If a = 1 9 ⎡ ⎢ ⎣ − 8 1 4 4 4 7 1 − 8 4 ⎤ ⎥ ⎦ , Prove that a − 1 = a 3 - Mathematics

प्रश्न

If \[A = \frac{1}{9}\begin{bmatrix}- 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & - 8 & 4\end{bmatrix}\],
prove that \[A^{- 1} = A^3\]

उत्तर

\[A = \frac{1}{9}\begin{bmatrix} - 8 & 1 & 4\\ 4 & 4 & 7\\ 1 & - 8 & 4 \end{bmatrix} = \begin{bmatrix} \frac{- 8}{9} & \frac{1}{9} & \frac{4}{9}\\\frac{4}{9} & \frac{4}{9} & \frac{7}{9}\\ \frac{1}{9} & \frac{- 8}{9} & \frac{4}{9} \end{bmatrix}\]
\[ \Rightarrow A^T = \begin{bmatrix} \frac{- 8}{9} & \frac{4}{9} & \frac{1}{9}\\\frac{1}{9} & \frac{4}{9} & \frac{- 8}{9}\\ \frac{4}{9} & \frac{7}{9} & \frac{4}{9} \end{bmatrix} = \frac{1}{9}\begin{bmatrix} - 8 & 4 & 1\\ 1 & 4 & - 8 \\ 4 & 7 & 4 \end{bmatrix} . . . (1)\]
\[\left| A \right| = \begin{vmatrix} \frac{- 8}{9} & \frac{1}{9} & \frac{4}{9}\\\frac{4}{9} & \frac{4}{9} & \frac{7}{9}\\ \frac{1}{9} & \frac{- 8}{9} & \frac{4}{9} \end{vmatrix} = \frac{1}{9 \times 9 \times 9}\begin{vmatrix} - 8 & 1 & 4\\ 4 & 4 & 7\\ 1 & - 8 & 4 \end{vmatrix}\]
\[ = \frac{1}{9 \times 9 \times 9}\left[ \left( - 8 \times 72 \right) - \left( 1 \times 9 \right) + \left\{ 4 \times \left( - 36 \right) \right\} \right]\]
\[ = \frac{1}{9 \times 9 \times 9} \times 9 \times \left\{ - 64 - 1 - 16 \right\} = - \frac{9 \times 81}{9 \times 9 \times 9} = - 1\]
\[\text{ If }C_{ij}\text{ is a cofactor of }a_{ij}\text{ such that A }= \left[ a_{ij} \right],\text{ then we have }\]
\[ C_{11} = \frac{8}{9} C {}_{12} = \frac{- 1}{9} C {}_{13} = \frac{- 4}{9}\]
\[ C_{21} = \frac{- 4}{9} C_{22} = \frac{- 4}{9} C_{23} = \frac{- 7}{9}\]
\[ C_{31} = \frac{- 1}{9} C_{32} = \frac{8}{9} C_{33} = \frac{- 4}{9}\]
Now,
\[adj A = \begin{bmatrix} \frac{8}{9} & \frac{- 1}{9} & \frac{- 4}{9}\\\frac{- 4}{9} & \frac{- 4}{9} & \frac{- 7}{9}\\\frac{- 1}{9} & \frac{8}{9} & \frac{- 4}{9} \end{bmatrix}^T = \begin{bmatrix} \frac{8}{9} & \frac{- 4}{9} & \frac{- 1}{9} \\\frac{- 1}{9} & \frac{- 4}{9} & \frac{8}{9}\\ \frac{- 4}{9} & \frac{- 7}{9} & \frac{- 4}{9} \end{bmatrix}\]
\[ \therefore A^{- 1} = \frac{1}{\left| A \right|}adj A = - 1 \times \frac{1}{9}\begin{bmatrix} 8 & - 4 & - 1\\ - 1 & - 4 & 8 \\ - 4 & - 7 & - 4 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} - 8 & 4 & 1\\1 & 4 & - 8 \\ 4 & 7 & 4 \end{bmatrix} = A^T [\text{ From } (1)]\]
\[ \Rightarrow A^{- 1} = A^T \]

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Inverse of Matrix - Inverse of a Square Matrix by the Adjoint Method

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Q 27 Q 26 Q 28

पाठ 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [पृष्ठ २४]

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आरडी शर्मा Mathematics [English] Class 12

पाठ 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 27 | पृष्ठ २४

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