose of – A
Answer Key:

A= 

-A= 

 Transpose of – A = (-A^T) =

 

6. If A = then verify (AT)T = A
Answer Key:

A^T=

(-A^T)^T=

(-A^T)^T=A 

Hence it is verified

 

7. Find the values of x, y and z from the following equations
(i)${\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo></mrow></math>}}^{}$

Answer Key:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3

 

(ii)${\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo></mrow></math>}}^{}$${\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo></mo></mrow></math>}}^{}$

Answer Key:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = $\frac{8}{x}$
Substitute the value of y =  $\frac{8}{x}$ in (1)
x + $\frac{8}{x}$ = 6
x² + 8 = 6x
x² – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = $\frac{8}{4}$= 2 or y = $\frac{8}{2}$ = 4

∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

 

(iii)${\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>=</mo></mrow></math>}}^{}$${\mathrm{<math\; xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo></mo></mrow></math>}}^{}$

Answer Key:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)

(1) - (2)  

$\frac{\underset{ \underset{(-) (-) (-)}{ x+0+z =5}}{ x+y+z=9}}{ y=4}$

Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3