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By using properties of determinant, show that \(\left| {\begin{array}{*{20}{c}} {x + y + 2z}&x&y \\ z&{y + z + 2x}&y \\ z&x&{z + x + 2y} \end{array}} \right| = 2{\left( {x + y + z} \right)^3}\)
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By using properties of determinant, show that \(\left| {\begin{array}{*{20}{c}} {x + y + 2z}&x&y \\ z&{y + z + 2x}&y \\ z&x&{z + x + 2y} \end{array}} \right| = 2{\left( {x + y + z} \right)^3}\)

Answer

\(L.H.S = \left| {\begin{array}{*{20}{c}} {x + y + 2z}&x&y \\ z&{y + z + 2x}&y \\ z&x&{z + x + 2y} \end{array}} \right|\)
\(\left[ {{C_1} \to {C_1} + {C_2} + {C_3}} \right]\)
\(= \left| {\begin{array}{*{20}{c}} {2\left( {x + y + z} \right)}&x&y \\ {2\left( {x + y + z} \right)}&{y + z + 2x}&y \\ {2\left( {x + y + z} \right)}&x&{z + x + 2y} \end{array}} \right|\)
Taking 2(x+y+z) common from C1
\(= 2\left( {x + y + z} \right)\left| {\begin{array}{*{20}{c}} 1&x&y \\ 1&{y + z + 2x}&y \\ 1&x&{z + x + 2y} \end{array}} \right|\)
\(\left[ {{R_2} \to {R_2} - {R_1}\,\,and\,\,{R_3} \to {R_3} - {R_1}} \right]\)
\(= 2\left( {x + y + z} \right)\left| {\begin{array}{*{20}{c}} 1&x&y \\ 0&{x + y + z}&0 \\ 0&0&{x + y + z} \end{array}} \right|\)
Expanding along Ist column
\(= 2\left( {x + y + z} \right).1\left| {\begin{array}{*{20}{c}} {x + y + z}&0 \\ 0&{x + y + z} \end{array}} \right|\)
\(= 2\left( {x + y + z} \right)\left[ {{{\left( {x + y + z} \right)}^2} - 0} \right]\)
\(= 2{\left( {x + y + z} \right)^3}\) = R.H.S.
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