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NCERT Exemplar · Q72

Q.If AA and BB are square matrices of the same order, then

(i) (AB)′=(AB)' = _________.
(ii) (kA)′=(kA)' = _________. (kk is any scalar)
(iii) [k(A−B)]′=[k(A - B)]' = _________.
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The transpose of a product reverses the order: (AB)′=B′A′(AB)' = B'A'. The transpose of a scalar multiple is the scalar times the transpose: (kA)′=kA′(kA)' = kA'. Combining these, [k(A−B)]′=k(A′−B′)[k(A - B)]' = k(A' - B').

Why this works — the core idea

Matrix transpose is like flipping a matrix over its main diagonal: rows become columns and columns become rows. The key property for products is that when you transpose ABAB, you must reverse the multiplication order. Why? Because the (i,j)(i,j) entry of ABAB comes from row ii of AA and column jj of BB. After transposing, that entry moves to position (j,i)(j,i), which now comes from row jj of B′B' and column ii of A′A' — hence B′A′B'A'.

For scalar multiplication, each entry is simply multiplied by kk, so transposing just carries that kk along.

Let's apply these ideas step by step.

  1. For (AB)′(AB)' The transpose of a product of two matrices equals the product of their transposes in reverse order.

(AB)′=B′A′(AB)' = B'A'

This is a standard result — you can verify it by checking the (i,j)(i,j) entry on both sides.

  1. For (kA)′(kA)' If every entry of AA is multiplied by kk, then transposing gives a matrix where every entry is still kk times the corresponding entry of A′A'. So

(kA)′=kA′(kA)' = kA'

The scalar simply factors out.

  1. For [k(A−B)]′[k(A - B)]' Work from the inside out. First, A−BA - B is just entry-wise subtraction. Then multiply by kk. Then transpose.
    • Inside the brackets: k(A−B)k(A - B) means each entry of (A−B)(A - B) is scaled by kk. …

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