If a = ⎡ ⎢ ⎣ 3 − 3 4 2 − 3 4 0 − 1 1 ⎤ ⎥ ⎦ , Show that a − 1 = a 3 - Mathematics

Question

If \[A = \begin{bmatrix}3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1\end{bmatrix}\] , show that \[A^{- 1} = A^3\]

Solution

\[\text{ We have, } A = \begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix}\]
\[ A^2 = \begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 6 + 0 & - 9 + 9 - 4 & 12 - 12 + 4\\6 - 6 + 0 & - 6 + 9 - 4 & 8 - 12 + 4\\0 - 2 + 0 & 0 + 3 - 1 & 0 - 4 + 1 \end{bmatrix} = \begin{bmatrix} 3 & - 4 & 4\\ 0 & - 1 & 0\\ - 2 & 2 & - 3 \end{bmatrix}\]
\[\text{ Now,} A^3 = A^2 \times A^{} = \begin{bmatrix} 3 & - 4 & 4\\ 0 & - 1 & 0\\ - 2 & 2 & - 3 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 9 - 8 & - 9 + 12 - 4 & 12 - 16 + 4\\ 0 - 2 + 0 & 0 + 3 + 0 & - 4\\ - 6 + 4 + 0 & 6 - 6 + 3 & - 8 + 8 - 3 \end{bmatrix} = \begin{bmatrix} 1 & - 1 & 0\\ - 2 & 3 & - 4\\ - 2 & 3 & - 3 \end{bmatrix}\]
\[\text{ Again, }A^3 \times A = \begin{bmatrix} 1 & - 1 & 0\\ - 2 & 3 & - 4\\ - 2 & 3 & - 3 \end{bmatrix}\begin{bmatrix} 3 & - 3 & 4\\2 & - 3 & 4\\0 & - 1 & 1 \end{bmatrix} = \begin{bmatrix} 3 - 2 + 0 & - 3 + 3 + 0 & 4 - 4 + 0\\ - 6 + 6 & 6 - 9 + 4 & - 8 + 12 - 4\\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = I_3 [\text{ Identity matrix of order 3 }]\]
\[ \Rightarrow A^3 \times A = I_3 \]
\[ \Rightarrow A^3 = A^{- 1}\]

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

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Q 28 Q 27 Q 29

Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 24]

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RD Sharma Mathematics [English] Class 12

Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 28 | Page 24

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