JEE Main 2021 — Mathematics Algebra

Given $A=[\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}]$A=[a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​]

Let, $X=[\begin{matrix}1 \\ 1 \\ 1\end{matrix}]$X=[111​]

Then, we have $AX=[\begin{matrix}{a}_{11} & {a}_{12} & {a}_{13} \\ {a}_{21} & {a}_{22} & {a}_{23} \\ {a}_{31} & {a}_{32} & {a}_{33}\end{matrix}][\begin{matrix}1 \\ 1 \\ 1\end{matrix}]$AX=[a11​a21​a31​​a12​a22​a32​​a13​a23​a33​​][111​]

$$\Rightarrow [\begin{matrix}{a}_{11}+{a}_{12}+{a}_{13} \\ {a}_{21}+{a}_{22}+{a}_{23} \\ {a}_{31}+{a}_{32}+{a}_{33}\end{matrix}]$$⇒[a11​+a12​+a13​a21​+a22​+a23​a31​+a32​+a33​​]

Given, ${a}_{i1}+{a}_{i2}+{a}_{i3}=1,$ai1​+ai2​+ai3​=1, for $i=1,2,3$i=1,2,3

$$\Rightarrow [\begin{matrix}{a}_{11}+{a}_{12}+{a}_{13} \\ {a}_{21}+{a}_{22}+{a}_{23} \\ {a}_{31}+{a}_{32}+{a}_{33}\end{matrix}]=[\begin{matrix}1 \\ 1 \\ 1\end{matrix}]$$⇒[a11​+a12​+a13​a21​+a22​+a23​a31​+a32​+a33​​]=[111​]

$$\Rightarrow AX=X...(i)$$⇒AX=X...(i)

Pre multiply by $A,$A, we get

$${A}^{2}X=AX...(ii)$$A2X=AX...(ii)

From (i)&(ii), we get ${A}^{2}X=X$A2X=X

Again, pre multiply by $A,$A, we get

$${A}^{3}X=AX...(iii)$$A3X=AX...(iii)

From (i)&(iii), we get ${A}^{3}X=X$A3X=X

Let ${A}^{3}=[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} \\ {y}_{1} & {y}_{2} & {y}_{3} \\ {z}_{1} & {z}_{2} & {z}_{3}\end{matrix}]$A3=[x1​y1​z1​​x2​y2​z2​​x3​y3​z3​​]

$$\Rightarrow {A}^{3}[\begin{matrix}1 \\ 1 \\ 1\end{matrix}]=[\begin{matrix}{x}_{1}+{x}_{2}+{x}_{3} \\ {y}_{1}+{y}_{2}+{y}_{3} \\ {z}_{1}+{z}_{2}+{z}_{3}\end{matrix}]=[\begin{matrix}1 \\ 1 \\ 1\end{matrix}]$$⇒A3[111​]=[x1​+x2​+x3​y1​+y2​+y3​z1​+z2​+z3​​]=[111​]

\Rightarrow {x}_{1}+{x}_{2}+{x}_{3}=1,{y}_{1}+{y}_{2}+{y}_{3}=1&{z}_{1}+{z}_{2}+{z}_{3}=1

$$\Rightarrow {x}_{1}+{x}_{2}+{x}_{3}+{y}_{1}+{y}_{2}+{y}_{3}+{z}_{1}+{z}_{2}+{z}_{3}=3.$$⇒x1​+x2​+x3​+y1​+y2​+y3​+z1​+z2​+z3​=3.

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