Scalar multiplication of matrices

Take a matrix $A = [a_{ij}]$ of order $m \times n$ — that is, $m$ rows and $n$ columns — and a real number $k$, called a scalar. The scalar multiple $kA$ is formed by multiplying every entry of $A$ by $k$:

where: $a_{ij}$ is the entry in row $i$, column $j$ of $A$, and $k$ is any real number. Notice the order of $kA$ is unchanged — a $2 \times 2$ matrix scaled by any $k$ is still $2 \times 2$. Unlike matrix multiplication, there is no compatibility condition to check first; scalar multiplication works for every matrix and every real $k$, because it never needs a second matrix's dimensions to line up. Three algebraic properties follow directly from the entry-by-entry definition, and they are used constantly when simplifying matrix expressions:

where: $A$ and $B$ are matrices of the same order, and $k$, $k_1$, $k_2$ are scalars. The first says a scalar distributes across a matrix sum; the second says two scalars added first act the same as scaling separately and adding; the third says two scalars applied one after another combine into a single product scalar. Setting $k = -1$ gives a special case worth naming: $(-1)A = -A$, the negative of $A$, obtained by flipping the sign of every entry. This is exactly the tool that defines matrix subtraction, $A - B = A + (-1)B$, so subtraction is not really a fourth operation but scalar multiplication and addition working together.

Given $A = \begin{bmatrix} 4 & -2 \\ 0 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, find $3A$, $-2A$ and then $3A - 2B$.

Step 1. Compute $3A$ by multiplying every entry of $A$ by $3$: $3(4) = 12$, $3(-2) = -6$, $3(0) = 0$, $3(5) = 15$, giving $3A = \begin{bmatrix} 12 & -6 \\ 0 & 15 \end{bmatrix}$.

Step 2. Compute $-2A$ by multiplying every entry of $A$ by $-2$: $-2(4) = -8$, $-2(-2) = 4$, $-2(0) = 0$, $-2(5) = -10$, giving $-2A = \begin{bmatrix} -8 & 4 \\ 0 & -10 \end{bmatrix}$.

Step 3. For $3A - 2B$, first scale $B$: $2B = \begin{bmatrix} 2 & 4 \\ 6 & 8 \end{bmatrix}$. This must be done before any combining, because $3A - 2B$ means "scale $A$ by 3, scale $B$ by 2, then subtract" — not "subtract $A$ and $B$ first."

Step 4. Subtract the two already-scaled matrices entry by entry: $3A - 2B = \begin{bmatrix} 12-2 & -6-4 \\ 0-6 & 15-8 \end{bmatrix}$.

Step 5. Simplify each entry to reach the final matrix. Answer: $3A - 2B = \begin{bmatrix} 10 & -10 \\ -6 & 7 \end{bmatrix}$.

Competitive stretch

If $kA$ is the zero matrix for some matrix $A$ that is not itself the zero matrix, what can you conclude about $k$? Since every entry of $kA$ is $k \cdot a_{ij}$, and at least one $a_{ij} \neq 0$, the product $k \cdot a_{ij}$ can only be zero for every entry if $k = 0$ — a real number times a nonzero number is never zero. So $kA = O$ with $A \neq O$ forces $k = 0$; this small fact is a favourite one-mark trap.

• Scaling only some entries of the matrix and leaving the rest unchanged — every single entry must be multiplied by $k$, with no exceptions.
• Confusing scalar multiplication (a number times a matrix) with matrix multiplication (matrix times matrix) — these are different operations governed by different rules, and only one of them has an order-compatibility condition.
• Mishandling negative scalars, especially forgetting to flip the sign correctly on entries that are already negative, so a double-negative silently becomes positive when it should stay negative or vice versa.
• Trying to combine matrices before distributing the scalar, e.g. computing $A - B$ first in an expression like $k(A - B)$ instead of scaling each matrix and then subtracting; or, symmetrically, applying $k$ to only the first matrix in a sum such as $k(A+B)$.

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