{x_1} + 2{x_2} + 3{x_3} = a2{x_1} + 3{x_2} + | vedclass

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(c) We have, ${x_1} + 2{x_2} + 3{x_3} = c$

$2a{x_1} + 3{x_2} + {x_3} = c$

$3b{x_1} + {x_2} + 2{x_3} = c$

Let $a = b = c = 1$.

Then $D = \l

Vedclass · Accepted Answer

Unique solution

Vedclass · Answer

(c) We have, ${x_1} + 2{x_2} + 3{x_3} = c$

$2a{x_1} + 3{x_2} + {x_3} = c$

$3b{x_1} + {x_2} + 2{x_3} = c$

Let $a = b = c = 1$.

Then $D = \left| {\,\begin{array}{*{20}{c}}1&2&3\2&3&1\3&1&2\end{array}\,} ight|$ = $1\,(5) - 2\,(1) + 3\,( - 7) = - 18 
e 0$

${D_x} = \left| {\,\begin{array}{*{20}{c}}1&2&3\1&3&1\1&1&2\end{array}\,} ight| = - 3$

Similarly ${D_y} = {D_z} = - 3$. Now, $x = \frac{{{D_z}}}{D}$ = $\frac{1}{6}$

Hence $D 
e 0$, $x = y = z$, i.e., unique solution.

${x_1} + 2{x_2} + 3{x_3} = a2{x_1} + 3{x_2} + {x_3} = $ $b3{x_1} + {x_2} + 2{x_3} = c$ this system of equations has

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