Q.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 $3A - 5B$.

Concept: Scalar Multiplication — multiplying each entry of a matrix by a scalar.

First, compute $3A$ by multiplying every entry of $A$ by 3:

Next, compute $5B$ similarly: …

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 $3$:

2. Compute $5B$ Multiply every entry of $B$ by $5$:

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

Method: Scalar Multiplication Followed by Matrix Subtraction

Use this for expressions like $pA-qB$ where scalars multiply matrices before they are combined. Scale first, then subtract entrywise.

Steps

Step 1: Multiply each matrix by its scalar, entry by entry.

$(kA)_{ij}=k\,a_{ij}$ — the scalar hits every entry, including fractions, so e.g. $3\cdot\tfrac{2}{3}=2$.

Step 2: Subtract the resulting matrices entrywise.

With both scaled matrices of the same order, $(pA-qB)_{ij}=p\,a_{ij}-q\,b_{ij}$. …

Common Mistakes

Mistake 1: Applying the scalar to only some entries.

Why it's wrong: $kA$ multiplies every entry by $k$, not just the diagonal or first row. Correct approach: scale all entries, fractions included.

Mistake 2: Arithmetic slips multiplying a scalar by a fraction.

Why it's wrong: e.g. $5\cdot\tfrac{7}{5}=7$; a slip here corrupts the entry. Correct approach: cancel carefully, $k\cdot\tfrac{p}{k}=p$. …