How to Multiply a Matrix by a Scalar?

When multiplying a matrix by a scalar, the resulting matrix will always have the same dimensions as the original matrix. In this step-by-step guide, you learn more about how to multiply a matrix by a scalar.

How to Multiply a Matrix by a Scalar?

The scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each input in the matrix is multiplied by the given scalar.

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Step by step guide to scalar multiplication of matrices

In matrix algebra, a real number is called a scalar. The scalar product of a real number, \(b\), and a matrix \(A\) is the matrix \(bA\). Each element of matrix \(bA\) is equal to \(b\) times its corresponding element in \(A\).

Given scalar \(\color {blue}{b}\) and matrix \(A= \begin{bmatrix}a_{11} & a_{12} \\a_{21} & a_{22} \end{bmatrix}\), \(\color{blue}{b}A=\begin{bmatrix}\color{blue}{b}a_{11} & \color{blue}{b}a_{12} \\\color{blue}{b}a_{21} &\color{blue}{b} a_{22} \end{bmatrix}\)

Properties of Scalar Multiplication:

Let \(A\) and \(B\) be \(m\times n\) matrices. Let \(O_{m\times n}\) be the \(m\times n\) zero matrix and let \(p\) and \(q\) be scalars.

  • Associative Property: \(\color{blue}{p(qA)=(pq)A}\)
  • Closure Property: \(\color{blue}{pA}\) is an \(\color{blue}{m×n}\) matrix
  • Commutative Property: \(\color{blue}{pA= Ap}\)
  • Distributive Property: \(\color{blue}{(p+q)A=pA+qA}\), \(\color{blue}{p(A+B)=pA+pB}\)
  • Identity Property:\(\color{blue}{1. A=A}\)
  • Multiplicative Property of \(-1\): \(\color{blue}{(-1)A=-A}\)
  • Multiplicative Property of \(0\):\(\color{blue}{\space0. A=O_{m×n}}\)

Scalar Multiplication of Matrices – Example 1:

If \(A=\) \(\begin{bmatrix}-5 & -5 \\-1 & 2 \end{bmatrix}\) , find \(3A\).

solution:

\(3A=3\) \(\begin{bmatrix}-5 & -5 \\-1 & 2 \end{bmatrix}\) \(=\) \(\begin{bmatrix}3(-5) & 3(-5) \\3(-1) &3(2) \end{bmatrix}\) \(=\) \(\begin{bmatrix}-15 & -15 \\-3& 6 \end{bmatrix}\)

Scalar Multiplication of Matrices – Example 2:

If \(A=\) \(\begin{bmatrix}6 & 4 & 24 \\1 & -9 & 8 \end{bmatrix}\), find \(4A\).

solution:

\(4A=\) \(\begin{bmatrix}6 & 4 & 24 \\1 & -9 & 8 \end{bmatrix}\) \(=\) \(\begin{bmatrix}4(6) & 4(4) & 4(24) \\4(1) & 4(-9) & 4(8) \end{bmatrix}\) \(=\) \(\begin{bmatrix}24 & 16 & 96 \\4 & -36 & 32\end{bmatrix}\)

Exercises for Scalar Multiplication of Matrices

Solve.

  1. \(\color{blue}{A=}\)\(\color{blue}{\begin{bmatrix}0 & 2 \\-2 & -5 \end{bmatrix}}\), \(\color{blue}{6A}\).
  2. \(\color{blue}{B=}\)\(\color{blue}{\begin{bmatrix}-5 & 0 &2 \\7& -3& 4 \\ -1& 3 & 2 \end{bmatrix}}\), \(\color{blue}{-3B}\).
  3. \(\color{blue}{D=}\)\(\color{blue}{\begin{bmatrix}6 & -2 \\3 & 7 \end{bmatrix}}\), \(\color{blue}{F=}\)\(\color{blue}{\begin{bmatrix}1 & -2 \\-3 & 4 \end{bmatrix}}\), \(\color{blue}{-2D+5F}\).
  4. \(\color{blue}{A=}\)\(\color{blue}{\begin{bmatrix}-5 & 2& 0 \\7 & -4 & 3 \\ -1 & 2 & 4 \end{bmatrix}}\), \(\color{blue}{B=}\)\(\color{blue}{\begin{bmatrix}0 & -1 & 7 \\6 & -12 & 2 \\ 9 & 5 & 1 \end{bmatrix}}\), \(\color{blue}{4A – 3B}\).
This image has an empty alt attribute; its file name is answers.png
  1. \(\color{blue}{\begin{bmatrix}0 & 12 \\-12 & -30 \end{bmatrix}}\)
  2. \(\color{blue}{\begin{bmatrix}15 & 0 & -6 \\-21 & 9 &-12 \\ 3 & -9 & -6 \end{bmatrix}}\)
  3. \(\color{blue}{\begin{bmatrix}-7 & -6 \\-21 & 6 \end{bmatrix}}\)
  4. \(\color{blue}{\begin{bmatrix}-20 & 11 & -21 \\10 & 20 & 6 \\ -31 & -7 & 13 \end{bmatrix}}\)

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