Question

Find the determinant of the matrix $\left( {\begin{array}{*{20}{c}}
  0&{{\text{p}} - {\text{q}}}&{{\text{p}} - {\text{r}}} \\
  {{\text{q}} - {\text{p}}}&0&{{\text{q}} - {\text{r}}} \\
  {{\text{r}} - {\text{p}}}&{{\text{r}} - {\text{q}}}&0
\end{array}} \right)$ .
 $({\text{A}})$ $({\text{p}} - {\text{q}})({\text{q}} - {\text{r}})({\text{r}} - {\text{p}})$
 $({\text{B}})$ $0$
 $({\text{C}})$ ${\text{pqr}}$
 $({\text{D}})$ ${\text{4pqr}}$

Answer 1

Hint:The determinant of the above given matrix can be calculated by taking the transpose of the given matrix. Then taking the minus outside from all the columns in the above given matrix. We will be getting the minus of the determinant of the above given matrix.

Complete step-by-step answer:
The data given in the question is,
Let the determinant of matrix ${\text{A = }}\left| {\begin{array}{*{20}{c}}
  0&{{\text{p}} - {\text{q}}}&{{\text{p}} - {\text{r}}} \\
  {{\text{q}} - {\text{p}}}&0&{{\text{q}} - {\text{r}}} \\
  {{\text{r}} - {\text{p}}}&{{\text{r}} - {\text{q}}}&0
\end{array}} \right|$

To find the determinant of the matrix A,
Taking transpose for the above given matrix we get,
 ${\text{|A| = }}\left| {\begin{array}{*{20}{c}}
  0&{{\text{q}} - {\text{p}}}&{{\text{r}} - {\text{p}}} \\
  {{\text{p}} - {\text{q}}}&0&{{\text{r}} - {\text{q}}} \\
  {{\text{p}} - {\text{r}}}&{{\text{q}} - {\text{r}}}&0
\end{array}} \right|$
Taking $( - 1)$ from all the columns of the above we get,
 ${\text{|A| = }}\dfrac{1}{{{{( - 1)}^3}}}\left| {\begin{array}{*{20}{c}}
  0&{{\text{p}} - {\text{q}}}&{{\text{p}} - {\text{r}}} \\
  {{\text{q}} - {\text{p}}}&0&{{\text{q}} - {\text{r}}} \\
  {{\text{r}} - {\text{p}}}&{{\text{r}} - {\text{q}}}&0
\end{array}} \right|$
By solving the above equation, we get,
 ${\text{|A|}} = ( - 1)\left| {\begin{array}{*{20}{c}}
  0&{{\text{p}} - {\text{q}}}&{{\text{p}} - {\text{r}}} \\
  {{\text{q}} - {\text{p}}}&0&{{\text{q}} - {\text{r}}} \\
  {{\text{r}} - {\text{p}}}&{{\text{r}} - {\text{q}}}&0
\end{array}} \right|$
Then the determinant of matrix A will be,
 $\Delta = - |{\text{A|}}$
The above can written as,
 $\Delta = - \Delta $
While bringing the determinant on the one side we get,
 $\Delta + \Delta = 0$
Then by adding the above we get,
 $2\Delta = 0$
At last the determinant of the matrix will be,
 $\Delta = 0$
 $\therefore $ The determinant of the above given matrix is $0$ .
Hence, the determinant of the above given matrix is zero.

So, the correct answer is “Option B”.

Note:If the two rows of the matrix are equal, then the determinant of that matrix will be zero. The determinant of the matrix can also be denoted as ${\text{det(A)}}$ , ${\text{del A}}$ .There are many properties of determinants. In the above question, we have used the switching property. If the matrix is said to be singular, then the determinant of the matrix is 0. If the matrix is said to be unimodular, then the determinant of the matrix is 1.