Question

Given the matrices \(J = \left[ {\begin{array}{*{20}{c}} 3&2&1\\ 2&4&2\\ 1&2&6 \end{array}} \right]\) and \(K = \left[ {\begin{array}{*{20}{c}} 1\\ 2\\ { - 1} \end{array}} \right]\), the product K^TJK is _________

Answer 1

This is simple matrix multiplication:

Given:

 \(J = \left[ {\begin{array}{*{20}{c}} 3&2&1\\ 2&4&2\\ 1&2&6 \end{array}} \right]\) and \(K = \left[ {\begin{array}{*{20}{c}} 1\\ 2\\ { - 1} \end{array}} \right]\),

∴ K^T = [1 2 -1]

Now,

\({K^T}\;J\;K = \left[ {1\;2\; - 1} \right]\left[ {\begin{array}{*{20}{c}} 3&2&1\\ 2&4&2\\ 1&2&6 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ { - 1} \end{array}} \right] = \left[ {6\;8\; - 1} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ { - 1} \end{array}} \right] = \left[ {23} \right]\)

∴ K^T J K = 23.