Question

Show that \[x = 2\] is a root of the equation \[\left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 3}&{2x}&{2 + x}
\end{array}} \right| = 0\]

Answer 1

Hint: * Roots of an equation \[a{x^n} + b{x^{n - 1}} + ........c = 0\]are those values of x which give the value of the equation equal to zero when substituted in the equation. Let y be a root of the equation, we can write \[(x - y) = 0\]is a factor of the equation. On equating the factor we write \[x = y\]is a root of the equation.
* Row transformation is a way of transforming elements of a row using operations like addition, multiplication etc with respect to any other row of the matrix.
* Determinant of a matrix \[\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)\]

Complete step-by-step answer:
We are given the determinant \[\left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 3}&{2x}&{2 + x}
\end{array}} \right|\]...............… (1)
We apply row transformations in order to make the terms of determinant simple.
Apply \[{R_3} \to {R_3} - {R_1}\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 3 - x}&{2x + 6}&{2 + x + 1}
\end{array}} \right|\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 3 - x}&{2x + 6}&{x + 3}
\end{array}} \right|\]
We can write elements of third row as
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - (x + 3)}&{2(x + 3)}&{x + 3}
\end{array}} \right|\]
Take \[(x + 3)\]common from every element in the third row.
 \[ \Rightarrow (x + 3)\left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\]
Apply \[{R_1} \to {R_1} - {R_2}\]
\[ \Rightarrow (x + 3)\left| {\begin{array}{*{20}{c}}
  {x - 2}&{ - 6 + 3x}&{ - 1 - x + 3} \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\]
\[ \Rightarrow (x + 3)\left| {\begin{array}{*{20}{c}}
  {x - 2}&{ - 6 + 3x}&{ - x + 2} \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\]
We can write elements of first row as
\[ \Rightarrow (x + 3)\left| {\begin{array}{*{20}{c}}
  {(x - 2)}&{3(x - 2)}&{ - 1(x - 2)} \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\]
Take \[(x - 2)\]common from every element in the first row.
 \[ \Rightarrow (x + 3)(x - 2)\left| {\begin{array}{*{20}{c}}
  1&3&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\]
Apply \[{R_1} \to {R_1} + {R_2}\]
\[ \Rightarrow (x + 3)(x - 2)\left| {\begin{array}{*{20}{c}}
  {1 - 1}&{3 + 2}&{ - 1 + 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\]
\[ \Rightarrow (x + 3)(x - 2)\left| {\begin{array}{*{20}{c}}
  0&5&0 \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right|\].............… (2)
Now we calculate the value of the determinant.
We know determinant of a matrix \[\left[ {\begin{array}{*{20}{c}}
  a&b&c \\
  d&e&f \\
  g&h&i
\end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  0&5&0 \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right| = 0\left( { - 3x - 2(x - 3)} \right) - 5\left( {2 - ( - 1)(x - 3)} \right) + 0\left( {4 - ( - 1)( - 3x)} \right)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  0&5&0 \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right| = - 5\left( {2 + x - 3} \right)\]
Solve the term inside the bracket
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  0&5&0 \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right| = - 5\left( {x - 1} \right)\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  0&5&0 \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right| = (5 - x)\]
Substitute the value of determinant back in equation (2)
\[ \Rightarrow (x + 3)(x - 2)\left| {\begin{array}{*{20}{c}}
  0&5&0 \\
  2&{ - 3x}&{x - 3} \\
  { - 1}&2&1
\end{array}} \right| = (x + 3)(x - 2)(5 - x)\]
Substitute the value in equation (1)
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  x&{ - 6}&{ - 1} \\
  2&{ - 3x}&{x - 3} \\
  { - 3}&{2x}&{2 + x}
\end{array}} \right| = (x + 3)(x - 2)(5 - x)\]
Since we are given the determinant is equal to zero
\[ \Rightarrow (x + 3)(x - 2)(5 - x) = 0\]
So, the equation has a factor \[(x - 2)\]
We know an equation has a factor, if we equate the factor to zero; we get the root of the equation.
\[ \Rightarrow x - 2 = 0\]
Shift the constant value to one side of the equation
\[ \Rightarrow x = 2\]

\[\therefore \]We can say \[x = 2\] is a root of the equation.

Note: Students many times try to solve the determinant directly without applying any row transformation. If we solve for determinants in such a way we will get a very complex equation, whose factors will be very difficult to find.