LA2.1.10

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Without expanding the determinant of the matrix A={\begin{bmatrix}0&p-q&p-r\\q-p&0&q-r\\r-p&r-q&0\end{bmatrix}}\, prove that |A|=0\,

Let A be the determinant of the given matrix. Then

A={\begin{vmatrix}0&p-q&p-r\\q-p&0&q-r\\r-p&r-q&0\end{vmatrix}}\,

Applying C_{1}\rightarrow C_{1}-C_{2},C_{2}\rightarrow C_{2}-C_{3}\, we have

A={\begin{vmatrix}0-(p-q)&(p-q)-(p-r)&p-r\\(q-p)-0&0-(q-r)&q-r\\(r-p)-(r-q)&(r-q)-0&0\end{vmatrix}}\,

A={\begin{vmatrix}q-p&r-q&p-r\\q-p&r-q&q-r\\q-p&r-q&0\end{vmatrix}}\,

Taking out the common from C1 and C2

(q-p)(r-q){\begin{vmatrix}1&1&p-r\\1&1&q-r\\1&1&0\end{vmatrix}}\,

A=(q-p)(r-q)(0)=0\,


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