1. **Problem statement:** Given a matrix $A = [a_{ij}]$ of size $3 \times 4$, show that $I_3 A = A$ where $I_3$ is the $3 \times 3$ identity matrix. 2. **Recall the identity matrix property:** The identity matrix $I_n$ of size $n \times n$ has 1's on the diagonal and 0's elsewhere. Multiplying any $n \times m$ matrix $A$ by $I_n$ on the left leaves $A$ unchanged. 3. **Matrix multiplication:** The product $I_3 A$ is defined since $I_3$ is $3 \times 3$ and $A$ is $3 \times 4$. The result is a $3 \times 4$ matrix. 4. **Calculate the $(i,j)$-th element of $I_3 A$:** $$ (I_3 A)_{ij} = \sum_{k=1}^3 (I_3)_{ik} a_{kj} $$ Since $(I_3)_{ik} = 1$ if $i=k$ and 0 otherwise, this sum reduces to: $$ (I_3 A)_{ij} = a_{ij} $$ 5. **Conclusion:** Each element of $I_3 A$ equals the corresponding element of $A$, so $$ I_3 A = A $$ This shows the identity matrix acts as a multiplicative identity on the left for matrices of compatible size.