Question

If $a \ne b \ne c$are value of x which satisfies the equation \left| {\begin{array}{*{20}{c}} 0&{x - a}&{x - b} \\\ {x + a}&0&{x - c} \\\ {x + b}&{x + c}&0 \end{array}} \right| = 0 is given by
a) $x = 0$
b) $x = c$
c) $x = b$
d) $x = a$

Answer 1

Here in this question, we have to find the value of the determinant of order $3 \times 3$. To solve this first we have to expand the determinant further and simplify using a basic arithmetical operation to get the required solution. Here the terms are in the form of algebraic expressions.

Complete answer:
Determinants are mathematical objects that are very useful in the analysis and solution of systems of linear equations. As shown by Cramer’s rule, a nonhomogeneous system of linear equations has a unique solution if and only if the determinant of the system's Matrix is non zero (i.e., the matrix is non-singular).
Now consider the given determinant of order $3 \times 3$:

0&{x - a}&{x - b} \\\ {x + a}&0&{x - c} \\\ {x + b}&{x + c}&0 \end{array}} \right| = 0$$ Now, expand the Determinant of a above $$3 \times 3$$ matrix by cofactor expansion theorem: The cofactor expansion theorem states that any determinant can be computed by adding the products of the elements of a column or row by their respective cofactors. $$ \Rightarrow \,\,0 \cdot \left| {\begin{array}{*{20}{c}} 0&{x - c} \\\ {x + c}&0 \end{array}} \right| - (x - a) \cdot \left| {\begin{array}{*{20}{c}} {x + a}&{x - c} \\\ {x + b}&0 \end{array}} \right| + (x - b) \cdot \left| {\begin{array}{*{20}{c}} {x + a}&0 \\\ {x + b}&{x + c} \end{array}} \right|$$ On simplifying the determinant, we have $$ \Rightarrow \,\,0 - (x - a) \cdot \left( {(x + a)(0) - (x + b)(x - c)} \right) + (x - b) \cdot \left( {(x + a)(x + c) - (x + b)(0)} \right) = 0$$ On simplifying the terms we have $$ \Rightarrow \,\,(x - a)(x + b)(x - c) + (x - b)(x + a)(x + c) = 0$$ We will consider the first option, $$x = 0$$ So we have, $$ \Rightarrow \,\,(0 - a)(0 + b)(0 - c) + (0 - b)(0 + a)(0 + c) = 0$$ On simplifying we get

\Rightarrow ,,abc - abc = 0 \\
\Rightarrow 0 = 0 \\

$$ \Rightarrow LHS = RHS$$ Hence, the value of x in the determinant $$\left| {\begin{array}{*{20}{c}} 0&{x - a}&{x - b} \\\ {x + a}&0&{x - c} \\\ {x + b}&{x + c}&0 \end{array}} \right|$$ is 0 **Therefore, the correct option is A** **Note:** When considering the determinant that should be in the square matrix it means the determinant should have a equal number of rows and column otherwise it can be solve by using a determinant method and remember that to find the determinant of a$$2 \times 2$$ matrix, you have to multiply the elements on the main diagonal and subtract the product of the elements on the secondary diagonal.