If $A = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&2&{ - 4}\\{ - 4}&2&{ - 4}\\2&{ - 1}&5\end{array}} \right|$ and $B = \left| {\begin{array}{cccccccccccccccccccc}1&{ - 1}&0\\2&3&4\\0&1&2\end{array}} \right|$, then find BA and use this to solve the system of equations $y + 2z = 7,x - y = 3$ and $2x + 3y + 4z = 17$.

We have, $A = \left| {\begin{array}{cccccccccccccccccccc}2&2&{ - 4}\\{ - 4}&2&{ - 4}\\2&{ - 1}&5\end{array}} \right|$

and $B = \left| {\begin{array}{cccccccccccccccccccc}1&{ - 1}&0\\2&3&4\\0&1&2\end{array}} \right|$

$\therefore$ $BA = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\2&3&4\\0&1&2\end{array}} \right|\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&2&{ - 4}\\{ - 4}&2&{ - 4}\\2&{ - 1}&5\end{array}} \right| = \left| {\begin{array}{llllllllllllllllllll}6&0&0\\0&6&0\\0&0&6\end{array}} \right| = 6I$

$\therefore$ ${B^{ - 1}} = \frac{A}{6} = \frac{1}{6}A = \frac{1}{6}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&2&{ - 4}\\{ - 4}&2&{ - 4}\\2&{ - 1}&5\end{array}} \right|$

….(i)
Also, $x - y = 3,$ $2x + 3y + 4z = 17$ and $y + 2z = 7$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\2&3&4\\0&1&2\end{array}} \right]\left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3\\{17}\\7\end{array}} \right]$

$\therefore$ $\left[ {\begin{array}{llllllllllllllllllll}x\\y\\z\end{array}} \right] = {\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{ - 1}&0\\2&3&4\\0&1&2\end{array}} \right]^{ - 1}}\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}3\\{17}\\7\end{array}} \right]$

$= \frac{1}{6}\left[ {\begin{array}{cccccccccccccccccccc}2&2&{ - 4}\\{ - 4}&2&{ - 4}\\2&{ - 1}&5\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}3\\{17}\\7\end{array}} \right]$

[using Eq. (i)]
$= \frac{1}{6}\left[ {\begin{array}{cccccccccccccccccccc}{6 + 34 - 28}\\{ - 12 + 34 - 28}\\{6 - 17 + 35}\end{array}} \right] = \frac{1}{6}\left[ {\begin{array}{cccccccccccccccccccc}{12}\\{ - 6}\\{24}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}2\\{ - 1}\\4\end{array}} \right]$

$\therefore$ $x = 2,$ $y = - 1$ and $z = 4$