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

**10th Maths Exercise 3.16 Solution Question 1.**

In the matrix A = \(\left[\begin{array}{cccc}{8} & {9} & {4} & {3} \\ {-1} & {\sqrt{7}} & {\frac{\sqrt{3}}{2}} & {5} \\ {1} & {4} & {3} & {0} \\ {6} & {8} & {-11} & {1}\end{array}\right]\)

(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}

Solution:

(i) 16

(ii) 4 × 4

(iii) \(\sqrt{7}, \frac{\sqrt{3}}{2}\), 5, 0, -11, 1

**10th Maths Exercise 3.16 Solutions Question 2.**

If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?

Solution:

1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1 and 1 × 6, 2 × 3, 3 × 2, 6 × 1

**10th Maths Exercise 3.16 Answers Question 3.**

Construct a 3 × 3 matrix whose elements are given by then find the transpose of A.

(i) a_{ij} = |i – 2j|

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

Solution:

(i) a_{ij} = |i – 2j|

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

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

∴ \(\left[\begin{array}{lll}{1} & {3} & {5} \\ {0} & {2} & {4} \\ {1} & {1} & {3}\end{array}\right]\) is the required 3 × 3 matrix

**Ex 3.16 Class 10 Samacheer Question 4.**

Solution:

**10th Maths Exercise 3.16 Question 5.**

Solution:

**10th Maths 3.16 Question 6.**

Solution:

**Exercise 3.16 Class 10 Question 7.**

Find the values of x,y and z from the following equations

Solution: