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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.17

**10th Maths Exercise 3.17 Answers Question 1.**

If A = \(\left[\begin{array}{cc}{1} & {9} \\ {3} & {4} \\ {8} & {-3}\end{array}\right]=\left[\begin{array}{ll}{5} & {7} \\ {3} & {3} \\ {1} & {0}\end{array}\right]\) then verify that

(i) A + B = B + A

(ii) A + (-A) = (-A) + A = 0

Solution:

**10th Maths Exercise 3.17 Question 2.**

Solution:

**Ex 3.17 Class 10 Question 3.**

Find X and Y if X + Y = \(\left[\begin{array}{ll}{7} & {0} \\ {3} & {5}\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}{3} & {0} \\ {0} & {4}\end{array}\right]\)

Solution:

**10th New Syllabus Maths Exercise 3.17 Question 4.**

If A = \(\left[\begin{array}{lll}{0} & {4} & {9} \\ {8} & {3} & {7}\end{array}\right]\), B = \(\left[\begin{array}{lll}{7} & {3} & {8} \\ {1} & {4} & {9}\end{array}\right]\) find the value of

(i) B – 5A

(ii) 3A – 9B

Solution:

**Exercise 3.17 Class 10 Question 5.**

Find the values of x, y, z if

Solution:

(i) \(\left(\begin{array}{cc}{x-3} & {3 x-z} \\ {x+y+7} & {x+y+z}\end{array}\right)=\left(\begin{array}{ll}{1} & {0} \\ {1} & {6}\end{array}\right)\)

x – 3 = 1 ⇒ x = 4

3x – z = 0

3(4) – z = 0

-z = -12 ⇒ z = 12

x + y + 7 = 1

x + y = -6

4 + y = -6

y = -10

x = 4, y = -10, z = 12

(ii) \(\left[\begin{array}{ccc}{x} & {y-z} & {z+3}\end{array}\right]+\left[\begin{array}{lll}{y} & {4} & {3}\end{array}\right]=\left[\begin{array}{lll}{4} & {8} & {16}\end{array}\right]\)

x + y = 4 ……………. (1)

y – z + 4 = 8 ………….. (2)

z + 3 + 3 = 16 ………….. (3)

From (3), we get z = 10

From (2), we get y – 10 + 4 = 8

From (2), we get y = 14

From (1) we get x + 14 = 4

x = -10

x = -10, y = 14, z = 10

**10th Maths Exercise 3.17 Answers 2021 Question 6.**

Solution:

**10th Maths Exercise 3.17 Samacheer Kalvi Question 7.**

Find the non-zero values of x satisfying the matrix equation

Solution:

**Exercise 3.17 Question 8.**

Solution:

x^{2} – 4x = 5

y^{2} – 2y = 8

y^{2} – 2y – 8 = 0

(y – 4)(y + 2) = 0

y = 4, -2

x^{2} – 4x – 5 = 0

(x – 5)(x + 1) = 0

x = 5, -1

x = -1, 5, y = 4, -2