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# If A and B are square matrices of the same order,then $(iii)\quad [k(A-B)']=\text{________}.$

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Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
Given
If A and B are square matrices of the same order then
$[k(A-B)']=k(A'-B')$
Proof:
Let A=$[a_{ij}]$
A'= $[a_{ji}]$
B=$[b_{ij}]$
B'= $[b_{ji}]$
(A-B)=$[a_{ij}$ - $b_{ij}]$
Then (A-B)'= $[a_{ji}$ - $b_{ji}]$
A'-B' = $[a_{ji}$ - $b_{ji}]$
The scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
Hence
$[k(A-B)']=k(A'-B')$