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# Using the property of determinants and without expanding, prove that determinant 0 a -b -a 0 -c b c 0 = 0 Ex 4.2 Q : 6

Q : 6          Using the property of determinants and without expanding, prove that

$\dpi{100} \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}=0$

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We have given determinant

$\dpi{100} \triangle = \begin{vmatrix} 0 &a &-b \\-a &0 & -c\\b &c &0 \end{vmatrix}$

Applying transformation, $\dpi{100} R_{1} \rightarrow cR_{1}$ we have then,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} 0 &ac &-bc \\-a &0 & -c\\b &c &0 \end{vmatrix}$

We can make the first row identical to the third row so,

Taking another row transformation: $\dpi{100} R_{1} \rightarrow R_{1}-bR_{2}$ we have,

$\dpi{100} \triangle = \frac{1}{c}\begin{vmatrix} ab &ac &0 \\-a &0 & -c\\b &c &0 \end{vmatrix} = \frac{a}{c} \begin{vmatrix} b &c &0 \\-a &0 & -c\\b &c &0 \end{vmatrix}$

So, determinant has two rows $\dpi{100} R_{1}\ and\ R_{3}$ identical.

Hence $\dpi{100} \triangle = 0$.

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