Question

Find the value of $x$ if $\left| {\begin{array}{*{20}{c}}
  {15 - x}&{11}&{10} \\
  {11 - 3x}&{17}&{16} \\
  {7 - x}&{14}&{13}
\end{array}} \right| = 0$

Answer 1

Hint: We will simplify the given determinant by the rows ${R_1},{R_2}and{C_2}$.Thereafter, we will convert a $1$${a_{12}} = 0,{a_{22}} = 0and{a_{32}} = 1$, and then we will find the value of $x$.


Complete step by step solution:
The given determinant is $\left| {\begin{array}{*{20}{c}}
  {15 - x}&{11}&{10} \\
  {11 - 3x}&{17}&{16} \\
  {7 - x}&{14}&{13}
\end{array}} \right| = 0$

Applying ${R_1} \to {R_1} - {R_2},\,\,{R_2} \to {R_2} - {R_3}$
$\left| {\begin{array}{*{20}{c}}
  {15 - x(11 - 3x)}&{11 - 17}&{10 - 16} \\
  {11 - 3x - (7 - x)}&{17 - 14}&{16 - 13} \\
  {7 - x}&{14}&{13}
\end{array}} \right| = 0$
$\left| {\begin{array}{*{20}{c}}
  {15 - x - 11 - 3x}&{6 - }&{ - 6} \\
  {11 - 3x - 7 - x}&{ + 3}&4 \\
  {7 - x}&{14}&{13}
\end{array}} \right| = 0$
$\left| {\begin{array}{*{20}{c}}
  {4 + 2x}&{ - 6}&{ - 6} \\
  {4 - 2x}&{{T^3}}&3 \\
  {7 - x}&{14}&{13}
\end{array}} \right| = 0$
Now ${C_2} \to {C_2} - {C_3}$
\[\left| {\begin{array}{*{20}{c}}
  {4 + 2x}&{ - 6 + ( - 6)}&{ - 6} \\
  {4 - 2x}&{3 - 3}&3 \\
  {7 - x}&{14 - 13}&{13}
\end{array}} \right| = 0\]
\[\left| {\begin{array}{*{20}{c}}
  {4 + 2x}&0&{ - 6} \\
  {4 - 2x}&0&3 \\
  {7 - x}&1&{13}
\end{array}} \right| = 0\]
Now, ${a_{12}} = 0,\,{a_{22}} = 0$ and ${a_{32}} = 1$
So, we will take${a_{32}}$.
Now ${a_{32}}$ is an odd term i.e. $3 + 2$is an odd,
${( - 1)^{3 + 2}} = {( - 1)^5} = {( - 1)^4} - 1$
\[ = 1 \times - 1\]
$ = - 1$
Now,
$ - [(4 + 2x)3 + ( - 6)(4 - 2x)] = 0$
$ - 1[3(4 + 2x) - 6(4 - 2x)] = 0$
$ - 3[4 + 2x - 2(4 - 2x)] = 0$
$ - 3[4 + 2x - 8 + 4x] = 0$
$[4 + 2x - 6 + 4x] = 0$
$[6x - 4] = 0$
$6x - 4 = 0$
$6x = 4$
$x = \dfrac{4}{6}$
$x = \dfrac{2}{3}$
Thus, the value of $x = \dfrac{2}{3}$


Note: Students we should keep in mind that in the determinant the coefficient of the even number like ${a_{22}}$ gives $1$ answer and the coefficient of the odd number like ${a_{32}}$ gives-1answer.