Question

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

8. \(\left( {\begin{array}{{}}0\\1\\{ - 2}\\1\end{array}} \right),\left( {\begin{array}{{}}1\\1\\0\\2\end{array}} \right),\left( {\begin{array}{{}}1\\4\\{ - 6}\\5\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}{ - 1}\\1\\{ - 4}\\0\end{array}} \right)\)

Short answer

The barycentric coordinates are \(\left( {2, - 1,0} \right)\).

Step-by-step solution

The barycentric coordinates

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 \({\mathop{\rm p}\nolimits} \).

Write the augmented matrix 

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

Write the augmented matrix as shown below:

\(\left( {\begin{array}{{}}{\widetilde {{{\bf{v}}_1}}}&{\widetilde {{{\bf{v}}_2}}}&{\widetilde {{{\bf{v}}_3}}}&{\widetilde {\bf{p}}}\end{array}} \right) \sim \left( {\begin{array}{{}}0&1&1&{ - 1}\\1&1&4&1\\{ - 2}&0&{ - 6}&{ - 4}\\1&2&5&0\\1&1&1&1\end{array}} \right)\)

Apply row operations

Interchange row 1 and row 2.

\( \sim \left( {\begin{array}{{}}1&1&4&1\\0&1&1&{ - 1}\\{ - 2}&0&{ - 6}&{ - 4}\\1&2&5&0\\1&1&1&1\end{array}} \right)\)

At row 3, multiply row 1 by 2 and add it to row 3.

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

At row 4, subtract row 1 from row 4. At row 5, subtract row 1 from row 5. At row 1, subtract row 2 from row 1.

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

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

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

At row 4, subtract row 2 from row 4.

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

Interchange row 3 and row 5.

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

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

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

Determine the barycentric coordinates of p

Convert the matrix into the system of equations as shown below

\(\begin{array}{}{{\mathop{\rm x}\nolimits} _1} + 3{{\mathop{\rm x}\nolimits} _3} = 2\\{{\mathop{\rm x}\nolimits} _2} + {{\mathop{\rm x}\nolimits} _3} = - 1\\{{\mathop{\rm x}\nolimits} _3} = 0\end{array}\)

Solve the system of the equation to obtain as shown below

\({{\mathop{\rm x}\nolimits} _1} = 2,{\rm{ }}{{\mathop{\rm x}\nolimits} _2} = - 1,{\rm{ }}{{\mathop{\rm x}\nolimits} _3} = 0\)

The coordinates are \(2, - 1,\,\,\,{\mathop{\rm and}\nolimits} \,\,0\), so \({\bf{p}} = 2{{\bf{v}}_1} - {{\bf{v}}_2} + 0{{\bf{v}}_3}\).

Thus, the barycentric coordinates are \(\left( {2, - 1,0} \right)\).