Question

Let M be a $3$ $\times$ $3$ matrix satisfying $M\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right],M\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$, and M $\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$ Then the sum of the diagonal entries of M is __________

Answer 1

Let M be $\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$.
Now from the first condition $\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ 2\\ 3\end{array}\right]$
we get the values of $b,e,h$ as$-1,2,3$ respectively.
Newly formed matrix M =$\left[\begin{array}{ccc}a& -1& c\\ d& 2& f\\ g& 3& i\end{array}\right]$
From the second condition
$\left[\begin{array}{ccc}a& -1& c\\ d& 2& f\\ g& 3& i\end{array}\right]\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$
Solving the matrix, we get three equations
$\Rightarrow a+1=1,d-2=1,g-3=-1$
$\Rightarrow a=0,d=3,g=2$
Modified matrix M =$\left[\begin{array}{ccc}0& -1& c\\ 3& 2& f\\ 2& 3& i\end{array}\right]$
Now the third and the last condition, $\left[\begin{array}{ccc}0& -1& c\\ 3& 2& f\\ 2& 3& i\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 12\end{array}\right]$
$\Rightarrow 2+3+i=12\Rightarrow i=7$
Sum of the diagonal entries of $M=a+e+i=0+2+7=9$