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Learn scalar multiplication: multiply every matrix element by a number. Learn properties, solve examples, and apply distributive laws.
---
## Understanding Matrix Scalar Multiplication
In the world of matrices, we not only deal with operations between matrices but also operations between a matrix and a single number. This single number is commonly referred to as a **scalar**.
Matrix scalar multiplication is one of the fundamental operations that is important to understand. Imagine you have a cake recipe, and you want to make twice as much. You would naturally multiply each ingredient's measurement by the number $$2$$, right?
Visible text: Matrix scalar multiplication is one of the fundamental operations that is important to understand. Imagine you have a cake recipe, and you want to make twice as much. You would naturally multiply each ingredient's measurement by the number , right?
A similar concept applies to matrix scalar multiplication.
## What is Matrix Scalar Multiplication?
Matrix scalar multiplication is the operation of multiplying every element in a matrix by a scalar number.
If we have a matrix $$A$$ and a scalar $$k$$, then the result of the scalar multiplication of $$k$$ by matrix $$A$$ (written as $$kA$$) is a new matrix where each element is the product of the corresponding element of matrix $$A$$ and the scalar $$k$$.
Visible text: If we have a matrix and a scalar , then the result of the scalar multiplication of by matrix (written as ) is a new matrix where each element is the product of the corresponding element of matrix and the scalar .
Mathematically, if matrix $$A$$ has an order of $$m \times n$$:
Visible text: Mathematically, if matrix has an order of :
```math
A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}
```
Then the multiplication of matrix $$A$$ by scalar $$k$$ is:
Visible text: Then the multiplication of matrix by scalar is:
```math
kA = k \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} = \begin{bmatrix} k \cdot a_{11} & k \cdot a_{12} & \cdots & k \cdot a_{1n} \\ k \cdot a_{21} & k \cdot a_{22} & \cdots & k \cdot a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ k \cdot a_{m1} & k \cdot a_{m2} & \cdots & k \cdot a_{mn} \end{bmatrix}
```
The resulting matrix, $$kA$$, will have the same order as matrix $$A$$.
Visible text: The resulting matrix, , will have the same order as matrix .
This concept is similar to repeated addition. For example, $$2A$$ is the same as $$A + A$$. If we add matrix $$A$$ $$k\text{ times}$$, the result is $$kA$$.
Visible text: This concept is similar to repeated addition. For example, is the same as . If we add matrix , the result is .
```math
\underbrace{A + A + \cdots + A}_{k \text{ times}} = kA
```
## Matrix Scalar Multiplication Examples
To better understand this concept, let's look at some examples.
**Example** $$1$$:
Visible text: **Example** :
Suppose we have matrix $$P$$ as in the following example:
Visible text: Suppose we have matrix as in the following example:
```math
P = \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix}
```
Determine $$2P$$!
Visible text: Determine !
**Solution:**
To calculate $$2P$$, we multiply each element of matrix $$P$$ by the scalar $$2$$.
Visible text: To calculate , we multiply each element of matrix by the scalar .
```math
2P = 2 \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 2 \times 9 & 2 \times 7 \\ 2 \times 3 & 2 \times 1 \end{bmatrix} = \begin{bmatrix} 18 & 14 \\ 6 & 2 \end{bmatrix}
```
So, the result of $$2P$$ is $$\begin{bmatrix} 18 & 14 \\ 6 & 2 \end{bmatrix}$$.
Visible text: So, the result of is .
**Example** $$2$$:
Visible text: **Example** :
Given matrix $$Q = \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}$$ and scalar $$k=4$$. Determine $$4Q$$!
Visible text: Given matrix and scalar . Determine !
**Solution:**
We will multiply each element in matrix $$Q$$ by the scalar $$4$$.
Visible text: We will multiply each element in matrix by the scalar .
Component: MathContainer
Children:
```math
4Q = 4 \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}
```
```math
= \begin{bmatrix} 4 \times (-1) & 4 \times \frac{1}{2} & 4 \times 1 \\ 4 \times 4 & 4 \times (-\frac{1}{4}) & 4 \times 2 \\ 4 \times (-6) & 4 \times 1 & 4 \times (-4) \end{bmatrix}
```
```math
= \begin{bmatrix} -4 & 2 & 4 \\ 16 & -1 & 8 \\ -24 & 4 & -16 \end{bmatrix}
```
Thus, $$4Q = \begin{bmatrix} -4 & 2 & 4 \\ 16 & -1 & 8 \\ -24 & 4 & -16 \end{bmatrix}$$.
Visible text: Thus, .
## Properties of Matrix Scalar Multiplication
Matrix scalar multiplication has several important properties to be aware of. Let $$A$$ and $$B$$ be matrices of the same order, $$h$$ and $$k$$ be scalars, and $$O$$ be the zero matrix.
