using elementary row transformation find inverse of it following matrx
[3 0 2
1 5 9
6 4 7]

Matrix A = [\begin{matrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7 \end{matrix}] Now, for inverse matrix by transformation A = IA [\begin{matrix} 3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{2} \leftrightarrow R_{1} [\begin{matrix} 1 & 5 & 9 \\ 3 & 0 & 2 \\ 6 & 4 & 7 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}] A R_{2} = R_{2} - 3 R_{1}, R_{3} = R_{3} - 6 R_{1} [\begin{matrix} 1 & 5 & 9 \\ 0 & - 15 & - 25 \\ 0 & - 26 & - 47 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 1 & - 3 & 0 \\ 0 & - 6 & 1 \end{matrix}] A R_{2} = \frac{R_{2}}{- 15} [\begin{matrix} 1 & 5 & 9 \\ 0 & 1 & \frac{5}{3} \\ 0 & - 26 & - 47 \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ \frac{- 1}{15} & \frac{1}{5} & 0 \\ 0 & - 6 & 1 \end{matrix}] A R_{1} = R_{1} - 5 R_{2}, R_{3} = R_{3} + 26 R_{2} [\begin{matrix} 1 & 0 & \frac{2}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & \frac{- 11}{3} \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ \frac{- 1}{15} & \frac{1}{5} & 0 \\ \frac{- 26}{15} & \frac{- 4}{5} & 1 \end{matrix}] A R_{3} = \frac{R_{3}}{- \frac{11}{3}} [\begin{matrix} 1 & 0 & \frac{2}{3} \\ 0 & 1 & \frac{5}{3} \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{3} & 0 & 0 \\ \frac{- 1}{15} & \frac{1}{5} & 0 \\ \frac{26}{55} & \frac{12}{55} & \frac{- 3}{11} \end{matrix}] A R_{2} = R_{2} - \frac{5}{3} R_{3}, R_{1} = R_{1} - \frac{2}{3} R_{3} [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{1}{55} & \frac{- 8}{55} & \frac{2}{11} \\ \frac{- 47}{55} & \frac{- 9}{55} & \frac{5}{11} \\ \frac{26}{55} & \frac{12}{55} & \frac{- 3}{11} \end{matrix}] A I = A^{- 1} A so, A^{- 1} = [\begin{matrix} \frac{1}{55} & \frac{- 8}{55} & \frac{2}{11} \\ \frac{- 47}{55} & \frac{- 9}{55} & \frac{5}{11} \\ \frac{26}{55} & \frac{12}{55} & \frac{- 3}{11} \end{matrix}]

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​using elementary row transformation find inverse of it following matrx ​ [3 0 2 1 5 9 6 4 7]

using elementary row transformation find inverse of it following matrx
[3 0 2
1 5 9
6 4 7]