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Question: Show that \[x = 2\] is a root of the equation \[\left| {\begin{array}{*{20}{c}} x&{ - 6}&{ - 1} ...

Show that x=2x = 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

Explanation

Solution

  • Roots of an equation axn+bxn1+........c=0a{x^n} + b{x^{n - 1}} + ........c = 0are 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 (xy)=0(x - y) = 0is a factor of the equation. On equating the factor we write x=yx = yis 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 R3R3R1{R_3} \to {R_3} - {R_1}

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.