Given A = \left[\begin{cases}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{cases}\right]
Let, X = \left[111\right]
Then, we have AX = \left[\begin{cases}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{cases}\right]\left[111\right]
⇒ \left[a11+a12+a13a21+a22+a23a31+a32+a33\right]
Given, ai1+ai2+ai3=1, for i=1,2,3
⇒ \left[a11+a12+a13a21+a22+a23a31+a32+a33\right] = \left[111\right]
⇒AX=X...(i)
Pre multiply by A, we get
A2X=AX...(ii)
From \left(i\right) \text{ and } \left(ii\right),weget{A}^{2}X = X</p><p>Again,premultiplybyA,weget</p><p>{A}^{3}X = AX . . . \left(iii\right)</p><p>From(i) and (iii), we get A3X=X
Let {A}^{3} = \left[\begin{cases}{x}_{1} & {x}_{2} & {x}_{3} \\ {y}_{1} & {y}_{2} & {y}_{3} \\ {z}_{1} & {z}_{2} & {z}_{3}\end{cases}\right]
⇒ {A}^{3}\left[111\right] = \left[x1+x2+x3y1+y2+y3z1+z2+z3\right] = \left[111\right]
⇒x1+x2+x3=1,y1+y2+y3=1 and z1+z2+z3=1
⇒x1+x2+x3+y1+y2+y3+z1+z2+z3=3.
Thus, the sum of all the elements is=3.
