Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

Students can download Maths Chapter 3 Algebra Ex 3.16 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

1. In the matrix
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 1
write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a22, a23, a24, a34, a43, a44.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 3
The possible orders of the matrix having 6 elements are
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 4

Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) aij = |i – 2j|
Answer:
aij = |i – 2j|
The general 3 × 3 matrices is
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 5
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
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 6

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

(ii) aij = \(\frac{(i+j)^{3}}{3}\)
Answer:
a11 = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)
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
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 7

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

Question 4.
If  Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 8 then find the tranpose of A.
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 9
transpose of A = (AT)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 10

Question 5.
If Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 11then find the tranpose of – A
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 12
Transpose of – A = (-AT) = Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 13

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

Question 6.
If A = Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 14then verify (AT)T = A
Answer:
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 15
Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 16
Answer:
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

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

(ii)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 17
Answer:
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
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 18
∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

(iii)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 19
Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16 20
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

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