Scalar Multiplication of Matrices

2.3. Scalar Multiplication of Matrices#

Let \(A =\left[a_{ij}\right]\) be a matrix, and \(k\) is a scalar (a number). The scalar multiplication of the matrix \(A\) by the scalar \(k\), denoted as \(kA\), is a new matrix obtained by multiplying each entry of \(A\) by the scalar \(k\).

Mathematically, if \(A\) is an \(m\times n\) matrix, then \(kA\) is also an \(m\times n\) matrix, and its entries are calculated as follows:

\[\begin{align*} (kA)_{ij} = k \cdot a_{ij} \end{align*}\]

Where:

\((kA)_{ij}\) is the entry in the \(i\)-th row and \(j\)-th column of the matrix \(kA\).
\(a_{ij}\) is the entry in the \(i\)-th row and \(j\)-th column of the original matrix \(A\).
\(k\) is the scalar (a constant number).

To put it simply, to get the entry in the resulting matrix \(kA\) at position \((i,j)\), we multiply the corresponding entry in matrix \(A\) by the scalar \(k\).

Example: For example:

Let’s consider the following matrix:

\[\begin{align*} A = \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} \end{align*}\]

If we multiply this matrix by the scalar \(k = 3\), the resulting matrix \(kA\) will be:

\[\begin{align*} kA = 3 \cdot \begin{bmatrix} 2 & 4 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 3\cdot2 & 3\cdot4 \\ 3\cdot1 & 3\cdot3 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 3 & 9 \end{bmatrix} \end{align*}\]

So, the resulting matrix \(kA\) is a \(2\times 2\) matrix, and each entry is obtained by multiplying the corresponding entry in matrix \(A\) by the scalar \(k = 3\).

Example: If \(A=\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\), find \(cA\) and \(-A\).

Solution:

\(3A\): We have the matrix \(A\) as:

\[\begin{align*} A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \end{align*}\]

To find \(3A\), we multiply each entry of matrix \(A\) by the scalar \(3\):

\[\begin{align*} 3A = 3 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} (3)(1) & (3)(2) \\ (3)(3) & (3)(4) \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix} \end{align*}\]

In this case, every entry in the resulting matrix \(3A\) is obtained by multiplying the corresponding entry in matrix \(A\) by the scalar \(3\).
\(-A\): We have the same matrix \(A\) as above:

\[\begin{align*} A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \end{align*}\]

To find \(-A\), we multiply each entry of matrix \(A\) by the scalar \(-1\) (which is the same as negating each entry):

\[\begin{align*} -A = (-1) \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} (-1)(1) & (-1)(2) \\ (-1)(3) & (-1)(4) \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -3 & -4 \end{bmatrix} \end{align*}\]

In this case, every entry in the resulting matrix \(-A\) is obtained by multiplying the corresponding entry in matrix \(A\) by the scalar \(-1\) (i.e., negating each entry).

Scalar multiplication of a matrix involves multiplying each entry of the matrix by the scalar, resulting in a new matrix with the same dimensions. It’s a straightforward operation and is a fundamental concept in linear algebra.

Remark

The subtraction of matrices is similar to the addition of matrices, except that we add the opposite of the second matrix to the first matrix. That is, \(A-B=A+(-B)\). Unlike addition, subtraction is not commutative, which means that the order of the matrices matters [Kuttler and Farah, 2020, Nicholson, 2018]. In general, for two matrices \(A\) and \(B\), we have

\[\begin{align*} A-B\neq B-A. \end{align*}\]

Proposition - Properties of Scalar Multiplication:

Some of the properties of scalar multiplication [Kuttler and Farah, 2020, Nicholson, 2018]:

Distributive Law over Matrix Addition: \(k(A+B) = kA+kB\),
Distributive Law over Scalar Addition: \((k+ p)A = kA+ pA\),
Associative Law for Scalar Multiplication: \(k(pA) = (kp)A\),
Rule for Multiplication by 1: \(1A = A\).