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Scalar Multiplication of Matrices

Core Concepts

Focuses on the operation of multiplying an entire matrix by a single number (a scalar). This stack covers the procedure and the key distributive and associative properties related to scalar multiplication.

CATEGORY
High School (Grades 9-12) / High School Advanced Math / Algebra II and Advanced Functions / Matrix Operations and Properties
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The Process of Matrix Multiplication
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Properties of Matrix Operations
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Scalar Multiplication of Matrices

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In the context of matrices, what is a 'scalar'?
A scalar is a single real number (not a matrix). It is used to scale a matrix up or down by multiplication.
How do you perform scalar multiplication with a scalar \(k\) and a matrix \(A\)?
To find the product \(kA\), you multiply every single element of the matrix \(A\) by the scalar \(k\).
$$ 5 \times \begin{pmatrix} 1 & -2 \\ 3 & 0 \end{pmatrix} = ? $$
$$ \begin{pmatrix} 5 \times 1 & 5 \times (-2) \\ 5 \times 3 & 5 \times 0 \end{pmatrix} = \begin{pmatrix} 5 & -10 \\ 15 & 0 \end{pmatrix} $$
$$ -2 \times \begin{pmatrix} 4 & 1 \\ -5 & 7 \\ 2 & -1 \end{pmatrix} = ? $$
$$ \begin{pmatrix} -8 & -2 \\ 10 & -14 \\ -4 & 2 \end{pmatrix} $$
What does the distributive property \(c(A+B)\) expand to?
The scalar \(c\) distributes to both matrices: \(c(A+B) = cA + cB\). This is valid as long as \(A\) and \(B\) have the same dimensions.
What does the distributive property \((c+d)A\) expand to?
The matrix \(A\) distributes to both scalars: \((c+d)A = cA + dA\).
Does the associative property hold for scalar multiplication? That is, is \((cd)A = c(dA)\)?
Yes. You can either multiply the scalars first and then by the matrix, or multiply the matrix by one scalar and then the resulting matrix by the second scalar. The result is the same.
If \(A = \begin{pmatrix} 8 & 12 \\ -4 & 0 \end{pmatrix}\), what is \(\frac{1}{4}A\)?
$$ \frac{1}{4} \begin{pmatrix} 8 & 12 \\ -4 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ -1 & 0 \end{pmatrix} $$
What is the result of multiplying any matrix \(A\) by the scalar \(1\)?
Multiplying by \(1\) does not change the matrix. \(1A = A\).
What is the relationship between scalar multiplication by \(-1\) and the additive inverse of a matrix \(A\)?
They are the same. Multiplying matrix \(A\) by the scalar \(-1\) produces its additive inverse, \(-A\). So, \(-1A = -A\).
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