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If $\left| {\,\begin{array}{*{20}{c}}{3x - 8}&3&3\\3&{3x - 8}&3\\3&3&{3x - 8}\end{array}\,} \right| = 0,$ then the values of $x$ are
$0, 2/3$
$2/3, 11/3$
$1/2, 1$
$11/3, 1$
Solution
(b) $\left| {\,\begin{array}{*{20}{c}}{3x – 8}&3&3\\3&{3x – 8}&3\\3&3&{3x – 8}\end{array}\,} \right| = 0$
${C_1} \to {C_1} + {C_2} + {C_3}$, we get
$(3x – 2)\left| {\,\begin{array}{*{20}{c}}1&3&3\\1&{3x – 8}&3\\1&3&{3x – 8}\end{array}\,} \right| = 0$
${R_1} \to {R_1} – {R_2}$and ${R_2} \to {R_2} – {R_3}$, we get
$(3x – 2)\left| {\,\begin{array}{*{20}{c}}0&{ – 3x + 11}&0\\0&{3x – 11}&{ – 3x + 11}\\1&3&{3x – 8}\end{array}\,} \right| = 0$
==>$(3x – 2)\left[ {{{( – 3x + 11)}^2}} \right] = 0$
==>$x = \frac{2}{3}$ or $x = \frac{{11}}{3}\,\,\,\, \Rightarrow x = \frac{2}{3},\frac{{11}}{3}$.