Visible text: Matrix scalar multiplication has several important properties to be aware of. Let and be matrices of the same order, and be scalars, and be the zero matrix.
1. **Distributive over Matrix Addition:**
```math
k(A + B) = kA + kB
```
This means multiplying a scalar by the sum of two matrices is the same as summing the
products of the scalar with each matrix.
2. **Distributive over Scalar Addition:**
```math
(h + k)A = hA + kA
```
This means multiplying the sum of two scalars by a matrix is the same as
summing the products of each scalar with the matrix.
3. **Associative with Scalar Multiplication:**
```math
(hk)A = h(kA) = k(hA)
```
This means multiplying a matrix by the product of two scalars is the same as
multiplying the first scalar by the product of the second scalar and the matrix.
4. **Scalar Multiplication Identity:**
```math
1A = A
```
Multiplying a matrix by the scalar $$1$$ does not change the matrix.
5. **Multiplication by Zero Scalar:**
```math
0A = O
```
Multiplying a matrix by the scalar $$0$$ results in the zero matrix ($$O$$), which is a matrix where all elements are $$0$$.
6. **Multiplication of Zero Matrix by a Scalar:**
```math
kO = O
```
Multiplying the zero matrix by any scalar results in the zero matrix.
7. **Multiplication by Scalar $$-1$$:**
```math
(-1)A = -A
```
Multiplying a matrix by the scalar $$-1$$ results in the negative of the matrix.
Visible text: 1. **Distributive over Matrix Addition:**
This means multiplying a scalar by the sum of two matrices is the same as summing the
products of the scalar with each matrix.
2. **Distributive over Scalar Addition:**
This means multiplying the sum of two scalars by a matrix is the same as
summing the products of each scalar with the matrix.
3. **Associative with Scalar Multiplication:**
This means multiplying a matrix by the product of two scalars is the same as
multiplying the first scalar by the product of the second scalar and the matrix.
4. **Scalar Multiplication Identity:**
Multiplying a matrix by the scalar does not change the matrix.
5. **Multiplication by Zero Scalar:**
Multiplying a matrix by the scalar results in the zero matrix (), which is a matrix where all elements are .
6. **Multiplication of Zero Matrix by a Scalar:**
Multiplying the zero matrix by any scalar results in the zero matrix.
7. **Multiplication by Scalar :**
Multiplying a matrix by the scalar results in the negative of the matrix.
These properties help simplify calculations and provide a deeper understanding of matrix algebra.
## Exercises
1. Given matrix $$X = \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}$$. Calculate $$3X$$!
2. If $$Y = \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}$$, determine $$\frac{1}{10}Y$$!
3. Given matrices $$A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$ and $$B = \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix}$$. Show that $$3(A+B) = 3A + 3B$$!
Visible text: 1. Given matrix . Calculate !
2. If , determine !
3. Given matrices and . Show that !
### Answer Key
1. Solution:
```math
3X = 3 \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}
```
```math
= \begin{bmatrix} 3 \times 5 & 3 \times (-2) \\ 3 \times 0 & 3 \times 4 \\ 3 \times (-1) & 3 \times 7 \end{bmatrix}
```
```math
= \begin{bmatrix} 15 & -6 \\ 0 & 12 \\ -3 & 21 \end{bmatrix}
```
2. Solution:
```math
\frac{1}{10}Y = \frac{1}{10} \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}
```
```math
= \begin{bmatrix} \frac{1}{10} \times 10 & \frac{1}{10} \times 20 \\ \frac{1}{10} \times (-30) & \frac{1}{10} \times 0 \end{bmatrix}
```
```math
= \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix}
```
3. To show $$3(A+B) = 3A + 3B$$:
First, calculate the left side of the equation, $$3(A+B)$$.
```math
A+B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 2+0 & 1+5 \\ 3+6 & 4+(-2) \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 9 & 2 \end{bmatrix}
```
Then,
```math
3(A+B) = 3 \begin{bmatrix} 2 & 6 \\ 9 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}
```
Next, calculate the right side of the equation, $$3A + 3B$$.
```math
3A = 3 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix}
```
```math
3B = 3 \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 15 \\ 18 & -6 \end{bmatrix}
```
Then,
```math
3A + 3B = \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix} + \begin{bmatrix} 0 & 15 \\ 18 & -6 \end{bmatrix} = \begin{bmatrix} 6+0 & 3+15 \\ 9+18 & 12+(-6) \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}
```
Since the result of the left side calculation ($$\begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}$$) is the same as the result of the right side calculation ($$\begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}$$), it is proven that $$3(A+B) = 3A + 3B$$.
Visible text: 1. Solution:
2. Solution:
3. To show :
First, calculate the left side of the equation, .
Then,
Next, calculate the right side of the equation, .
Then,
Since the result of the left side calculation () is the same as the result of the right side calculation (), it is proven that .