Question

Given $A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 4& 1\\ 2& 3& 1\end{array}\right],B=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$, Let $P$ such that $BPA=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\end{array}\right]$.

If $P=$$\left[\begin{array}{ccc}k& m& n\\ p& -q& r\end{array}\right]$.

Find $(k+m+n+p+q+r)$ ?

Answer 1

$A=\left[\begin{array}{ccc}1& 1& 1\\ 2& 4& 1\\ 2& 3& 1\end{array}\right],B=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right],P=\left[\begin{array}{ccc}k& m& n\\ p& -q& r\end{array}\right]$

Now $BPA=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\end{array}\right]\Rightarrow \left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left[\begin{array}{ccc}k& m& n\\ p& -q& r\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 2& 4& 1\\ 2& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{ccc}2k+3p& 2m-3q& 2n+3r\\ 3k+4p& 3m-4q& 3n+4r\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 2& 4& 1\\ 2& 3& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\end{array}\right]$

$\Rightarrow 2k+3p+4m-6q+4n+6r=1$ ...(1)

and $2k+3p+8m-12q+6n+9r=0$ ...(2)

and $2k+3p+2m-3q+2n+3r=1$ ...(3)

and $3k+4p+6m-8q+6n+8r=0$ ...(4)

and $3k+4p+12m-12q+9n+12r=1$ ...(5)

and $3k+4p+3m-4q+3n+4r=0$ ...(6)

Solving $\left(1\right),\left(2\right),\left(3\right),\left(4\right),\left(5\right)$ and $\left(6\right)$, we get

$k+m+n+p+q+r=1$