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

Hint: 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 \[3 \times 3\]:
\[ \Rightarrow \left| {\begin{array}{*{20}{c}}
  0&{x - a}&{x - b} \\
  {x + a}&0&{x - c} \\
  {x + b}&{x + c}&0
\end{array}} \right| = 0\]
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.