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Using the properties of determinants in prove that:

\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|=4 x y z

Answers (1)

LHS given,

\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|

Apply, R\textsubscript{1}$ \rightarrow $ R\textsubscript{1} + R\textsubscript{2} + R\textsubscript{3},\\

\\\begin{aligned} &=\left|\begin{array}{ccc} \mathrm{y}+\mathrm{z}+\mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{z}+\mathrm{x}+\mathrm{x} & \mathrm{y}+\mathrm{x}+\mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ &=\left|\begin{array}{ccc} 2 z+2 y & 2 z+2 x & 2 x+2 y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &2 \text { can be taken common from } \mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{x}+\mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right|\\ \end{aligned}

\\\begin{aligned} \\&\text { Apply, } R_{1} \rightarrow R_{1}-R_{2}\\ &=2\left|\begin{array}{ccc} \mathrm{z}+\mathrm{y}-\mathrm{z} & \mathrm{z}+\mathrm{x}-(\mathrm{z}+\mathrm{x}) & \mathrm{x}+\mathrm{y}-\mathrm{x} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \end{array}\right| \end{aligned}

\\\begin{aligned} &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|\\ &\text { Apply, } R_{3} \rightarrow R_{3}-R_{1},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z} & \mathrm{z}+\mathrm{x} & \mathrm{x} \\ \mathrm{y}-\mathrm{y} & \mathrm{x}-0 & \mathrm{x}+\mathrm{y}-\mathrm{y} \end{array}\right|\\ &=2\left|\begin{array}{ccc} y & 0 & y \\ z & z+x & x \\ 0 & x & x \end{array}\right|\\ &\text { Now, Apply } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3},\\ &=2\left|\begin{array}{ccc} \mathrm{y} & 0 & \mathrm{y} \\ \mathrm{z}-0 & \mathrm{z}+\mathrm{x}-\mathrm{x} & \mathrm{x}-\mathrm{x} \\ 0 & \mathrm{x} & \mathrm{x} \end{array}\right|\\ &=2\left|\begin{array}{lll} y & 0 & y \\ z & z & 0 \\ 0 & x & x \end{array}\right| \end{aligned}

Take y,z,x common from the R1, R2 and R3 respectively

\begin{aligned} &\text { Expand along Column } 1\\ &\begin{array}{r} |A|=a_{11}(-1)^{1+1}\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|+a_{21}(-1)^{2+1}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right| \\ +a_{31}(-1)^{3+1}\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array}\right| \end{array} \end{aligned}

\\\vspace{\baselineskip}= 2xyz [(1)$ \{ $ (1) -0$ \} $ -(1)$ \{ $ 0 -1$ \} $ + 0$ \} $ ]\\ \\ \vspace{\baselineskip}= 2xyz [1 + 1]\\ \\ \vspace{\baselineskip}= 4xyz\\ \\ \vspace{\baselineskip}= RHS\\ $ \therefore $ LHS = RHS\\ \\

Hence Proved

 

 

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infoexpert22

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