Math, asked by keva, 10 months ago

solve this matrix........​

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Answered by MissSolitary
0

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: {  \underline{ \huge{ \boxed{ \boxed{ \underline{{{ \textbf{ \textsf{MATRIX}}}}}}}}}}

 \underline{ \underline{{ \huge{ \blue{ \mathfrak{S}}}}  {\blue{\bold{OLUTION -  }}}}}

{ \mathtt{A = \left[ \begin{array}{c c c } \sf  2& \sf 0  \:  \:   \:  \:   \: \sf \: 1\\\\ \sf \:  - 1& \sf \: 1  \:  \:  \:  \:  \:  \sf \: 5\end{array} \right]}} \\  \\  \\{ \mathtt{B =  \left[ \begin{array}{c c c } \sf   - 1& \sf 1 \:  \:  \:  \:  \:  \sf \: 0 \\\\ \sf \: 0& \sf \: 1 \:  \:  \:  \:  \:  \sf \:  - 2\end{array} \right] }}

{ \mathtt{Transpose  \: of \:  B  \implies B' = \left[ \begin{array}{c c  } \sf   - 1& \sf 0  \\\\ \sf 1& \sf \: 1 \\  \\  \sf 0 & \sf  - 2 \end{array} \right]}}

Now,

 \implies{ \mathtt{(AB') = \left[ \begin{array}{c c c }  \sf  2& \sf 0  \:  \:   \:  \:   \: \sf \: 1\\\\ \sf \:  - 1& \sf \: 1  \:  \:  \:  \:  \:  \sf \: 5\end{array} \right].\left[ \begin{array}{c c  } \sf   - 1& \sf 0  \\\\ \sf 1& \sf \: 1 \\  \\  \sf 0 & \sf  - 2 \end{array} \right]}} \\  \\  \\  \implies{ \mathtt{\left[ \begin{array}{c c  } \sf 2 \times  - 1+ 0 \times 1 + 1 \times 0& & \sf 2 \times 0 + 0 \times 1 + 1 \times  - 2  \\\\ \sf  - 1 \times  - 1 + 1 \times 1 + 5 \times 0&&  \sf \:  - 1 \times 0 + 1 \times 1 + 5 \times  - 2 \end{array} \right]}} \\  \\  \\  \implies{ \mathtt{\left[ \begin{array}{c c }  \sf   - 2 + 0 + 0&& \sf0 + 0 - 2 \\\\ \sf 1 + 1 + 0&& \sf  - 0 + 1 + 10\end{array} \right]}} \\  \\  \\  \implies{ \mathtt{ \red{\left[ \begin{array}{c c  } \sf   - 2& \sf  - 2   \\\\ \sf 2& \sf 9 \end{array} \right] \:  \:  \:  \:  \:  \: ans... \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: }}}

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