Find the inverse by adjoining method of coefficient matrix of the following equations and solve them
$2 x + 2 y + z = 3, 3 x + 4 y - 2 z + 5 = 0, 5 x - 3 y + 3 z = 1$

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Solution

$\begin{matrix} 2 x + 2 y + z = 3, 3 x + 4 y - 2 z = - 5, 5 x - 2 y + 3 z = 1 \Rightarrow A = ⎡ ⎢ ⎣ \begin{matrix} 231 34 - 2 5 - 33 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ a n d B = ⎡ ⎢ ⎣ \begin{matrix} 3 - 5 1 \end{matrix} ⎤ ⎥ ⎦ A X = B \Rightarrow | A | = [2 (12 - 6) - 2 (9 + 10) + 1 (- 9 - 20)] \Rightarrow | A | = [12 - 38 - 29] = - 55 \neq 0 A i s i n v e r t i b l e X = A^{- 1} B C o f a c t o r o f | A | a r e \Rightarrow A_{11} = 6, A_{12} = - 19, A_{13} = - 29 \Rightarrow A_{21} = - 9, A_{22} = 1, A_{23} = + 16 \Rightarrow A_{31} = - 8, A_{32} = + 7, A_{33} = 2 \Rightarrow a \frac{d}{A} = ⎡ ⎢ ⎣ \begin{matrix} 6 - 9 - 8 - 1917 - 29162 \end{matrix} ⎤ ⎥ ⎦ (X = A^{1} B) \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x y z \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{- 55} ⎡ ⎢ ⎣ \begin{matrix} 6 - 9 - 8 - 1917 - 29162 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 3 - 5 1 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} x y z \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{- 55} ⎡ ⎢ ⎣ \begin{matrix} 55 - 55 - 165 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} - 1 13 \end{matrix} ⎤ ⎥ ⎦ ∴ x = - 1, y = 1 a n d 2 = 3 A n s . \end{matrix}$

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