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અહી વાસ્તવિક શ્રેણિક $A=\left[a_{i j}\right]$ ની કક્ષા $3 \times 3$ છે કે જેથી $i=1,2,3$ માટે $a_{i 1}+a_{i 2}+a_{i 3}=1$ થાય તો શ્રેણિક $A^{3}$ ના બધાજ ઘટકોનો સરવાળો મેળવો.
$1$
$2$
$3$
$9$
Solution
$A=\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]$
Let $x=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
$A X=\left[\begin{array}{l}a_{11}+a_{12}+a_{13} \\ a_{21}+a_{22}+a_{23} \\ a_{31}+a_{32}+a_{33}\end{array}\right]=\left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]$
$\Rightarrow \mathrm{AX}=\mathrm{X}$
Replace $\mathrm{X}$ by $\mathrm{AX}$
$\mathrm{A}^{2} \mathrm{X}=\mathrm{AX}=\mathrm{X}$
Replace $\mathrm{X}$ by $\mathrm{AX}$
$\mathrm{A}^{3} \mathrm{X}=\mathrm{AX}=\mathrm{X}$
Let $A^{3}=\left[\begin{array}{lll}x_{1} & x_{2} & X_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3}\end{array}\right]$
$A^{3}\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{l}x_{1}+x_{2}+x_{3} \\ y_{1}+y_{2}+y_{3} \\ z_{1}+z_{2}+z_{3}\end{array}\right]=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
Sum of all the element $=3$
