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 = ? 

Options

• A^k

•  k + A 

• kA 

• kS

Answer 1

kA

\[S = \left[ S_{ij} \right]\]

\[ \Rightarrow S = \begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix} \left[ \because S_{ij} = k \right] \]

\[Let A = \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} \left[ \because \text{A is square matrix} \right]\]

\[Now, \]

\[AS = \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix} = \begin{bmatrix}k a_{11} & k a_{12} \\ k a_{21} & k a_{22}\end{bmatrix} = k\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = kA\]

\[SA = \begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = \begin{bmatrix}k a_{11} & k a_{12} \\ k a_{21} & k a_{22}\end{bmatrix} = k\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = kA\]

\[ \therefore AS = SA = kA\]