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# Class 12 RD Sharma Solutions- Chapter 5 Algebra of Matrices – Exercise 5.4

• Last Updated : 11 Feb, 2021

### Question 1: Let A = and B = verify that

(i) (2A)T = 2AT

(ii) (A + B)T = AT + BT

(iii) (A − B)T = AT − BT

(iv) (AB)T = BT AT

Solution:

(i) Given: A = and B = Assume,

(2A)T = 2AT

Substitute the value of A L.H.S = R.H.S

Hence, proved.

(ii) Given: A = and B = Assume,

(A+B)T = AT + BT L.H.S = R.H.S

Hence, proved.

(iii) Given: A= and B= Assume,

(A − B)T = AT − BT L.H.S = R.H.S

Hence, proved

(iv) Given: A = and B = Assume,

(AB)T = BTAT Therefore, (AB)T = BTAT

Hence, proved.

### Question 2: A = and B = Verify that (AB)T = BTAT

Solution:

Given: A = and B = Assume,

(AB)T = BTAT L.H.S = R.H.S

Hence proved

### Find AT, BT and verify that

(i) (A + B)T = AT + BT

(ii) (AB)T = BTAT

(iii) (2A)T = 2AT

Solution:

(i) Given: A = and B = Assume

(A + B)T = AT + BT L.H.S = R.H.S

Hence proved

(ii) Given: A = and B = Assume, (AB)T = BTAT L.H.S =R.H.S

Hence proved

(iii) Given: A = and B = Assume,

(2A)T = 2AT L.H.S = R.H.S

Hence proved

### Question 4: if A = , B = , verify that (AB)T = BTAT

Solution:

Given: A = and B = Assume,

(AB)T = BTAT L.H.S = R.H.S

Hence proved

### Question 5: If A = and B = , find (AB)T

Solution:

Given: A = and B = Here we have to find (AB)T Hence,

(AB)T ### (i) For two matrices A and B, verify that (AB)T = BTAT

Solution:

Given, (AB)T = BTAT

⇒ ⇒ ⇒ ⇒ ⇒ L.H.S = R.H.S

Hence,

(AB)T = BTAT

### (ii) For the matrices A and B, verify that (AB)T = BTAT, where Solution:

Given, (AB)T = BTAT

⇒ ⇒ ⇒ ⇒ ⇒ L.H.S = R.H.s

So,

(AB)T = BTAT

### Question 7: Find , AT – BT

Solution:

Given that We need to find AT – BT.

Given that,  Let us find AT – BT

⇒ ⇒ ⇒ ### Question 8: If , then verify that A’A = 1

Solution:   ⇒ ⇒ ⇒ Hence,we have verified that A’A = I

### Question 9: , then verify that A’A = I

Solution:    Hence, we have verified that A’A = I

### Where Solution:

Given,

li, mi, ni are direction cosines of three mutually perpendicular vectors

⇒ And, Given,    = I

Hence,

AAT = I

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