Question

If S = [S_ij] is a scalar matrix such that s_ij = k and A is a square matrix of the same order, then AS = SA = ?
(a) A^k
(b) k + A
(c) kA
(d) kS

Answer 1

(c) kA

Here,

$$\begin{gathered} S=\left[S_{ij}\right] \\ \Rightarrow S=\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]\left[\because S_{ij}=k\right] \\ \mathrm{Let}A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\because A\mathrm{is}\mathrm{square}\mathrm{matrix}\right] \\ \mathrm{Now}, \\ AS=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]=\left[\begin{array}{cc}k{a}_{11}& k{a}_{12}\\ k{a}_{21}& k{a}_{22}\end{array}\right]=k\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=kA \\ SA=\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=\left[\begin{array}{cc}k{a}_{11}& k{a}_{12}\\ k{a}_{21}& k{a}_{22}\end{array}\right]=k\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=kA \\ \therefore AS=SA=kA \end{gathered}$$