3 and 4 .Determinants and Matrices
hard

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

A

$0, 2/3$

B

$2/3, 11/3$

C

$1/2, 1$

D

$11/3, 1$

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 $x = \frac{{11}}{3}\,\,\,\, \Rightarrow x = \frac{2}{3},\frac{{11}}{3}$.

Standard 12
Mathematics

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