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

1. In the matrix

write (i) The number of elements

(ii) The order of the matrix

(iii) Write the elements a_{22}, a_{23}, a_{24}, a_{34}, a_{43}, a_{44}.

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

a_{11} = |1 – 2(1)| = |1 – 2| = | – 1| = 1

a_{12} = |1 – 2(2)| = |1 – 4| = | – 3| = 3

a_{13} = |1 – 2(3)| = |1 – 6| = | – 5| = 5

a_{21} = |2 – 2(1)| = |2 – 2| = 0 = 0

a_{22} = |2 – 2(2)| = |2 – 4| = | – 2| = 2

a_{23} = |2 – 2(3)| = |2 – 6| = | – 4| = 4

a_{31} = |3 – 2(1)| = |3 – 2| = | 1 | = 1

a_{32} = |3 – 2(2)| = |3 – 4| = | – 1 | = 1

a_{33} = |3 – 2(3)| = |3 – 6| = | – 3 | = 3

The required matrix

(ii) a_{ij} = \(\frac{(i+j)^{3}}{3}\)

Answer:

a_{11} = \(\frac{(1+1)^{3}}{3}\) = \(\frac{2^{3}}{3}\) = \(\frac { 8 }{ 3 } \)

a_{12} = \(\frac{(1+2)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9

a_{13} = \(\frac{(1+3)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)

a_{21} = \(\frac{(2+1)^{3}}{3}\) = \(\frac { 27 }{ 3 } \) = 9

a_{22} = \(\frac{(2+2)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)

a_{23} = \(\frac{(2+3)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)

a_{31} = \(\frac{(3+1)^{3}}{3}\) = \(\frac { 64 }{ 3 } \) = \(\frac { 64 }{ 3 } \)

a_{32} = \(\frac{(3+2)^{3}}{3}\) = \(\frac { 125 }{ 3 } \) = \(\frac { 125 }{ 3 } \)

a_{33} = \(\frac{(3+3)^{3}}{3}\) = \(\frac { 216 }{ 3 } \) = 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

x^{2} + 8 = 6x

x^{2} – 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