Solving for the Sum of Entries in Matrix $A^3$
We are given a $3\times3$ real matrix $A = [a_{ij}]$ with the property that the sum of the entries in each row is 1. Mathematically, this means:
$ a_{i1} + a_{i2} + a_{i3} = 1 \quad \text{for } i=1, 2, 3 $
We need to find the sum of all entries in the matrix $A^3$.
Matrix Representation of Row Sum Property
Let $J$ be the $3\times1$ column vector where every element is 1:
$ J = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} $The condition that the sum of entries in each row of $A$ is 1 can be represented using matrix multiplication:
$ AJ = \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} = \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} = J $So, we have the property $AJ = J$.
Calculating the Sum of Entries of $A^3$
The sum of all entries of any $3\times3$ matrix $M$ can be calculated as $J^T M J$, where $J^T = [1, 1, 1]$ is the transpose of $J$. We want to calculate the sum of entries for $A^3$, which is $J^T A^3 J$.
Using the property $AJ = J$, we can find $A^3 J$:
- Multiply by $A$ on the left: \( A(AJ) = AJ \) which implies \( A^2 J = J \).
- Multiply by $A$ again: \( A(A^2 J) = AJ \) which implies \( A^3 J = J \).
Final Calculation
Now, substitute $A^3 J = J$ into the expression for the sum of entries:
Sum = \( J^T A^3 J = J^T (A^3 J) = J^T J \)
Calculate $J^T J$:
$ J^T J = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = (1 \times 1) + (1 \times 1) + (1 \times 1) = 1 + 1 + 1 = 3 $Therefore, the sum of all the entries of the matrix $A^3$ is 3.
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