If $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \ \frac{7}{5} & \frac{6}{5} & \frac{

Question

If $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$, then compute $3A - 5B$.

Accepted Answer

> [!TLDR]
> Scalar multiplication means multiplying every entry of a matrix by a constant. Compute $3A$ and $5B$ entrywise, then subtract corresponding entries. The result is a $3 \times 3$ matrix with all entries zero: the zero matrix.

The core idea here is **scalar multiplication of a matrix** — you simply multiply each element inside the matrix by the number outside. It’s just like scaling a vector, but applied to every entry. Once you’ve scaled both $A$ and $B$, you subtract them element by element, exactly as you would with any two matrices of the same size.

Let’s walk through it.

1. **Compute $3A$**  
   Multiply every entry of $A$ by $3$:
   $$
   3A = \begin{bmatrix}
   3 \cdot \frac{2}{3} & 3 \cdot 1 & 3 \cdot \frac{5}{3} \[4pt]
   3 \cdot \frac{1}{3} & 3 \cdot \frac{2}{3} & 3 \cdot \frac{4}{3} \[4pt]
   3 \cdot \frac{7}{3} & 3 \cdot 2 & 3 \cdot \frac{2}{3}
   \end{bmatrix}
   = \begin{bmatrix}
   2 & 3 & 5 \
   1 & 2 & 4 \
   7 & 6 & 2
   \end{bmatrix}
   $$

2. **Compute $5B$**  
   Multiply every entry of $B$ by $5$:
   $$
   5B = \begin{bmatrix}
   5 \cdot \frac{2}{5} & 5 \cdot \frac{3}{5} & 5 \cdot 1 \[4pt]
   5 \cdot \frac{1}{5} & 5 \cdot \frac{2}{5} & 5 \cdot \frac{4}{5} \[4pt]
   5 \cdot \frac{7}{5} & 5 \cdot \frac{6}{5} & 5 \cdot \frac{2}{5}
   \end{bmatrix}
   = \begin{bmatrix}
   2 & 3 & 5 \
   1 & 2 & 4 \
   7 & 6 & 2
   \end{bmatrix}
   $$

3. **Subtract $5B$ from $3A$**  
   Since both matrices are now identical, subtracting gives: …

Concept understanding — Scalar Multiplication

Scalar Multiplication of a Matrix

The Idea

An Example

Properties (all inherited from ordinary numbers)

Concept understanding — Scalar Multiplication

Scalar Multiplication of a Matrix

The Idea

An Example

Properties (all inherited from ordinary numbers)

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