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Using the property of determinant and without expanding prove that $$\left| {\begin{array}{*{20}{c}} x&a&{x + a} \\ y&b&{y + b} \\ z&c&{z + c} \end{array}} \right| = 0$$
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## QuestionMathsClass 12

Using the property of determinant and without expanding prove that $$\left| {\begin{array}{*{20}{c}} x&a&{x + a} \\ y&b&{y + b} \\ z&c&{z + c} \end{array}} \right| = 0$$

See the analysis below.
4.6
4.6

## Solution

$$L.H.S=$$$$\left| {\begin{array}{*{20}{c}} x&a&{x + a} \\ y&b&{y + b} \\ z&c&{z + c} \end{array}} \right|$$
Applying, $${C_1} \to {C_1} + {C_2}$$
$$=$$$$\left| {\begin{array}{*{20}{c}} {x + a}&a&{x + a} \\ {y + b}&b&{y + b} \\ {z + c}&c&{z + c} \end{array}} \right|$$
$$=$$$$0$$  [$$C _{1}$$ and $$C _{2}$$are identical]
$$= R.H.S.$$
As $$L.H.S= R.H.S.$$
Hence, proved.