prove $$ K_{i j}^{\prime} A_{j}^{\prime}=B_{i}^{\prime}=a_{i k} B_{k} $$ Where aik $$ =\frac{\partial x_{i}^{\prime}}{\partial x_{k}} $$

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The equation $$ K_{ij}' A_j' = B_i' = a_{ik} B_k $$ is proven by showing that the transformation of vector $$ A_j' $$ through matrix $$ K_{ij}' $$ is equivalent to transforming vector $$ B_k $$ through the Jacobian $$ a_{ik} $$.

prove \( K_{i j}^{\prime} A_{j}^{\prime}=B_{i}^{\prime}=a_{i k} B_{k} \) Where aik \( =\frac{\partial x_{i}^{\prime}}{\partial x_{k}} \)

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prove \( K_{i j}^{\prime} A_{j}^{\prime}=B_{i}^{\prime}=a_{i k} B_{k} \) Where aik \( =\frac{\partial x_{i}^{\prime}}{\partial x_{k}} \)

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