Question

Absolute value of sum of roots of the equation
\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {3x + 5}&{5x + 8}&{10x + 17} \end{array}} \right| = 0 is____.

Answer 1

Hint : Firstly evaluate the given determinant by using the row transformations and equate it to zero and then find the roots of the equation. Then add all the roots and take its absolute value.

Complete step-by-step answer :
Given: The equation is given is
\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {3x + 5}&{5x + 8}&{10x + 17} \end{array}} \right| = 0 .
Firstly evaluate the determinant by using the row transformations. Firstly use the transformation as ${R_3} \to {R_3} - {R_2}$ it gives:
\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {3x + 5}&{5x + 8}&{10x + 17} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {x + 2}&{2\left( {x + 2} \right)}&{6\left( {x + 2} \right)} \end{array}} \right|
Now use the transformation ${R_2} \to {R_2} - {R_1}$ on the determinant:
\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {x + 2}&{2\left( {x + 2} \right)}&{6\left( {x + 2} \right)} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {x + 1}&{x + 1}&{x + 1} \\\ {x + 2}&{2\left( {x + 2} \right)}&{6\left( {x + 2} \right)} \end{array}} \right|
Now take $x + 1$ and $x + 2$ common from second and third row respectively.
\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {x + 1}&{x + 1}&{x + 1} \\\ {x + 2}&{2\left( {x + 2} \right)}&{6\left( {x + 2} \right)} \end{array}} \right| = \left( {x + 1} \right)\left( {x + 2} \right)\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ 1&1&1 \\\ 1&2&6 \end{array}} \right|
Now evaluate the determinant:
\left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ 1&1&1 \\\ 1&2&6 \end{array}} \right| = 1\left( {\left( {2x + 3} \right) \times 1 - \left( {3x + 4} \right) \times 1} \right) - 2\left( {\left( {x + 2} \right) \times 1 - \left( {3x + 4} \right) \times 1} \right) + 6\left( {\left( {x + 2} \right) \times 1 - \left( {2x + 3} \right) \times 1} \right) \\\ = 1\left( {2x + 3 - 3x - 4} \right) - 2\left( {x + 2 - 3x - 4} \right) + 6\left( {x + 2 - 2x - 3} \right) \\\ = - x - 1 - 2\left( { - 2x - 2} \right) + 6\left( { - x - 1} \right) \\\ = - x - 1 + 4x + 4 - 6x - 6 \\\ = - 3x - 3 \\\
So, the value of \left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {3x + 5}&{5x + 8}&{10x + 17} \end{array}} \right| is equal to $\left( {x + 1} \right)\left( {x + 2} \right)\left( { - 3x - 3} \right)$ .
Now solve the equation $\left( {x + 1} \right)\left( {x + 2} \right)\left( { - 3x - 3} \right) = 0$ . It gives the roots of the equation as $- 1, - 2, - 1$ . The sum of the roots is equal to $- 1 - 2 - 1 = - 4$ .
The absolute value of the sum of roots of the equation is equal to $4$ .
So, the absolute value of sum of roots of the equation \left| {\begin{array}{*{20}{c}} {x + 2}&{2x + 3}&{3x + 4} \\\ {2x + 3}&{3x + 4}&{4x + 5} \\\ {3x + 5}&{5x + 8}&{10x + 17} \end{array}} \right| = 0 is equal to $4$ .
So, the correct answer is “4”.

Note : Please note that the question is for the absolute value of the sum of roots, don’t confuse it with the sum of the absolute values of the roots of the equation. Also the determinant of a matrix doesn’t change by operating its rows or columns as per the property of determinants