With A = [[9,1,1],[1,2,1],[1,18,1]] and C = [[10,2,2],[2,3,2],[2,19,2]], compute (A + C)^2.

Step-by-step solution

1. 1

   Add before squaring, and do not expand the binomial. It is tempting to write $(A+C)^2=A^2+2AC+C^2$, but that identity requires $AC=CA$, which matrices generally do not satisfy. Adding first avoids the issue entirely and is also far less arithmetic.

2. 2

   Add the matrices entrywise. Both are $3\times 3$, so the sum is defined:

   $S=A+C=\begin{pmatrix}19&3&3\\3&5&3\\3&37&3\end{pmatrix}$

3. 3

   Set up the squaring as S·S. Each entry of $S^2$ is the dot product of a row of $S$ with a column of $S$. The columns of $S$ are $(19,3,3)$, $(3,5,37)$ and $(3,3,3)$.

4. 4

   Compute the first row.

   $(19,3,3)\cdot(19,3,3)=361+9+9=379$

   $(19,3,3)\cdot(3,5,37)=57+15+111=183$

   $(19,3,3)\cdot(3,3,3)=57+9+9=75$

5. 5

   Compute the second and third rows.

   $(3,5,3)\cdot(19,3,3)=57+15+9=81,\quad (3,5,3)\cdot(3,5,37)=9+25+111=145,\quad (3,5,3)\cdot(3,3,3)=33$

   $(3,37,3)\cdot(19,3,3)=57+111+9=177,\quad (3,37,3)\cdot(3,5,37)=9+185+111=305,\quad (3,37,3)\cdot(3,3,3)=129$

6. 6

   Assemble the result.

   $\boxed{(A+C)^{2}=\begin{pmatrix}379&183&75\\81&145&33\\177&305&129\end{pmatrix}}$

7. 7

   Check the third column quickly. Every entry of the third column of $S$ equals $3$, so the third column of $S^2$ must be $3\times$(row sums of $S$): the row sums are $25$, $11$ and $43$, giving $75$, $33$ and $129$ — matching the computed values and confirming the arithmetic.