We are given matrices A and B, such that

In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as A^T or A’.

So, in transpose of a matrix,

Rows of matrix becomes columns of the same matrix.

(i). We need to verify that, (2A + B)’ = 2A’ + B’.

Take L.H.S: (2A + B)’

Substitute the matrices A and B, in (2A + B)’.

For transpose of (2A + B),

(3, 6), (14, 6) and (17, 15) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Take R.H.S: 2A’ + B’

If ,

(1, 2), (4, 1) and (5, 6) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Multiply both sides by 2,

Also,

If .

(1, 2), (6, 4) and (7, 3) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Now, add 2A’ and B’.

Since, L.H.S = R.H.S

Thus, (2A + B)’ = 2A’ + B’.

(ii). We need to verify that, (A – B)’ = A’ – B’.

Take L.H.S: (A – B)’

Substitute the matrices A and B in (A – B)’.

To find transpose of (A – B),

(0, 0), (-2, -3) and (-2, 3) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Take R.H.S: A’ – B’

If ,

(1, 2), (4, 1) and (5, 6) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Also,

If ,

(1, 2), (6, 4) and (7, 3) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Subtract B’ from A’,

Since, L.H.S = R.H.S

Thus, (A – B)’ = A’ – B’.