Key Concept: The Power of a Special Vector

This problem elegantly demonstrates how understanding matrix-vector multiplication can simplify complex matrix power problems. The core idea revolves around finding a special column vector, say $\mathbf{x}$, such that when multiplied by the given matrix $A$, it reveals useful information about the matrix's properties or its entries. In this specific problem, the given condition $a_{i1} + a_{i2} + a_{i3} = 1$ for each row $i$ directly points to such a vector.

Step 1: Translating the Given Condition into Matrix Form

Let the given matrix $A$ be: $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ The problem states that for $i = 1, 2, 3$, the sum of entries in the $i$-th row is 1: $a_{11} + a_{12} + a_{13} = 1$ $a_{21} + a_{22} + a_{23} = 1$ $a_{31} + a_{32} + a_{33} = 1$

Now, consider a column vector $\mathbf{x}$ where all its entries are 1: $\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ Let's perform the matrix-vector multiplication $A\mathbf{x}$: $A\mathbf{x} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ According to the rules of matrix multiplication, the first entry of the resulting column vector will be $(a_{11} \times 1) + (a_{12} \times 1) + (a_{13} \times 1) = a_{11} + a_{12} + a_{13}$. Similarly for the other rows. $A\mathbf{x} = \begin{bmatrix} a_{11} + a_{12} + a_{13} \\ a_{21} + a_{22} + a_{23} \\ a_{31} + a_{32} + a_{33} \end{bmatrix}$ Using the given condition, we can substitute the sums: $A\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ Thus, we have established a crucial relationship: $A\mathbf{x} = \mathbf{x}$.

Explanation: This step is vital because it converts the verbal description of row sums into a compact and powerful matrix equation. The vector $\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ is essentially a "summation vector" because multiplying a matrix by it yields a column vector whose entries are the row sums of the original matrix.

Step 2: Extending the Property to Higher Powers of A

We found that $A\mathbf{x} = \mathbf{x}$. Let's see what happens when we multiply by $A$ again: $A^2\mathbf{x} = A(A\mathbf{x})$ Since we know $A\mathbf{x} = \mathbf{x}$, we can substitute it into the expression: $A^2\mathbf{x} = A(\mathbf{x})$ And again, $A\mathbf{x} = \mathbf{x}$: $A^2\mathbf{x} = \mathbf{x}$ We can generalize this pattern. For any positive integer $n$: $A^n\mathbf{x} = \mathbf{x}$ In this problem, we are interested in $A^3$, so we have: $A^3\mathbf{x} = \mathbf{x}$

Explanation: This step uses the associative property of matrix multiplication. If a vector is an eigenvector with eigenvalue 1 (which $\mathbf{x}$ is, in this case), then it remains an eigenvector with eigenvalue 1 for any positive integer power of the matrix. This property significantly simplifies the problem, as we don't need to explicitly calculate $A^3$.

Step 3: Calculating the Sum of all Entries of $A^3$

Let $A^3$ be a new matrix, say $B$, with entries $b_{ij}$: $A^3 = B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$ We want to find the sum of all entries of $A^3$, which is $b_{11} + b_{12} + b_{13} + b_{21} + b_{22} + b_{23} + b_{31} + b_{32} + b_{33}$.

Now, let's use the result from Step 2: $A^3\mathbf{x} = \mathbf{x}$. Substituting $A^3 = B$: $B\mathbf{x} = \mathbf{x}$ Writing this out with the matrix $B$ and vector $\mathbf{x}$: $\begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ Performing the matrix multiplication on the left side: $\begin{bmatrix} b_{11} + b_{12} + b_{13} \\ b_{21} + b_{22} + b_{23} \\ b_{31} + b_{32} + b_{33} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ By comparing the corresponding entries of the column vectors, we get: Sum of entries in the first row of $A^3$: $b_{11} + b_{12} + b_{13} = 1$ Sum of entries in the second row of $A^3$: $b_{21} + b_{22} + b_{23} = 1$ Sum of entries in the third row of $A^3$: $b_{31} + b_{32} + b_{33} = 1$

The question asks for the sum of all the entries of the matrix $A^3$. This is simply the sum of the row sums: Sum of all entries of $A^3 = (b_{11} + b_{12} + b_{13}) + (b_{21} + b_{22} + b_{23}) + (b_{31} + b_{32} + b_{33})$ Sum of all entries of $A^3 = 1 + 1 + 1 = 3$.

Explanation: This final step directly uses the definition of matrix-vector multiplication with the all-ones vector to extract the row sums of $A^3$. Since we already knew $A^3\mathbf{x}$ evaluates to the all-ones vector, we immediately get that each row sum of $A^3$ must be 1. Adding these individual row sums gives us the total sum of all entries.

Tips and Common Mistakes:

1. Recognize the "Summation Vector": Always be on the lookout for problems involving row sums or column sums. The vector $\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}$ (or its transpose for column sums) is a powerful tool.
   • $A\mathbf{x}$ gives a column vector where each entry is the sum of the corresponding row of $A$.
   • $\mathbf{x}^T A$ gives a row vector where each entry is the sum of the corresponding column of $A$.
2. Eigenvalue/Eigenvector Connection: The relation $A\mathbf{x} = \mathbf{x}$ means that $\mathbf{x}$ is an eigenvector of $A$ with an eigenvalue of 1. This is a very important concept in linear algebra and often appears in JEE problems in disguise. This property immediately implies $A^n\mathbf{x} = \mathbf{x}$ for any positive integer $n$.
3. Avoid Unnecessary Calculation: Do not try to find $A^3$ explicitly. It would be a lengthy and error-prone process. The elegance of this solution lies in avoiding such calculations.
4. Careful with Matrix vs. Vector: A common mistake (as seen in the provided initial solution) is to confuse a matrix $A^3$ with the column vector $A^3\mathbf{x}$. They are different entities. $A^3\mathbf{x}$ is a column vector whose entries are the row sums of $A^3$, not the matrix $A^3$ itself.

Summary and Key Takeaway:

This problem beautifully illustrates that sometimes, direct calculation is not the most efficient or intended method. By identifying the special property of the matrix $A$ (that its row sums are all 1), we could represent this property using a specific column vector $\mathbf{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ such that $A\mathbf{x} = \mathbf{x}$. This relationship holds for any positive integer power of $A$, meaning $A^3\mathbf{x} = \mathbf{x}$. Interpreting this final matrix-vector multiplication showed that each row sum of $A^3$ is 1. Therefore, the sum of all entries of $A^3$ is simply the sum of these row sums, which is $1+1+1=3$. This technique is a valuable tool for solving problems involving sums of matrix entries, especially for powers of matrices.

The final answer is $\boxed{3}$.