Mathematics - Matrices Question with Solution | TestHub

Given $A=\left[\begin{array}{lll}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]$

Let, $X=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

Then, we have $AX=\left[\begin{array}{lll}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{l}{a}_{11}+a_{12}+a_{13}\\ a_{21}+a_{22}+a_{23}\\ a_{31}+a_{32}+a_{33}\end{array}\right]$

Given, $a_{i1}+a_{i2}+a_{i3}=1,$ for $i=1, 2, 3$

$\Rightarrow \left[\begin{array}{l}{a}_{11}+a_{12}+a_{13}\\ a_{21}+a_{22}+a_{23}\\ a_{31}+a_{32}+a_{33}\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

$\Rightarrow AX=X ...\left(i\right)$

Pre multiply by $A,$ we get

$A^{2}X=AX ...\left(ii\right)$

From $\left(i\right) \& \left(ii\right),$ we get $A^{2}X=X$

Again, pre multiply by $A,$ we get

$A^{3}X=AX ...\left(iii\right)$

From $\left(i\right) \& \left(iii\right),$ we get $A^{3}X=X$

Let $A^3=\left[\begin{array}{lll}{x}_1& x_2& x_3\\ y_1& y_2& y_3\\ z_1& z_2& z_3\end{array}\right]$

$\Rightarrow A^3\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=\left[\begin{array}{l}{x}_1+x_2+x_3\\ y_1+y_2+y_3\\ z_1+z_2+z_3\end{array}\right]=\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

$\Rightarrow x_1+x_2+x_3=1, y_1+y_2+y_3=1 \& z_1+z_2+z_3=1$

$\Rightarrow x_1+x_2+x_3+y_1+y_2+y_3+z_1+z_2+z_3=3.$

Thus, the sum of all the elements is$=3.$