A nice,unconventional Algebra question.

Sum of $\text{Row A}$ is $\frac { 9 }{ 2 } \left( 1+9 \right) =45$ .

Sum of $\text{Row B}$ is $45+9(1)=54$ , sum of $\text{Row C}$ is $54+9(1)=63$ , and so on.

The sums form an AP with ${ a }_{ 1 }=54$ and ${ a }_{ 10 }=126$ .

Hence, $\frac { 10 }{ 2 } \left( 45+126 \right) =855$ .

You can look at the rows diagonally :)

The sum of all numbers of Row A to Row J is given by:

$\begin{aligned} S & = \sum_{k=1}^{10} {\dfrac{9(k+k+8)}{2}} = \sum_{k=1}^{10} {9(k+4)} = 9\sum_{k=1}^{10} {k} + 9\sum_{k=1}^{10} {4} \\ & = \dfrac {9(10)(11)}{2} + 9(10)(4) = 495 + 360 = \boxed{855} \end{aligned}$