Given, $A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}]$ compute A^-1 and show that 2A^- = 9I-A

Solution

$A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}] C_{11} = (- 1)^{1 + 1} (7) = 7, C_{12} = (- 1)^{1 + 2} (- 4) = 4 C_{21} = (- 1)^{2 + 1} (- 3) = 3 C_{22} = (- 1)^{2 + 2} 2 = 2 Adj A = [\begin{matrix} 7 & 4 \\ 3 & 2 \end{matrix}]' = [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] | A | = 2 x 7 - (- 3) x (- 4) = 14 - 12 \neq 0 ∴ A^{- 1} = \frac{1}{| A |} adj A = \frac{1}{2} [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] = [\begin{matrix} 7 / 2 & 3 / 2 \\ 2 & 1 \end{matrix}] L . H . S = 2 A^{- 1} = 2 [\begin{matrix} 7 / 2 & 3 / 2 \\ 2 & 1 \end{matrix}] = [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] R . H . S = 9 I - A = 9 [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}] = [\begin{matrix} 9 & 0 \\ 0 & 9 \end{matrix}] - [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}] = [\begin{matrix} 9 - 2 & 3 \\ 4 & 9 - 7 \end{matrix}] = [\begin{matrix} 7 & 3 \\ 4 & 2 \end{matrix}] ∴ L . H . S = R . H . S Hence, verified that 2 A^{- 1} = 9 I - A$

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Given, $A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}]$ compute A^-1 and show that 2A^- = 9I-A

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Given, $A = [\begin{matrix} 2 & - 3 \\ - 4 & 7 \end{matrix}]$ compute A^-1 and show that 2A^- = 9I-A