Question #65686, Math / Other
Find the inverse of the matrix { ( 3 , 1 , 2 ) , ( − 2 , 3 , − 5 ) , ( 1 , 2 , 4 ) } \{(3, 1, 2), (-2, 3, -5), (1, 2, 4)\} {( 3 , 1 , 2 ) , ( − 2 , 3 , − 5 ) , ( 1 , 2 , 4 )}
Answer.
A = ( 3 1 2 − 2 3 − 5 1 2 4 ) A = \left( \begin{array}{ccc} 3 & 1 & 2 \\ -2 & 3 & -5 \\ 1 & 2 & 4 \end{array} \right) A = ⎝ ⎛ 3 − 2 1 1 3 2 2 − 5 4 ⎠ ⎞ ( 3 1 2 1 0 0 − 2 3 − 5 0 1 0 1 2 4 0 0 1 ) → ( 1 1 3 2 3 1 3 0 0 − 2 3 − 5 0 1 0 1 2 4 0 0 1 ) → \left( \begin{array}{cccc} 3 & 1 & 2 & 1 & 0 & 0 \\ -2 & 3 & -5 & 0 & 1 & 0 \\ 1 & 2 & 4 & 0 & 0 & 1 \end{array} \right) \to \left( \begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 0 & 0 \\ -2 & 3 & -5 & 0 & 1 & 0 \\ 1 & 2 & 4 & 0 & 0 & 1 \end{array} \right) \to ⎝ ⎛ 3 − 2 1 1 3 2 2 − 5 4 1 0 0 0 1 0 0 0 1 ⎠ ⎞ → ⎝ ⎛ 1 − 2 1 3 1 3 2 3 2 − 5 4 3 1 0 0 0 1 0 0 0 1 ⎠ ⎞ → \left( \begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{11}{3} &td><td> \frac{2}{3} }{ 3 } \end{array} \right| \frac{1}{3} \quad 0 \quad 0 ( 0 1 3 − 11 3 1 2 4 ) ( 0 0 1 ) → ( 1 1 3 2 3 0 11 3 − 11 3 0 5 3 − 10 3 ) → \begin{array}{l} \left( \begin{array}{c c c} 0 & \frac {1}{3} & - \frac {11}{3} \\ 1 & 2 & 4 \end{array} \right) \end{array} \quad \left( \begin{array}{c c c} 0 & 0 & 1 \end{array} \right) \to \left( \begin{array}{c c c} 1 & \frac {1}{3} & \frac {2}{3} \\ 0 & \frac {11}{3} & - \frac {11}{3} \\ 0 & \frac {5}{3} & - \frac {10}{3} \end{array} \right) \to ( 0 1 3 1 2 − 3 11 4 ) ( 0 0 1 ) → ⎝ ⎛ 1 0 0 3 1 3 11 3 5 3 2 − 3 11 − 3 10 ⎠ ⎞ → ( 1 1 3 2 3 0 1 − 1 0 5 3 − 10 3 ) → ( 1 1 3 2 3 0 1 − 1 0 0 5 ) → ( 1 1 3 0 3 3 55 2 11 1 5 − 7 11 − 5 11 1 ) → \left( \begin{array}{c c c} 1 & \frac {1}{3} & \frac {2}{3} \\ 0 & 1 & -1 \\ 0 & \frac {5}{3} & - \frac {10}{3} \end{array} \right) \to \left( \begin{array}{c c c} 1 & \frac {1}{3} & \frac {2}{3} \\ 0 & 1 & -1 \\ 0 & 0 & 5 \end{array} \right) \to \left( \begin{array}{c c c} 1 & \frac {1}{3} & \frac {0}{3} \\ \frac {3}{55} & \frac {2}{11} & \frac {1}{5} \\ - \frac {7}{11} & - \frac {5}{11} & 1 \end{array} \right) \to ⎝ ⎛ 1 0 0 3 1 1 3 5 3 2 − 1 − 3 10 ⎠ ⎞ → ⎝ ⎛ 1 0 0 3 1 1 0 3 2 − 1 5 ⎠ ⎞ → ⎝ ⎛ 1 55 3 − 11 7 3 1 11 2 − 11 5 3 0 5 1 1 ⎠ ⎞ → ( 1 1 3 0 0 1 0 0 0 5 ∣ 23 55 2 33 − 2 15 1 3 55 1 2 11 1 5 → ( 1 0 0 0 1 0 0 0 1 ) → ( 2 0 − 1 5 3 55 2 11 1 5 − 7 55 − 1 11 1 5 ) \left( \begin{array}{c c c} 1 & \frac {1}{3} & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{array} \right| \frac { \begin{array}{c} 23 \\ \frac {55}{} \end{array} } \quad \frac { \begin{array}{c} 2 \\ \frac {33}{} \end{array} } \quad \frac { - \frac {2}{15} }{ \begin{array}{c} 1 \\ \frac {3}{55} \end{array} } \quad \frac { \begin{array}{c} 1 \\ \frac {2}{11} \end{array} } \quad \frac { \begin{array}{c} 1 \\ \frac {5}{} \end{array} } \quad \rightarrow \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \to \left( \begin{array}{c c c} 2 & 0 & - \frac {1}{5} \\ \frac {3}{55} & \frac {2}{11} & \frac {1}{5} \\ - \frac {7}{55} & - \frac {1}{11} & \frac {1}{5} \end{array} \right) ⎝ ⎛ 1 0 0 3 1 1 0 0 0 5 ∣ ∣ 23 55 2 33 1 55 3 − 15 2 1 11 2 1 5 → ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ → ⎝ ⎛ 2 55 3 − 55 7 0 11 2 − 11 1 − 5 1 5 1 5 1 ⎠ ⎞
So A − 1 = ( 2 3 0 − 1 5 3 55 2 11 1 5 − 7 55 − 1 11 1 5 ) . A^{-1} = \left( \begin{array}{ccc} \frac{2}{3} & 0 & -\frac{1}{5} \\ \frac{3}{55} & \frac{2}{11} & \frac{1}{5} \\ -\frac{7}{55} & -\frac{1}{11} & \frac{1}{5} \end{array} \right). A − 1 = ⎝ ⎛ 3 2 55 3 − 55 7 0 11 2 − 11 1 − 5 1 5 1 5 1 ⎠ ⎞ .
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#340153 on Dec 2023