If `S=[S_(i j)]` is a scalar matrix such that `s_(i i)=ka n dA` is a square matrix of the same order, then `A S=S A=?`
(a)`A^k`
(b) `k+A`
(c) `k A`
(d) `k S`

Explanation

`S=[S_{i j}]`

`Rightarrow S=[[k,0],[0,k]]` [because `S_{i j}=k`]


Let A`=[[a_{11} ,a_{12} ] ,[a_{21} , a_{22}]]` [because A is square matrix ]

Now,


`A S=[[a_{11} , a_{12} ],[a_{21} , a_{22}]] [[k,0],[0,k]]= [[k a_{11} , k a_{12} ],[ k a_{21} , ka_{22}]] = k[[a_{11} ,a_{12}],[ a_{21} , a_{22}]]=k A`

`SA =[[k,0],[0,k]] [[a_{11} , a_{12} ],[a_{21} , a_{22}]] = [[k a_{11} , k a_{12} ],[ k a_{21} , ka_{22}]] = k[[a_{11} ,a_{12}],[ a_{21} , a_{22}]]=k A`

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