Transpose of a Matrix

Transpose of a Matrix

Transpose of a Matrix

Let A be a matrix. Then a new matrix obtained by interchanging the corresponding rows and columns of A is called the transpose of A. It is denoted by A’ or At.

If A = (■(a_11&a_12&a_13@a_21&a_22&a_23 )), then At = (■(a_11&a_21@a_12&a_22@a_13&a_23 )).

Here, the order of matrix A is 2 × 3 and that of At is 3 × 2. Hence if the order of a matrix A is m × n, then the order of transpose of matrix A i.e. At will be n × m. If A is a square matrix of order n, then its transpose At is also a square matrix of order n. If A is a row matrix, then its transpose At is a column matrix.

For example:

i.   If A = (■(2&4&6@3&6&9)), then At = (■(2&3@4&6@6&9)) ii.   If A = (■(1&2@3&4)), then At = (■(1&3@2&4)) iii.  If A = (■(1&2&3)), then At = (■(1@2@3))

 

Properties of Transpose of a Matrix

1.   The transpose of the transpose of a matrix is the matrix itself, i.e. (At)t = A.

Let A = (■(2&4&6@1&0&3)) …………………..(i) Then, At = (■(2&1@4&0@6&3)) And, (At)t = (■(2&4&6@1&0&3)) ……….…..(ii) From (i) and (ii), we get (At)t = A.

2.   The transpose of the sum of two matrices is equal to the sum of their transposes, i.e. (A + B)t = At + Bt.

Let A = (■(2&8@-6&4)) and B = (■(1&3@4&6)). Then, At = (■(2&-6@8&4)) and Bt = (■(1&4@3&6)).  Now, A + B = (■(2&8@-6&4)) + (■(1&3@4&6)) = (■(3&11@-2&10)).  And, (A + B)t = (■(3&-2@11&10)) ………………………….(i) Again, At + Bt = (■(2&-6@8&4)) + (■(1&4@3&6)) = (■(3&-2@11&10)) …...(ii) From (i) and (ii), we get (A + B)t = At + Bt

3.   If A is any matrix and k is any number, then (kA)t = kAt.

Let k = 2, A = (■(2&3&5@4&-4&8)). ∴ At = (■(2&4@3&-4@5&8)) Now, kA = 2(■(2&3&5@4&-4&8)) = (■(4&6&10@8&-8&16)) ∴ (kA)t = (■(4&8@6&-8@10&16)) ………………..………(i) Again, kAt = 2(■(2&4@3&-4@5&8)) = (■(4&8@6&-8@10&16)) ……...….(ii) From (i) and (ii), we get (kA)t = kAt

4.   If A and B are two matrices conformable for multiplication, then (AB)t = BtAt.

Let A = (■(1&0&-1@2&0&3)) and B = (■(1&0@3&1@0&2)). Then, At = (■(1&2@0&0@-1&3)) and Bt = (■(1&3&0@0&1&2)) Now, AB = (■(1&0&-1@2&0&3))(■(1&0@3&1@0&2)) = (■(1&-2@2&6)) And, (AB)t = (■(1&2@-2&6))………………………………..(i) Again, BtAt = (■(1&3&0@0&1&2))(■(1&2@0&0@-1&3)) = (■(1&2@-2&6))……..…(ii) From (i) and (ii), we get (AB)t = BtAt


Worked Out Examples

Example 1: Let A = (■(1&2@-3&6@0&1)), B = (■(0&3@5&7@1&-4)) and k = 3. Compute At, Bt and verify: 	(At)t = A 	(A + B)t = At + Bt 	(kA)t = kAt  Solution: Here, A = (■(1&2@-3&6@0&1)), B = (■(0&3@5&7@1&-4)) and k = 3. ∴ At = (■(1&-3&0@2&6&1)) and Bt = (■(0&5&1@3&7&-4)) Now, i)    (At)t = (■(1&2@-3&6@0&1)) = A. Hence verified. ii)   Here, A + B = (■(1&2@-3&6@0&1)) + (■(0&3@5&7@1&-4)) = (■(1&5@2&13@1&-3))        ∴ (A + B)t = (■(1&2&1@5&13&-3)) …………...(i)        At + Bt = (■(1&-3&0@2&6&1)) + (■(0&5&1@3&7&-4)) = (■(1&2&1@5&13&-3)) ……………….(ii)        From (i) and (ii), we get        (A + B)t = At + Bt. Hence verified. iii)  Here, kA = 3(■(1&2@-3&6@0&1)) = (■(3&6@-9&18@0&3))        ∴ (kA)t = (■(3&-9&0@6&18&3)) ……………..…..(i)        And,        kAt = 3(■(1&-3&0@2&6&1)) = (■(3&-9&0@6&18&3)) …..(ii)        From (i) and (ii), we get        (kA)t = kAt. Hence verified.


Example 2: If A = (■(6&-5@9&10)) and B = (■(1&0@0&1)). Find At, Bt, (AB)t and BtAt. Solution: Here, A = (■(6&-5@9&10)) and B = (■(1&0@0&1)) ∴ At = (■(6&9@-5&10)) and Bt = (■(1&0@0&1)) AB = (■(6&-5@9&10))(■(1&0@0&1)) = (■(6&-5@9&10)) ∴ (AB)t = (■(6&9@-5&10)) And, BtAt = (■(1&0@0&1))(■(6&9@-5&10)) = (■(6&9@-5&10))


Example 3: Given that A = (■(x&5@3&y)) and B = (■(4&3@5&-2)). Find x and y if At = B. Solution: We have A = (■(x&5@3&y)) and B = (■(4&3@5&-2)) By question, At = B i.e. (■(x&3@5&y)) = (■(4&3@5&-2)) Equating the corresponding elements, we have x = 4 and y = -2

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