If S = [Sij] is a scalar matrix such that sij = k and A is a square matrix of the same order, then AS = SA = ?

(c) $k A$

Here,

$S=\left[S_{i j}\right]$

$\Rightarrow S=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right] \quad\left[\because S_{i j}=k\right]$

Let $A=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right] \quad[\because A$ is square matrix $]$'

Now,

$A S=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]=\left[\begin{array}{ll}k a_{11} & k a_{12} \\ k a_{21} & k a_{22}\end{array}\right]=k\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=k A$

$S A=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=\left[\begin{array}{ll}k a_{11} & k a_{12} \\ k a_{21} & k a_{22}\end{array}\right]=k\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]=k A$

$\therefore A S=S A=k A$