Find the inverse of each of the following matrices and verify that $A^{- 1} A = I_{3}$ .

(i) $[\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}]$

(ii) $[\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}]$

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Solution

$(i) A = [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}] Now, C_{11} = |\begin{matrix} 4 & 3 \\ 3 & 4 \end{matrix}| = 7, C_{12} = - |\begin{matrix} 1 & 3 \\ 1 & 4 \end{matrix}| = - 1 and C_{13} = |\begin{matrix} 1 & 4 \\ 1 & 3 \end{matrix}| = - 1 C_{21} = - |\begin{matrix} 3 & 3 \\ 3 & 4 \end{matrix}| = - 3, C_{22} = |\begin{matrix} 1 & 3 \\ 1 & 4 \end{matrix}| = 1 and C_{23} = - |\begin{matrix} 1 & 3 \\ 1 & 3 \end{matrix}| = 0 C_{31} = |\begin{matrix} 3 & 3 \\ 4 & 3 \end{matrix}| = - 3, C_{32} = - |\begin{matrix} 1 & 3 \\ 1 & 3 \end{matrix}| = 0 and C_{33} = |\begin{matrix} 1 & 3 \\ 1 & 4 \end{matrix}| = 1 adj A = {[\begin{matrix} 7 & - 1 & - 1 \\ - 3 & 1 & 0 \\ - 3 & 0 & 1 \end{matrix}]}^{T} = [\begin{matrix} 7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] and |A| = 1 A^{- 1} = [\begin{matrix} 7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] Now, A^{- 1} A = [\begin{matrix} 7 & - 3 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I_{3} (ii) B = [\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}] Now, C_{11} = |\begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix}| = 1, C_{12} = - |\begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix}| = - 3 and C_{13} = |\begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix}| = 9 C_{21} = - |\begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix}| = 1, C_{22} = |\begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix}| = 1 and C_{23} = - |\begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix}| = - 5 C_{31} = |\begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix}| = - 1, C_{32} = - |\begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix}| = 1 and C_{33} = |\begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix}| = - 1 adj B = {[\begin{matrix} 1 & - 3 & 9 \\ 1 & 1 & - 5 \\ - 1 & 1 & - 1 \end{matrix}]}^{T} = [\begin{matrix} 1 & 1 & - 1 \\ - 3 & 1 & 1 \\ 9 & - 5 & - 1 \end{matrix}] and |B| = 2 B^{- 1} = \frac{1}{2} [\begin{matrix} 1 & 1 & - 1 \\ - 3 & 1 & 1 \\ 9 & - 5 & - 1 \end{matrix}] Now, B^{- 1} B = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = I_{3}$

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