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Consistency of Linear System of Equations

Given two mat...

Question

Given two matrices $A$ and $B$
$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 3 1 & 4 & 1 1 & - 3 & 2 \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} 11 & - 5 & - 14 - 1 & - 1 & 2 - 7 & 1 & 6 \end{matrix} ⎤ ⎥ ⎦$
Find $A B$ and use this result to solve the following system of equations:
$x - 2 y + 3 z = 6, x + 4 y + z = 12, x - 3 y + 2 z = 1$

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Solution

Given,
$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 3 1 & 4 & 1 1 & - 3 & 2 \end{matrix} ⎤ ⎥ ⎦$ and $B = ⎡ ⎢ ⎣ \begin{matrix} 11 & - 5 & - 14 - 1 & - 1 & 2 - 7 & 1 & 6 \end{matrix} ⎤ ⎥ ⎦$

$A B = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 3 1 & 4 & 1 1 & - 3 & 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 11 & - 5 & - 14 - 1 & - 1 & 2 - 7 & 1 & 6 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} 11 + 2 - 21 & - 5 + 2 + 3 & - 14 - 4 + 18 11 - 4 - 7 & - 5 - 4 + 1 & - 14 + 8 + 6 11 + 3 - 14 & - 5 + 3 + 2 & - 14 - 6 + 12 \end{matrix} ⎤ ⎥ ⎦$

$= ⎡ ⎢ ⎣ \begin{matrix} - 8 & 0 & 0 0 & - 8 & 0 0 & 0 & - 8 \end{matrix} ⎤ ⎥ ⎦$

$∴ A B = - 8 I_{3}$
$\Rightarrow - \frac{1}{8} (B) = A^{- 1}$

The given system of equations can be written as,
$⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 3 1 & 4 & 1 1 & - 3 & 2 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 6121 \end{matrix} ⎤ ⎥ ⎦$
Or, $A X = C$
Where,
$X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ and $C = ⎡ ⎢ ⎣ \begin{matrix} 6121 \end{matrix} ⎤ ⎥ ⎦$

As A exists, the given system of equations has a unique solution
$∴ X = A^{- 1} C$
$\Rightarrow X = (\begin{matrix} - \frac{1}{8} B \end{matrix}) C$

$= - \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} 11 & - 5 & - 14 - 1 & - 1 & 2 - 7 & 1 & 6 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 6121 \end{matrix} ⎤ ⎥ ⎦$

$= - \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} 66 & - 60 & - 14 - 6 & - 12 & + 2 - 42 & + 12 & + 6 \end{matrix} ⎤ ⎥ ⎦$

$⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = - \frac{1}{8} ⎡ ⎢ ⎣ \begin{matrix} - 8 - 16 - 24 \end{matrix} ⎤ ⎥ ⎦$

$⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow x = 1, y = 2, z = 3$

Hence the solution of equations is,
$x = 1, y = 2, z = 3$

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