If each element of a row is expressed as sum of two elements then verify for a third order determinant that the determinant can be expressed as sum of two determinants.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{following} \mathrm{determinant}. \\ \left|\begin{array}{ccc}{\mathrm{a}}_1+\mathrm{k}& {\mathrm{a}}_2+\mathrm{k}& {\mathrm{a}}_3+\mathrm{k}\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right| \\ \mathrm{We} \mathrm{need} \mathrm{to} \mathrm{prove} \mathrm{that}, \\ \left|\begin{array}{ccc}{\mathrm{a}}_1+\mathrm{k}& {\mathrm{a}}_2+\mathrm{k}& {\mathrm{a}}_3+\mathrm{k}\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right|=\left|\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right|+\left|\begin{array}{ccc}\mathrm{k}& \mathrm{k}& \mathrm{k}\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right| \\ \mathrm{Consder} \mathrm{LHS} \mathrm{and} \mathrm{expand} \mathrm{it} \mathrm{to} \mathrm{get} \mathrm{RHS} \\ \mathrm{LHS}=\left|\begin{array}{ccc}{\mathrm{a}}_1+\mathrm{k}& {\mathrm{a}}_2+\mathrm{k}& {\mathrm{a}}_3+\mathrm{k}\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right| \\ \mathrm{Expanding} \mathrm{along} {\mathrm{R}}_1 \\ =\left({\mathrm{a}}_1+\mathrm{k}\right) \left|\begin{array}{cc}{\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right|-\left({\mathrm{a}}_2+\mathrm{k}\right)\left|\begin{array}{cc}{\mathrm{b}}_1& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_3\end{array}\right|+\left({\mathrm{a}}_3+\mathrm{k}\right)\left|\begin{array}{cc}{\mathrm{b}}_1& {\mathrm{b}}_2\\ {\mathrm{c}}_1& {\mathrm{c}}_2\end{array}\right| \\ =\left({\mathrm{a}}_1+\mathrm{k}\right) \left({\mathrm{b}}_2{\mathrm{c}}_3-{\mathrm{c}}_2{\mathrm{b}}_3\right)-\left({\mathrm{a}}_2+\mathrm{k}\right)\left({\mathrm{b}}_1{\mathrm{c}}_3-{\mathrm{c}}_1{\mathrm{b}}_3\right)+\left({\mathrm{a}}_3+\mathrm{k}\right)\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{c}}_1{\mathrm{b}}_2\right) \\ ={\mathrm{a}}_1 \left({\mathrm{b}}_2{\mathrm{c}}_3-{\mathrm{c}}_2{\mathrm{b}}_3\right)+\mathrm{k} \left({\mathrm{b}}_2{\mathrm{c}}_3-{\mathrm{c}}_2{\mathrm{b}}_3\right)-{\mathrm{a}}_2\left({\mathrm{b}}_1{\mathrm{c}}_3-{\mathrm{c}}_1{\mathrm{b}}_3\right)-\mathrm{k}\left({\mathrm{b}}_1{\mathrm{c}}_3-{\mathrm{c}}_1{\mathrm{b}}_3\right) \\ +{\mathrm{a}}_3\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{c}}_1{\mathrm{b}}_2\right)+\mathrm{k}\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{c}}_1{\mathrm{b}}_2\right) \\ =\left[{\mathrm{a}}_1 \left({\mathrm{b}}_2{\mathrm{c}}_3-{\mathrm{c}}_2{\mathrm{b}}_3\right)-{\mathrm{a}}_2\left({\mathrm{b}}_1{\mathrm{c}}_3-{\mathrm{c}}_1{\mathrm{b}}_3\right)+{\mathrm{a}}_3\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{c}}_1{\mathrm{b}}_2\right)\right] \\ +\left[\mathrm{k} \left({\mathrm{b}}_2{\mathrm{c}}_3-{\mathrm{c}}_2{\mathrm{b}}_3\right)-\mathrm{k}\left({\mathrm{b}}_1{\mathrm{c}}_3-{\mathrm{c}}_1{\mathrm{b}}_3\right)+\mathrm{k}\left({\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{c}}_1{\mathrm{b}}_2\right)\right] \\ =\left|\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_3\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right|+\left|\begin{array}{ccc}\mathrm{k}& \mathrm{k}& \mathrm{k}\\ {\mathrm{b}}_1& {\mathrm{b}}_2& {\mathrm{b}}_3\\ {\mathrm{c}}_1& {\mathrm{c}}_2& {\mathrm{c}}_3\end{array}\right| \\ =\mathrm{RHS} \\ \mathrm{Hence} \mathrm{it} \mathrm{is} \mathrm{proved} \mathrm{that}, \\ \left|\begin{array}{ccc}{\mathbf{a}}_{\mathbf{1}}\mathbf{+}\mathbf{k}& {\mathbf{a}}_{\mathbf{2}}\mathbf{+}\mathbf{k}& {\mathbf{a}}_{\mathbf{3}}\mathbf{+}\mathbf{k}\\ {\mathbf{b}}_{\mathbf{1}}& {\mathbf{b}}_{\mathbf{2}}& {\mathbf{b}}_{\mathbf{3}}\\ {\mathbf{c}}_{\mathbf{1}}& {\mathbf{c}}_{\mathbf{2}}& {\mathbf{c}}_{\mathbf{3}}\end{array}\right|\mathbf{=}\left|\begin{array}{ccc}{\mathbf{a}}_{\mathbf{1}}& {\mathbf{a}}_{\mathbf{2}}& {\mathbf{a}}_{\mathbf{3}}\\ {\mathbf{b}}_{\mathbf{1}}& {\mathbf{b}}_{\mathbf{2}}& {\mathbf{b}}_{\mathbf{3}}\\ {\mathbf{c}}_{\mathbf{1}}& {\mathbf{c}}_{\mathbf{2}}& {\mathbf{c}}_{\mathbf{3}}\end{array}\right|\mathbf{+}\left|\begin{array}{ccc}\mathbf{k}& \mathbf{k}& \mathbf{k}\\ {\mathbf{b}}_{\mathbf{1}}& {\mathbf{b}}_{\mathbf{2}}& {\mathbf{b}}_{\mathbf{3}}\\ {\mathbf{c}}_{\mathbf{1}}& {\mathbf{c}}_{\mathbf{2}}& {\mathbf{c}}_{\mathbf{3}}\end{array}\right| \end{gathered}$$

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