If and , then verify that:

(i) (A’)’ = A

(ii) (AB)’ = B’A’

(iii) (kA)’ = (kA’).

We are given with matrices A and B, such that

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

Take L.H.S: (A’)’

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.

So,

If ,

(0, -1, 2) and (4, 3, -4) are 1^st and 2^nd rows respectively, will become 1^st and 2^nd columns respectively.

Then

Also, if ,

Similarly, (0, 4), (-1, 3) and (2, -4) are 1^st, 2^nd and 3^rd rows respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Then

Note, that

Thus, verified that (A’)’ = A.

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

Take L.H.S: (AB)’

Compute AB.

Order of A = 2 × 3

Order of B = 3 × 2

Then, order of AB = 2 × 2

Multiplying 1^st row of matrix A by matching members of 1^st column of matrix B, then sum them up.

(0, -1, 2)(4, 1, 2) = (0 × 4) + (-1 × 1) + (2 × 2)

⇒ (0, -1, 2)(4, 1, 2) = 0 – 1 + 4

⇒ (0, -1, 2)(4, 1, 2) = 3

Similarly, repeat the process to find the other elements.

Transpose of AB is (AB)’.

(3, 9) and (11, -15) are 1^st and 2^nd rows respectively, will become 1^st and 2^nd columns respectively.

Take R.H.S: B’A’

If ,

(4, 0), (1, 3) and (2, 6) are 1^st, 2^nd and 3^rd rows of matrix B respectively, will become 1^st, 2^nd and 3^rd columns respectively.

Also, if ,

(0, -1, 2) and (4, 3, -4) are 1^st and 2^nd rows respectively, will become 1^st and 2^nd columns respectively.

Multiply B’ by A’.

Order of B’ = 2 × 3

Order of A’ = 3 × 2

Then, order of B’A’ = 2 × 2

Multiply 1^st row of matrix B’ by matching members of 1^st column of matrix A’, then sum them up.

(4, 1, 2)(0, -1, 2) = (4 × 0) + (1 × -1) + (2 × 2)

⇒ (4, 1, 2)(0, -1, 2) = 0 – 1 + 4

⇒ (4, 1, 2)(0, -1, 2) = 3

Similarly, repeat the same steps to find out other elements.

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

Thus, (AB)’ = B’A’.

(iii). We need to verify that, (kA)’ = kA’.

Take L.H.S: (kA)’

We know that,

Multiply k on both sides, (k is a scalar quantity)

Now, to find transpose of kA,

(0, -k, 2k) and (4k, 3k, -4k) are 1^st and 2^nd rows of matrix kA respectively, will become 1^st and 2^nd columns respectively.

Take R.H.S: kA’

If

Then, for transpose of A,

(0, -1, 2) and (4, 3, -4) are 1^st and 2^nd rows of matrix A respectively, will become 1^st and 2^nd columns respectively.

Multiply k on both sides,

Note that, L.H.S = R.H.S.

Thus, (kA)’ = kA’.