Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.16

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

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:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to

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

The possible orders of the matrix having 6 elements are

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

Question 4.
If  then find the tranpose of A.
Answer:

transpose of A = (A^T)

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

Transpose of – A = (-A^T) =

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

Hence it is verified

Question 7.
Find the values of x, y and z from the following equations
(i)

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

(ii)

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

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

(iii)

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

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