AcademyMatrix Algebra
Academy
Scalar Multiplication
Level 1 - Linear Algebra topic page in Matrix Algebra.
Principle
Scalar multiplication multiplies every entry of a matrix by the same real number. It stretches the matrix entrywise and is compatible with matrix addition.
Notation
\(\lambda,\mu\)
real scalars
\(A,B\)
matrices
\(a_{ij}\)
entry of A in row i and column j
\(\lambda A\)
the scalar multiple of A by \lambda
\(0\)
the zero scalar or a zero matrix, depending on context
The Core Method
If \(A=(a_{ij})\), then
Scalar multiplication
\[(\lambda A)_{ij}=\lambda a_{ij}\]
That is, multiply every entry by \(\lambda\). Scalar multiplication distributes over matrix addition:
Distribute over matrix addition
\[\lambda(A+B)=\lambda A+\lambda B\]
and over scalar addition:
Distribute over scalar addition
\[(\lambda+\mu)A=\lambda A+\mu A\]
Multiplying by \(0\) gives the zero matrix of the same size as \(A\). Multiplying by \(-1\) gives the additive inverse \(-A\).
Worked Cases
Question
Compute \(-3\begin{pmatrix}2&-1\\0&4\end{pmatrix}\).
Answer
Multiply every entry by \(-3\). The entries become \(-3\cdot2=-6\), \(-3\cdot(-1)=3\), \(-3\cdot0=0\), and \(-3\cdot4=-12\). Therefore \(-3\begin{pmatrix}2&-1\\0&4\end{pmatrix}=\begin{pmatrix}-6&3\\0&-12\end{pmatrix}\).
Examples
Question
What is \(0A\) when \(A\) is a \(3\) by \(2\) matrix?
Answer
Multiplying every entry of \(A\) by the scalar \(0\) gives the \(3\) by \(2\) zero matrix. The size is preserved; only the entries become zero.