যদি E=[[1,2,3], [4,5,6], [7,8,9]] এবং F=[[1,0,0], [0,1,0], [0,0,1]] হয়, তাহলেঃ

Explanation:

Another Explanation (5):

সমাধান

যদি \(E = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) এবং \(F = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) হয়, তাহলে \(EF\) এবং \(FE\) নির্ণয় করে দেখা যাক তারা সমান কিনা।

EF নির্ণয়

\(EF = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) \(EF = \begin{bmatrix} (1\times1 + 2\times0 + 3\times0) & (1\times0 + 2\times1 + 3\times0) & (1\times0 + 2\times0 + 3\times1) \\ (4\times1 + 5\times0 + 6\times0) & (4\times0 + 5\times1 + 6\times0) & (4\times0 + 5\times0 + 6\times1) \\ (7\times1 + 8\times0 + 9\times0) & (7\times0 + 8\times1 + 9\times0) & (7\times0 + 8\times0 + 9\times1) \end{bmatrix}\) \(EF = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\)

FE নির্ণয়

\(FE = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) \(FE = \begin{bmatrix} (1\times1 + 0\times4 + 0\times7) & (1\times2 + 0\times5 + 0\times8) & (1\times3 + 0\times6 + 0\times9) \\ (0\times1 + 1\times4 + 0\times7) & (0\times2 + 1\times5 + 0\times8) & (0\times3 + 1\times6 + 0\times9) \\ (0\times1 + 0\times4 + 1\times7) & (0\times2 + 0\times5 + 1\times8) & (0\times3 + 0\times6 + 1\times9) \end{bmatrix}\) \(FE = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\)

ফলাফল

যেহেতু \(EF = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\) এবং \(FE = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\), তাই \(EF = FE\)। 🥳