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Matrix

Matrix Scalar Multiplication

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?

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 AA and a scalar kk, then the result of the scalar multiplication of kk by matrix AA (written as kAkA) is a new matrix where each element is the product of the corresponding element of matrix AA and the scalar kk.

Mathematically, if matrix AA has an order of m×nm \times n:

A=[a11a12a1na21a22a2nam1am2amn]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 AA by scalar kk is:

kA=k[a11a12a1na21a22a2nam1am2amn]=[ka11ka12ka1nka21ka22ka2nkam1kam2kamn]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, kAkA, will have the same order as matrix AA.

This concept is similar to repeated addition. For example, 2A2A is the same as A+AA + A. If we add matrix AA kk times, the result is kAkA.

A+A++Ak times=kA\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:

Suppose we have matrix PP as in the reference image:

P=[9731]P = \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix}

Determine 2P2P!

Solution:

To calculate 2P2P, we multiply each element of matrix PP by the scalar 2.

2P=2[9731]=[2×92×72×32×1]=[181462]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 2P2P is [181462]\begin{bmatrix} 18 & 14 \\ 6 & 2 \end{bmatrix}.

Example 2:

Given matrix Q=[11214142614]Q = \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix} and scalar k=4k=4. Determine 4Q4Q!

Solution:

We will multiply each element in matrix QQ by the scalar 4.

4Q=4[11214142614]4Q = 4 \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}
=[4×(1)4×124×14×44×(14)4×24×(6)4×14×(4)]= \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}
=[424161824416]= \begin{bmatrix} -4 & 2 & 4 \\ 16 & -1 & 8 \\ -24 & 4 & -16 \end{bmatrix}

Thus, 4Q=[424161824416]4Q = \begin{bmatrix} -4 & 2 & 4 \\ 16 & -1 & 8 \\ -24 & 4 & -16 \end{bmatrix}.

Properties of Matrix Scalar Multiplication

Matrix scalar multiplication has several important properties to be aware of. Let AA and BB be matrices of the same order, hh and kk be scalars, and OO be the zero matrix.

  1. Distributive over Matrix Addition:

    k(A+B)=kA+kBk(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:

    (h+k)A=hA+kA(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:

    (hk)A=h(kA)=k(hA)(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:

    1A=A1A = A

    Multiplying a matrix by the scalar 1 does not change the matrix.

  5. Multiplication by Zero Scalar:

    0A=O0A = O

    Multiplying a matrix by the scalar 0 results in the zero matrix (OO), which is a matrix where all elements are 0.

  6. Multiplication of Zero Matrix by a Scalar:

    kO=OkO = O

    Multiplying the zero matrix by any scalar results in the zero matrix.

  7. Multiplication by Scalar -1:

    (1)A=A(-1)A = -A

    Multiplying a matrix by the scalar -1 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=[520417]X = \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}. Calculate 3X3X!
  2. If Y=[1020300]Y = \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}, determine 110Y\frac{1}{10}Y!
  3. Given matrices A=[2134]A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} and B=[0562]B = \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix}. Show that 3(A+B)=3A+3B3(A+B) = 3A + 3B!

Answer Key

  1. Solution:

    3X=3[520417]3X = 3 \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}
    =[3×53×(2)3×03×43×(1)3×7]= \begin{bmatrix} 3 \times 5 & 3 \times (-2) \\ 3 \times 0 & 3 \times 4 \\ 3 \times (-1) & 3 \times 7 \end{bmatrix}
    =[156012321]= \begin{bmatrix} 15 & -6 \\ 0 & 12 \\ -3 & 21 \end{bmatrix}
  2. Solution:

    110Y=110[1020300]\frac{1}{10}Y = \frac{1}{10} \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}
    =[110×10110×20110×(30)110×0]= \begin{bmatrix} \frac{1}{10} \times 10 & \frac{1}{10} \times 20 \\ \frac{1}{10} \times (-30) & \frac{1}{10} \times 0 \end{bmatrix}
    =[1230]= \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix}
  3. To show 3(A+B)=3A+3B3(A+B) = 3A + 3B:

    First, calculate the left side of the equation, 3(A+B)3(A+B).

    A+B=[2134]+[0562]=[2+01+53+64+(2)]=[2692]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,

    3(A+B)=3[2692]=[618276]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+3B3A + 3B.

    3A=3[2134]=[63912]3A = 3 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix}
    3B=3[0562]=[015186]3B = 3 \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 0 & 15 \\ 18 & -6 \end{bmatrix}

    Then,

    3A+3B=[63912]+[015186]=[6+03+159+1812+(6)]=[618276]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 ([618276]\begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}) is the same as the result of the right side calculation ([618276]\begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}), it is proven that 3(A+B)=3A+3B3(A+B) = 3A + 3B.