First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.1

Solution

Let $A=[a_{ij}]_{3\times 3}$ and $I_3=[\delta_{ij}]_{3\times 3}$ .

The $(i,j)$ -entry of $I_3A$ is:

(I_3A)_{ij}=\sum_{k=1}^{3}(I_3)_{ik}a_{kj}=\sum_{k=1}^{3}\delta_{ik}a_{kj}=a_{ij}

Hence $I_3A$ and $A$ have the same entries, so:

\boxed{I_3A=A}

The $(i,j)$ -entry of $AI_3$ is:

(AI_3)_{ij}=\sum_{k=1}^{3}a_{ik}(I_3)_{kj}=\sum_{k=1}^{3}a_{ik}\delta_{kj}=a_{ij}

Therefore:

\boxed{AI_3=A}