In Exercises 7 and 8, find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.

7. \(\left( {\begin{array}{{}}1\\{ - 1}\\2\\1\end{array}} \right),\left( {\begin{array}{{}}2\\1\\0\\1\end{array}} \right),\left( {\begin{array}{{}}1\\2\\{ - 2}\\0\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}5\\4\\{ - 2}\\2\end{array}} \right)\)

Consider the set \(S = \left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},...,{{\mathop{\rm v}\nolimits} _k}} \right\}\)as an affinely independent.So, for every point \({\mathop{\rm p}\nolimits} \) in \({\mathop{\rm aff}\nolimits} S\), the coefficients \({c_1},...,{c_k}\) in the unique representation (7) of \({\mathop{\rm p}\nolimits} \) are referred to as barycentric coordinates(or sometimes, affine) of \({\mathop{\rm p}\nolimits} \).

Move the last row of ones to the top to simplify the arithmetic.

Write the augmented matrix as shown below:

\(\left( {\begin{array}{{}}{\widetilde {{{\bf{v}}_1}}}&{\widetilde {{{\bf{v}}_2}}}&{\widetilde {{{\bf{v}}_3}}}&{\widetilde {\bf{p}}}\end{array}} \right) \sim \left( {\begin{array}{{}}1&2&1&5\\{ - 1}&1&2&4\\2&0&{ - 2}&{ - 2}\\1&1&0&2\\1&1&1&1\end{array}} \right)\)

At row 2, add row 1 and row 2.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&3&3&9\\2&0&{ - 2}&{ - 2}\\1&1&0&2\\1&1&1&1\end{array}} \right)\)

At row 3, multiply row 1 by 2 and subtract it from row 3.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&3&3&9\\0&{ - 4}&{ - 4}&{ - 12}\\1&1&0&2\\1&1&1&1\end{array}} \right)\)

At row 4, subtract row 1 from row 4.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&3&3&9\\0&{ - 4}&{ - 4}&{ - 12}\\0&{ - 1}&{ - 1}&{ - 3}\\1&1&1&1\end{array}} \right)\)

At row 5, subtract row 1 from row 5.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&3&3&9\\0&{ - 4}&{ - 4}&{ - 12}\\0&{ - 1}&{ - 1}&{ - 3}\\0&{ - 1}&0&{ - 4}\end{array}} \right)\)

At row 2, multiply row 2 by \(\frac{1}{3}\).

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&1&1&3\\0&{ - 4}&{ - 4}&{ - 12}\\0&{ - 1}&{ - 1}&{ - 3}\\0&{ - 1}&0&{ - 4}\end{array}} \right)\)

At row 3, multiply row 2 by 4 and add it row 3.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&1&1&3\\0&0&0&0\\0&{ - 1}&{ - 1}&{ - 3}\\0&{ - 1}&0&{ - 4}\end{array}} \right)\)

At row 4, add row 4 and row 2.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&1&1&3\\0&0&0&0\\0&0&0&0\\0&{ - 1}&0&{ - 4}\end{array}} \right)\)

At row 5, add row 5 and row 2.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&1&1&3\\0&0&0&0\\0&0&0&0\\0&0&1&{ - 1}\end{array}} \right)\)

Interchange row 5 and row 3.

\( \sim \left( {\begin{array}{{}}1&2&1&5\\0&1&1&3\\0&0&1&{ - 1}\\0&0&0&0\\0&0&0&0\end{array}} \right)\)

Convert the matrix into the system of equations as shown below

\(\begin{array}{r}{{\mathop{\rm x}\nolimits} _1} + {{\mathop{\rm x}\nolimits} _2} + {{\mathop{\rm x}\nolimits} _3} = 5\\{{\mathop{\rm x}\nolimits} _2} + {{\mathop{\rm x}\nolimits} _3} = 3\\{{\mathop{\rm x}\nolimits} _3} = - 1\end{array}\)

Solve the system of the equation to obtain as shown below

\({{\mathop{\rm x}\nolimits} _1} = - 2,{\rm{ }}{{\mathop{\rm x}\nolimits} _2} = 4,{\rm{ }}{{\mathop{\rm x}\nolimits} _3} = - 1\)

The coordinates are \( - 2,4,\,\,\,{\mathop{\rm and}\nolimits} \,\, - 1\), so \({\bf{p}} = - 2{{\bf{v}}_1} + 4{{\bf{v}}_2} - {{\bf{v}}_3}\).

Thus, the barycentric coordinates are \(\left( { - 2,4, - 1} \right)\).