Q.
Let M be a 3×3 matrix satisfying
M⎣⎡010⎦⎤=⎣⎡−123⎦⎤M⎣⎡1−10⎦⎤=⎣⎡11−1⎦⎤, and M⎣⎡111⎦⎤=⎣⎡0012⎦⎤
then sum of the diagonal entries of M is
M⎣⎡100⎦⎤=M⎝⎛⎣⎡1−10⎦⎤+⎣⎡010⎦⎤⎠⎞ M=⎣⎡−123⎦⎤+⎣⎡11−1⎦⎤=⎣⎡032⎦⎤
and M⎣⎡001⎦⎤=M⎝⎛⎣⎡111⎦⎤−⎣⎡100⎦⎤−⎣⎡010⎦⎤⎠⎞ =⎣⎡0012⎦⎤−⎣⎡032⎦⎤−⎣⎡−123⎦⎤=⎣⎡1−57⎦⎤ M⎣⎡100⎦⎤=⎣⎡030⎦⎤⇒m11=0
Similarly, M⎣⎡010⎦⎤=⎣⎡−123⎦⎤⇒m22=2
and m33=7