Animated Solution for Mathematics - Matrices and Determinants: Let $A=[a_{ij}]$ be a real matrix of order $3\times 3$, such that $a_{i1}+a_{i2}+a_{i3}=1$, for $i=1,2,3$. Then, the sum of all the entries of the matrix $A^3$ is equal to:

Welcome, future engineer. Today, we are going to peel back the layers of a problem that, at first glance, looks like a tedious exercise in matrix multiplication. You might be tempted to write out the general form of a $3\times 3$ matrix, cube it, and then sum the entries. But stop. If you do that, you are fighting the math, not dancing with it.

Let us look at the soul of the matrix $A$. We are given that the sum of the entries in each row is $1$. This is not just a random condition; it is a geometric constraint.

Imagine the matrix $A$ as a linear transformation. When we multiply $A$ by a vector, we are transforming that vector. What happens if we multiply $A$ by a vector of ones? Let $X=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$.

When we compute $AX$, the first entry of the resulting vector is the dot product of the first row of $A$ and $X$, which is $a_{11}(1)+a_{12}(1)+a_{13}(1)$. This is exactly the sum of the first row! Since the problem guarantees that every row sums to $1$, the result of $AX$ is simply $\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$, which is $X$ itself.

We have discovered that $AX=X$. This is a profound realization. It means that the vector $X$ is an eigenvector of $A$ corresponding to the eigenvalue $1$.

Now, let us take this further. If $AX=X$, what happens when we look at $A^2$? We can write $A^{2}X$ as $A(AX)$.

Since we know $AX=X$, this becomes $A(X)$, which we already know is $X$. So, $A^{2}X=X$. Do you see the pattern emerging? It is beautiful, isn't it?

We can apply this logic again for $A^3$:

No matter how many times we multiply $A$ by itself, the vector $X$ remains unchanged. This is the power of understanding the transformation rather than just the numbers.

We are asked for the sum of all entries of $A^3$. Let $A^3=B$, where $B=[b_{ij}]$. The sum of all entries of $B$ is the sum of the row sums of $B$.

We know that:

This tells us that the sum of the first row of $B$ is $1$, the sum of the second row is $1$, and the sum of the third row is $1$. Therefore, the total sum of all entries in $A^3$ is simply $1+1+1=3$.

We did not need to calculate a single element of $A^3$. We used the structure of the matrix to bypass the calculation entirely. This is the mindset of a JEE Advanced topper: look for the invariant, look for the symmetry, and let the math do the heavy lifting for you. The final answer is 3.