The sum of three numbers is $6$ . If we multiply the third number by $3$ and add it to the second number to it, we get $11$ . By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

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Solution

Let the first , second and third number be $x, y, z$ respectively.
Then, according to the given condition, we have
$x + y + z = 6 y + 3 z = 11$
$x + z = 2 y$ or $x - 2 y + z = 0$

This system of equations can be written as $A X = B$ , where
$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 0 & 1 & 3 1 & - 2 & 1 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ & B = ⎡ ⎢ ⎣ \begin{matrix} 6110 \end{matrix} ⎤ ⎥ ⎦$

$A = 1 (1 + 6) - 0 + 1 (3 - 1)$
$= 9$
$\Rightarrow | A | \neq 0$
$∴$ The system of equation is consistent and has a unique solution.

Now, we find $a d j A$

$A_{11} = 7, A_{12} = 3, A_{13} = - 1$
$A_{21} = - 3, A_{22} = 0, A_{23} = 3$
$A_{31} = 2, A_{32} = - 3, A_{33} = 1$

Hence , $a d j (A) = ⎡ ⎢ ⎣ \begin{matrix} 7 & - 3 & 2 3 & 0 & - 3 - 1 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

Thus $A^{- 1} = \frac{1}{| A |} a d j (A)$
$= \frac{1}{9} ⎡ ⎢ ⎣ \begin{matrix} 7 & - 3 & 2 3 & 0 & - 3 - 1 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦$

Since, $A X = B$
$∴ X = A^{- 1} B$
$\Rightarrow X = \frac{1}{9} ⎡ ⎢ ⎣ \begin{matrix} 7 & - 3 & 2 3 & 0 & - 3 - 1 & 3 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 6110 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦ = \frac{1}{9} ⎡ ⎢ ⎣ \begin{matrix} 91827 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 123 \end{matrix} ⎤ ⎥ ⎦$
$\Rightarrow x = 1, y = 2, z = 3$

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