Question

If $ \left| {\begin{array}{*{20}{c}}
  {3x + 4}&{4x + 5}&{10x + 17} \\
  {x + 2}&{2x + 3}&{3x + 5} \\
  {2x + 3}&{3x + 4}&{5x + 8}
\end{array}} \right| = 0 $ . Find out $ x $ .

Answer 1

Hint:In order to deal with this question we will first do raw and column operation to convert the given determinant matrix in the simplest form further we will open it with any row or column and we will get the equation in $ x $ and by solving it we will get the required answer.

Complete step-by-step answer:
Given equation is $ \left| {\begin{array}{*{20}{c}}
  {3x + 4}&{4x + 5}&{10x + 17} \\
  {x + 2}&{2x + 3}&{3x + 5} \\
  {2x + 3}&{3x + 4}&{5x + 8}
\end{array}} \right| = 0 $

First we have to simplify above equation so for it we will evaluate row-column operations
Therefore first we have to do column operation in second column as $ {C_2} \to {C_2} - {C_1} $
So we have

 $ \left| {\begin{array}{*{20}{c}}
  {3x + 4}&{4x + 5 - (3x + 4)}&{10x + 17} \\
  {x + 2}&{2x + 3 - (x + 2)}&{3x + 5} \\
  {2x + 3}&{3x + 4 - (2x + 3)}&{5x + 8}
\end{array}} \right| = 0 $

 $ \Rightarrow \left| {\begin{array}{*{20}{c}}
  {3x + 4}&{x + 1}&{10x + 17} \\
  {x + 2}&{x + 1}&{3x + 5} \\
  {2x + 3}&{x + 1}&{5x + 8}
\end{array}} \right| = 0 $

By taking $ x + 1 $ common from second column , we have
 $ \Rightarrow (x + 1)\left| {\begin{array}{*{20}{c}}
  {3x + 4}&1&{10x + 17} \\
  {x + 2}&1&{3x + 5} \\
  {2x + 3}&1&{5x + 8}
\end{array}} \right| = 0 $
Now by row operation in second and third row as $ {R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1} $ , we get

 $ (x + 1)\left| {\begin{array}{*{20}{c}}
  {3x + 4}&1&{10x + 17} \\
  {x + 2 - (3x + 4)}&{1 - 1}&{3x + 5 - (10x + 17)} \\
  {2x + 3 - (3x + 4)}&{1 - 1}&{5x + 8 - (10x + 17)}
\end{array}} \right| = 0 $

 $ (x + 1)\left| {\begin{array}{*{20}{c}}
  {3x + 4}&1&{10x + 17} \\
  { - 2x - 2}&0&{ - 7x - 12} \\
  { - x - 1}&0&{ - 5x - 9}
\end{array}} \right| = 0 $

Now by expanding above matrix with the help of second column we have

 $
  (x + 1)\left[ { - 1\left| {\begin{array}{*{20}{c}}
  { - 2x - 2}&{ - 7x - 12} \\
  { - x - 1}&{ - 5x - 9}
\end{array}} \right| + 0 - 0} \right] = 0 \\
   \Rightarrow (x + 1)\left[ { - 1\{ ( - 2x - 2)( - 5x - 9) - ( - 7x - 12)( - x - 1)\} } \right] = 0 \\
   \Rightarrow (x + 1)[ - 1\{ ( - 2x - 2)( - 5x - 9) - (7x + 12)(x + 1)\} ] = 0 \\
   \Rightarrow (x + 1)[7{x^2} + 19x + 12 - 10{x^2} - 28x - 18] = 0 \\
   \Rightarrow (x + 1)[ - 3{x^2} - 9x - 6] = 0 \\
 $

Further solving quadratic equation by taking common as $ 3 $ , we get
 $
    \\
   \Rightarrow (x + 1)( - 3)[{x^2} + 3x - 2] = 0 \\
   \Rightarrow (x + 1)[{x^2} + 3x + 2] = \dfrac{0}{{( - 3)}} \\
   \Rightarrow (x + 1)[{x^2} + 2x + x + 2] = 0 \\
   \Rightarrow (x + 1)[x(x + 2) + 1(x + 2)] = 0 \\
   \Rightarrow {(x + 1)^2}(x + 2) = 0 \\
  {\text{Either }}{(x + 1)^2} = 0{\text{ or }}(x + 2) = 0 \\
  {\text{if }}{(x + 1)^2} = 0 \\
   \Rightarrow x = - 1 \\
  {\text{By taking }}(x + 2) = 0 \\
   \Rightarrow x = - 2 \\
 $
 Hence, we get two values of $ x $ are $ - 1{\text{ and - 2}} $ .

Note:A column vector or column matrix is an $ M \times 1 $ matrix, that is, a matrix consisting of a single column of $ M $ elements, Similarly, a row vector or row matrix is a $ 1 \times M $ matrix, that is, a matrix consisting of a single row of m elements Throughout, boldface is used for the row and column vectors.