Question

One root of the equation \left| {\begin{array}{*{20}{c}} {3x - 8}&3&3 \\\ 3&{3x - 8}&3 \\\ 3&3&{3x - 8} \end{array}} \right| is which of the following?
A.$\dfrac{8}{3}$
B.$\dfrac{2}{3}$
C.$\dfrac{1}{3}$
D.$\dfrac{{16}}{3}$

Answer 1

First we have to define what the terms we need to solve the problem:
To find the one root of the given equation, we apply the root finding condition here. The given equation in $3 \times 3$. So we solve the equation by row subtraction and then simplify it into $2 \times 2$ matrix. By solving the $2 \times 2$ matrix we will get the one root of the equation.
Formula to be used:
To find the roots of the equation we use the condition $p(x) = 0$, which means given equation is equal to zero.

Complete answer:
First, we take given determinant as $p(x)$
$p(x)$ = $$$$\left| {\begin{array}{*{20}{c}} {3x - 8}&3&3 \\\ 3&{3x - 8}&3 \\\ 3&3&{3x - 8} \end{array}} \right|
We know that to find the roots of the equation we take $p(x) = 0$
So we take the give equation is equal to zero.
\left| {\begin{array}{*{20}{c}} {3x - 8}&3&3 \\\ 3&{3x - 8}&3 \\\ 3&3&{3x - 8} \end{array}} \right|$$$$ = $$$$0
We are going to perform row reduction determinant rule here,
We are going to do the Row subtraction to simplify the given determinant,
${R_1} \to {R_1} - {R_2}$
Subtracting ${R_2}$ from ${R_1}$ , and then place the answer in ${R_1}$ ,
By subtracting we get,
\left| {\begin{array}{*{20}{c}} {3x - 5}&{ - (3x - 5)}&0 \\\ 3&{3x - 8}&3 \\\ 3&3&{3x - 8} \end{array}} \right| = $$$$0
Then we have ule that any number or equation common in one row it is taken outside as common for determinant
Taking common equation from ${R_1}$ , Taking $(3x - 5)$ as common,(3x - 5)\left| {\begin{array}{*{20}{c}} 1&{ - 1}&0 \\\ 3&{3x - 8}&3 \\\ 3&3&{3x - 8} \end{array}} \right| = 0
Again, doing same process between ${R_2}$ and ${R_3}$
${R_2} \to {R_2} - {R_3}$
By simplifying we get,
(3x - 5)\left| {\begin{array}{*{20}{c}} 1&{ - 1}&0 \\\ 0&{3x - 5}&{ - (3x - 5)} \\\ 3&3&{3x - 8} \end{array}} \right| = 0
Again taking $(3x - 5)$ as common, we get
{(3x - 5)^2}\left| {\begin{array}{*{20}{c}} 1&{ - 1}&0 \\\ 0&1&{ - 1} \\\ 3&3&{3x - 8} \end{array}} \right| = 0
We want to simplify the $3 \times 3$into roots, that’s why we are taking the determinant value.{(3x - 5)^2}1\left| {\begin{array}{*{20}{c}} 1&{ - 1} \\\ 3&{3x - 8} \end{array}} \right| - \left( { - 1\left| {\begin{array}{*{20}{c}} 0&{ - 1} \\\ 3&{3x - 8} \end{array}} \right|} \right) + 0\left| {\begin{array}{*{20}{c}} 0&1 \\\ 3&3 \end{array}} \right|
Simplifying each determinant we get,
${(3x - 5)^2}(3x - 8 + 3) - ( - 3) + 0 = 0$
Simplifying the algebraic equation to get one root of the equation
${(3x - 5)^2}(3x - 5 + 3) = 0$
Therefore,
$3x - 2 = 0$
$3x = 2$
Hence the value of $x$ will be
$x = \dfrac{2}{3}$
The one root of the equation is $x = \dfrac{2}{3}$

Hence the correct option is option B) $\dfrac{2}{3}$

Note:
For the given equation we not only have one root, by simplifying ${(3x - 5)^2}(3x - 5 + 3) = 0$into ${(3x - 5)^2} = 0$we get $x = \dfrac{5}{3}$, Hence this is another root of the equation. By checking the power of the equation, we can conclude the total number of roots we have. For linear equation we have one root, for quadratic equation we have two roots. For cubic equation we have three roots.