Question

The total number of distinct $x \in R$ for which $\left( {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + {x^3}} \\
  {2x}&{4{x^2}}&{1 + 8{x^3}} \\
  {3x}&{9{x^2}}&{1 + 27{x^3}}
\end{array}} \right) = 10$ is?

Answer 1

Hint: To solve the matrix problems, try to convert them in the linear equation and solve them. Prior to making the linear equation, make as many as columns and/ or rows a null values so that linear equations made after that will be easy to solve.

Complete step-by-step answer:
Firstly, write the expression given in the question as,
$\left( {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + {x^3}} \\
  {2x}&{4{x^2}}&{1 + 8{x^3}} \\
  {3x}&{9{x^2}}&{1 + 27{x^3}}
\end{array}} \right) = 10$
This can be re-written by taking out $x$from the first column and ${x^2}$from second column in the following way,
  $x.{x^2}\left( {\begin{array}{*{20}{c}}
  1&1&{1 + {x^3}} \\
  2&4&{1 + 8{x^3}} \\
  3&9&{1 + 27{x^3}}
\end{array}} \right) = 10$
Now doing the following operation in the matrix,
\[
  {R_2} \to {R_2} - 2{R_1} \\
  {R_3} \to {R_3} - 3{R_1} \\
 \]
We will get the following expression,
\[
  x.{x^2}\left( {\begin{array}{*{20}{c}}
  1&1&{1 + {x^3}} \\
  0&2&{1 + 8{x^3} - 2 - 2{x^3}} \\
  0&6&{1 + 27{x^3} - 3 - 3{x^3}}
\end{array}} \right) = 10 \\
  {x^3}\left( {\begin{array}{*{20}{c}}
  1&1&{1 + {x^3}} \\
  0&2&{6{x^3} - 1} \\
  0&6&{24{x^3} - 2}
\end{array}} \right) = 10 \\
 \]
Now taking out 2 as common from second column,
\[{x^3}.2\left( {\begin{array}{*{20}{c}}
  1&1&{1 + {x^3}} \\
  0&1&{6{x^3} - 1} \\
  0&3&{24{x^3} - 2}
\end{array}} \right) = 10\]
Further simplifying it and making a linear equation as below,
$2{x^3}[1.1.(24{x^3} - 2) - 3.(6{x^3} - 1)] = 10$
Divide the whole expression by 2,
$ \Rightarrow {x^3}[24{x^3} - 2 - 18{x^3} + 3] = 5$
Doing further mathematical operations, we will get the following expression,
$ \Rightarrow {x^3}[6{x^3} + 1] = 5$
Now, take 5 to LHS and make the equation as
$ \Rightarrow 6{x^6} + {x^3} - 5 = 0$
Factoring the above expression in terms of ${x^3}$ as
$ \Rightarrow 6{({x^3})^2} + {x^3} - 5 = 0$
Making the expression like $(A + B)(A - C) = {A^2} - AC + AB - BC$as,
$
   \Rightarrow 6{({x^3})^2} + 6{x^3} - 5{x^3} - 5 = 0 \\
   \Rightarrow 6{x^3}({x^3} + 1) - 5({x^3} + 1) = 0 \\
   \Rightarrow ({x^3} + 1)(6{x^3} - 5) = 0 \\
 $
In this way, we get two values for $x$as
First value: $
  {x^3} + 1 = 0 \\
   \Rightarrow {x^3} = - 1 \\
   \Rightarrow x = \sqrt[3]{{ - 1}} \\
   \Rightarrow x = - 1 \\
 $
And second value: $
  6{x^3} - 5 = 0 \\
   \Rightarrow {x^3} = \dfrac{5}{6} \\
   \Rightarrow x = \sqrt[3]{{\dfrac{5}{6}}} \\
 $
In this way, $x$ will have two values as follows,
$ - 1{\text{ and }}\sqrt[3]{{\dfrac{5}{6}}}$.

Note: There is an alternate method for simplification of the given matrix shown in the following steps. This method shows that breaking matrix into two matrices will make problem solving easier. Let us see how this simplification is done.
 Firstly, write the expression given in the question as,
$\left( {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + {x^3}} \\
  {2x}&{4{x^2}}&{1 + 8{x^3}} \\
  {3x}&{9{x^2}}&{1 + 27{x^3}}
\end{array}} \right) = 10$
The next step is to break the matrix into two, as below
\[
   \Rightarrow \left( {\begin{array}{*{20}{c}}
  x&{{x^2}}&1 \\
  {2x}&{4{x^2}}&1 \\
  {3x}&{9{x^2}}&1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  x&{{x^2}}&{{x^3}} \\
  {2x}&{4{x^2}}&{8{x^3}} \\
  {3x}&{9{x^2}}&{27{x^3}}
\end{array}} \right) = 10 \\
   \Rightarrow x.{x^2}.\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  2&{{2^2}}&1 \\
  3&{{3^2}}&1
\end{array}} \right) + x.{x^2}.{x^3}\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  2&{{2^2}}&{{2^3}} \\
  3&{{3^2}}&{{3^3}}
\end{array}} \right) = 10 \\
   \Rightarrow {x^3}.\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  2&{{2^2}}&1 \\
  3&{{3^2}}&1
\end{array}} \right) + 2.3.{x^6}\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&{{2^2}} \\
  1&3&{{3^2}}
\end{array}} \right) = 10 \\
   \Rightarrow {x^3}.\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  2&{{2^2}}&1 \\
  3&{{3^2}}&1
\end{array}} \right) + 6{x^6}\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&{{2^2}} \\
  1&3&{{3^2}}
\end{array}} \right) = 10 \\
   \Rightarrow {x^3}(1 + 6{x^3})\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&2&4 \\
  1&3&9
\end{array}} \right) = 10 \\
 \]
Now, using ${C_2} \to {C_2} - {C_1}{\text{ and }}{C_3} \to {C_3} - {C_1}$we get
\[
   \Rightarrow {x^3}(1 + 6{x^3})\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&3 \\
  1&2&8
\end{array}} \right) = 10 \\
   \Rightarrow {x^3}(1 + 6{x^3}).1.(8 - 6) = 10 \\
   \Rightarrow {x^3}(1 + 6{x^3}) = 5 \\
   \Rightarrow 6{({x^3})^2} + {x^3} - 5 = 0 \\
 \]
Making the expression like $(A + B)(A - C) = {A^2} - AC + AB - BC$as,
$
   \Rightarrow 6{({x^3})^2} + 6{x^3} - 5{x^3} - 5 = 0 \\
   \Rightarrow 6{x^3}({x^3} + 1) - 5({x^3} + 1) = 0 \\
   \Rightarrow ({x^3} + 1)(6{x^3} - 5) = 0 \\
 $
In this way, we get two values for $x$as
First value: $
  {x^3} + 1 = 0 \\
   \Rightarrow {x^3} = - 1 \\
   \Rightarrow x = \sqrt[3]{{ - 1}} \\
   \Rightarrow x = - 1 \\
 $
And second value: $
  6{x^3} - 5 = 0 \\
   \Rightarrow {x^3} = \dfrac{5}{6} \\
   \Rightarrow x = \sqrt[3]{{\dfrac{5}{6}}} \\
 $
In this way, $x$ will have two values as follows,
$ - 1{\text{ and }}\sqrt[3]{{\dfrac{5}{6}}}$.