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Consistency of Linear System of Equations

Find A-1, w...

Question

Find $A^{- 1}$ , where $A = ω ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 3 1 & 1 & 1 3 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦$
Hence, solve the following system of linear equations
$4 x + 2 y + 3 z = 2$
$x + y + z = 1$
$3 x + y - 2 z = 5$

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Solution

Here $| A | = ω ∣ ∣ ∣ ∣ \begin{matrix} 4 & 2 & 3 1 & 1 & 1 3 & 1 & - 2 \end{matrix} ∣ ∣ ∣ ∣$
$= 4 (- 2 - 1) - 2 (- 2 - 3) + (1 - 3)$
$= - 12 + 10 - 6 = - 8 \neq 0$
$\Rightarrow A^{- 1}$ exists and $A^{- 1} = \frac{1}{| A |}$ adj A.

$A_{11} = ∣ ∣ ∣ \begin{matrix} 1 & 1 1 & - 2 \end{matrix} ∣ ∣ ∣ = - 3, A_{12} = - ∣ ∣ ∣ \begin{matrix} 1 & 1 3 & - 2 \end{matrix} ∣ ∣ ∣ = 5, A_{13} = ∣ ∣ ∣ \begin{matrix} 1 & 1 3 & 1 \end{matrix} ∣ ∣ ∣ = - 2$

$A_{21} = - ∣ ∣ ∣ \begin{matrix} 2 & 3 1 & - 2 \end{matrix} ∣ ∣ ∣ = 7, A_{22} = ∣ ∣ ∣ \begin{matrix} 4 & 3 3 & - 1 \end{matrix} ∣ ∣ ∣ = - 17, A_{23} = - ∣ ∣ ∣ \begin{matrix} 4 & 2 3 & 1 \end{matrix} ∣ ∣ ∣ = 2$

$A_{31} = ∣ ∣ ∣ \begin{matrix} 2 & 3 1 & 1 \end{matrix} ∣ ∣ ∣ = - 1, A_{32} = - ∣ ∣ ∣ \begin{matrix} 4 & 3 1 & 1 \end{matrix} ∣ ∣ ∣ = - 1, A_{33} = ∣ ∣ ∣ \begin{matrix} 4 & 2 1 & 1 \end{matrix} ∣ ∣ ∣ = 2$

$a d j . A$ $= {⎡ ⎢ ⎣ \begin{matrix} - 3 & 5 & - 2 7 & - 17 & 2 - 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦}^{T} = ⎡ ⎢ ⎣ \begin{matrix} - 3 & 7 & - 1 5 & - 17 & - 1 - 2 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦$

$∴ A^{- 1} = \frac{1}{| A |} a d j . A = - \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} - 3 & 7 & - 1 5 & - 17 & - 1 - 2 & 2 & 2 \end{matrix} ⎤ ⎥ ⎦$

$= \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} 3 & - 7 & 1 - 5 & 17 & 1 2 & - 2 & - 2 \end{matrix} ⎤ ⎥ ⎦$
The given system of equation is
$4 x + 2 y + 3 z = 2$
$x + y + z = 1$
$3 x + y - 2 z = 5$
This system can be written as $A x = B$
Where $A = ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 3 1 & 1 & 1 3 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} 215 \end{matrix} ⎤ ⎥ ⎦$

As $| A | \neq 0$ , the given system has a unique solution $X = A^{- 1} B$
$\Rightarrow X = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 7 & 1 - 5 & 17 & 1 2 & - 2 & - 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 215 \end{matrix} ⎤ ⎥ ⎦$

$= \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} 6 & - 7 & + 5 - 10 & + 17 & + 5 4 & - 2 & - 10 \end{matrix} ⎤ ⎥ ⎦ = \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} 412 - 8 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} 13 - 2 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow x = \frac{1}{2}, y = \frac{3}{2}, z = 1$

Hence, the solution of the given system of equation is
$x = \frac{1}{2}, y = \frac{3}{2}, z = - 1$

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x - y + 2 z = 1

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A = ⎡ ⎢ ⎣ \begin{matrix} - 5 & 1 & 3 7 & 1 & - 5 1 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 2 3 & 2 & 1 2 & 1 & 3 \end{matrix} ⎤ ⎥ ⎦

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