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प्रश्न
If S = [Sij] is a scalar matrix such that sij = k and A is a square matrix of the same order, then AS = SA = ?
विकल्प
Ak
k + A
kA
kS
उत्तर
kA
\[S = \left[ S_{ij} \right]\]
\[ \Rightarrow S = \begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix} \left[ \because S_{ij} = k \right] \]
\[Let A = \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} \left[ \because \text{A is square matrix} \right]\]
\[Now, \]
\[AS = \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix} = \begin{bmatrix}k a_{11} & k a_{12} \\ k a_{21} & k a_{22}\end{bmatrix} = k\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = kA\]
\[SA = \begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix}\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = \begin{bmatrix}k a_{11} & k a_{12} \\ k a_{21} & k a_{22}\end{bmatrix} = k\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} = kA\]
\[ \therefore AS = SA = kA\]
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