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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.18

Question 1.

If A is of order p × q and B is of order q × r what is the order of AB and BA?

Solution:

If A is of order p × q [∵ p × q q × r = p × r]

the order of AB = p × r [∵ q × r p × q = r ≠ p]

Product of BA cannot be defined/found as the number of columns in B ≠. The number of rows in A.

Question 2.

If A is of order p × q and B is of order q × r what is the order of AB and BA?

Answer:

Order of A = a × (a + 3)

Order of B = b × (17 – b)

Given: Product of AB exist

a + 3 = b

a – b = – 3 ….(1)

Product of BA exist

17 – b = a

– a – b = -17

a + b = 17 ………(2)

(1) + (2) ⇒ 2a = 14

a = \(\frac { 14 }{ 2 } \) = 7

Substitute the value of a = 7 in (1)

7 – b = -3 ⇒ -b = -3 -7

-b = -10 ⇒ b = 10

The value of b = 7 and b = 10

Question 3.

Find the order of the product matrix AB if

Solution:

Question 4.

If A = \(\left[\begin{array}{ll}{2} & {5} \\ {4} & {3}\end{array}\right]\), B = \(\left[\begin{array}{cc}{1} & {-3} \\ {2} & {5}\end{array}\right]\) find AB, BA and check if AB = BA?

Solution:

Question 5.

Solution:

Question 6.

Show that the matrices A = \(\left[\begin{array}{ll}{1} & {2} \\ {3} & {1}\end{array}\right]\), B = \(\left[\begin{array}{cc}{1} & {-2} \\ {-3} & {1}\end{array}\right]\) satisfy commutative property AB = BA

Solution:

Question 7.

(i) A(BC) = (AB)C

(ii) (A – B)C = (AC – BC)

(iii) (A- B)^{T} = A^{T} – B^{T}

Solution:

(i) A(BC) = (AB)C

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Verify that A^{2} = I when A = \(\left(\begin{array}{cc}{5} & {-4} \\ {6} & {-5}\end{array}\right)\)

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.

If A = \(\) show that A^{2} – 5A + 7I_{2} = 0.

Solution: