Let M be a $3x3$ matrix satisfying $M[0;1;0] = [-1;2;3], M[1;-1;0] = [1;1;-1],$ and $M[1;1;1] = [0;0;12].$ Then the sum of the diagonal entries of M is

Let M be a 3 × 3 matrix satisfying M0 1 0=-1 2 3, M1 -1 0=1 1 -1, and M1 1 1=0 0 12. Then the sum of the diagonal entries of M is M0 1 0 = -1 2 3 M1 -1 0 = 1 1 -1 M1 1 1 = 0 0 12

$\text{To find: } \operatorname{tr}(M) = \text{sum of the diagonal entries of } M$

$c_1,\ c_2,\ c_3 = \text{the three columns of } M$

$M e_2 = c_2, \quad M(1,-1,0)^T = c_1 - c_2$

$M(1,1,1)^T = c_1 + c_2 + c_3$

c₂ = M e₂ = -1 2 3 c₁ - c₂ = 1 1 -1 c₁ = 1 1 -1 + -1 2 3 = 0 3 2 c₁ + c₂ + c₃ = 0 0 12 c₃ = 0 0 12 - 0 3 2 - -1 2 3 = 1 -5 7 M = 0 -1 1 3 2 -5 2 3 7

$\text{diagonal entries: } 0,\ 2,\ 7$

$\operatorname{tr}(M) = 0 + 2 + 7 = 9$

$\boxed{9}$