Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.16

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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 a₂₂, a₂₃, a₂₄, a₃₄, a₄₃, a₄₄
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|
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
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
∴ \(\left[\begin{array}{lll}{1} & {3} & {5} \\ {0} & {2} & {4} \\ {1} & {1} & {3}\end{array}\right]\) 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: