Question

Let $A=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ and $B=\left[\begin{array}{ccc}{9}^2& -{10}^2& {11}^2\\ {12}^2& {13}^2& -{14}^2\\ -{15}^2& {16}^2& {17}^2\end{array}\right]$, then the value of $A^{\prime }BA$ is

Answer 1

$A^{\prime }BA={\left[\begin{array}{ccc}1& 1& 1\end{array}\right]}_{1\times 3}{\left[\begin{array}{ccc}{9}^2& -{10}^2& {11}^2\\ {12}^2& {13}^2& -{14}^2\\ -{15}^2& {16}^2& {17}^2\end{array}\right]}_{3\times 3}{\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]}_{3\times 1}$

So final matrix will be of $1\times 1$ matrix

$\therefore A^{\prime }BA=[9^2-{10}^2+{11}^2+{12}^2+{13}^2-{14}^2-{15}^2+{16}^2+{17}^2]$
$=[-19+265+(-27)+31+289]$
$=[585-46]=\left[539\right]$