Answer :

We have two matrices A and B, such that

We need to verify whether (AB)’ = B’A’.

Let us understand what a transpose is.

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}.

Take L.H.S = (AB)’

So, let us compute AB.

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

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

⇒ (2, 4, 0)(1, 2, 1) = 2 + 8 + 0

⇒ (2, 4, 0)(1, 2, 1) = 10

Multiply 1^{st} row of matrix A by matching members of 2^{nd} column of matrix B, then sum them up.

(2, 4, 0)(4, 8, 3) = (2 × 4) + (4 × 8) + (0 × 3)

⇒ (2, 4, 0)(4, 8, 3) = 8 + 32 + 0

⇒ (2, 4, 0)(4, 8, 3) = 40

Similarly, let us do it for the rest of the elements.

So,

Now, for transpose of AB, rows will become columns.

Now, take R.H.S = B’A’

If

Then, if (1, 4) are the elements of 1^{st} row, it will become elements of 1^{st} column, and so on.

Also,

Then, if (2, 4, 0) are the elements of 1^{st} row, it will become elements of 1^{st} column, and so on.

Now, multiply B’A’.

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

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

⇒ (1, 2, 1)(2, 4, 0) = 2 + 8 + 0

⇒ (1, 2, 1)(2, 4, 0) = 10

Multiply 1^{st} row of matrix B’ by matching members of 2^{nd} column of matrix A’, then sum them up.

(1, 2, 1)(3, 9, 6) = (1 × 3) + (2 × 9) + (1 × 6)

⇒ (1, 2, 1)(3, 9, 6) = 3 + 18 + 6

⇒ (1, 2, 1)(3, 9, 6) = 27

Similarly, filling up for the rest of the elements.

⇒ L.H.S = R.H.S

Thus, (AB)’ = B’A’.

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