Lesson Explainer: Scalar Multiplication of Matrices | Nagwa Lesson Explainer: Scalar Multiplication of Matrices | Nagwa

Lesson Explainer: Scalar Multiplication of Matrices Mathematics • First Year of Secondary School

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In this explainer, we will learn how to multiply a scalar by a matrix and identify the properties of their multiplication.

A matrix contains an array of numbers organized by rows and columns. Before we can learn to use matrices for practical applications, we must first know how to perform operations with matrices. In this explainer, we will focus on scalar multiplication.

A scalar refers to a single number, for instance, 2 or 13. How do we multiply a matrix, an array of numbers, by a scaler 2 or by 13? Let us begin by defining this algebraic operation for general matrices.

Definition: Scalar Multiplication

For a matrix of order 𝑚×𝑛, defined as 𝐴=𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎, we can complete the “scalar multiplication” by a number 𝑘. This requires multiplying every entry of 𝐴 by 𝑘: 𝑘𝐴=𝑘×𝑎𝑘×𝑎𝑘×𝑎𝑘×𝑎𝑘×𝑎𝑘×𝑎𝑘×𝑎𝑘×𝑎𝑘×𝑎. This operation is often referred to as “scaling” the matrix by the constant 𝑘.

In other words, we can multiply a number (a scalar) by a matrix by multiplying the number by each entry in the matrix.

For instance, we consider the matrix 𝐴=53131106175427262638.

Suppose now that we wished to scale the matrix by the constant 2. We would then multiply every entry of 𝐴 by 2, hence finding that 2𝐴=(2)×5(2)×3(2)×(1)(2)×(3)(2)×1(2)×1(2)×0(2)×6(2)×1(2)×7(2)×5(2)×4(2)×(2)(2)×7(2)×(2)(2)×6(2)×(2)(2)×6(2)×(3)(2)×8.

Completing the operation for every entry gives 2𝐴=1062622012214108414412412616.

It is not the case that we are restricted to scaling a matrix by an integer and we could equally choose to scale by a fraction, an irrational number, or even a complex number if we are feeling adventurous. Although it is not strictly necessary, if scaling a matrix by a fraction, it is normally considered good practice to simplify any resultant fractions into their simplest form. For instance, we define the matrix 𝐵=536488306111 and decide to scale by the constant 13. We would find that 13𝐵=13×513×(3)13×613×413×813×813×313×013×613×(1)13×113×(1).

Reducing as many fractions as possible to their lowest form gives 13𝐵=5312438383102131313.

The scalar multiplication of a matrix multiplies a constant by each entry of the matrix, as we have seen. Based on this definition, we can see that the scalar multiplication by 1 preserves any matrix, and the scalar multiplication by 0 changes any matrix into a zero matrix.

Property: Multiplicative Identity and Multiplication by Zero in Scalar Multiplication

For any matrix 𝐴, 1𝐴=𝐴 and 0𝐴=𝑂, where 𝑂 is the zero matrix of the same order as 𝐴.

Before discussing some challenging problems which arise from the definition of scalar multiplication, we will first practice one question.

Example 1: Multiplying a Matrix by a Scalar

Given that 𝐴=(18), find 3𝐴.

Answer

Recall that we can multiply a number (a scalar) by a matrix by multiplying the number by each entry in the matrix.

To multiply 𝐴 by 3, we multiply every entry by this number and therefore we have 3𝐴=(3×(1)3×(8))=(324).

One of the key principles of scalar multiplication is that every single entry has the exact same process applied to it, namely, that every entry is multiplied by the same number. It is never the case that scalar multiplication multiplies different entries by different numbers. The following question gives an example of how this principle can be applied in terms of solving problems in linear algebra.

Example 2: Finding a Scalar Multiple of a Matrix

Given that 𝑥×2035=1402135, find the value of 𝑥.

Answer

Recall that we can multiply a number (a scalar) by a matrix by multiplying the number by each entry in the matrix.

By multiplying every entry by 𝑥 in the left-hand matrix, we are looking to find 𝑥 which solves the following equation: 𝑥×(2)𝑥×0𝑥×(3)𝑥×(5)=1402135, which can equivalently be written as 2𝑥03𝑥5𝑥=1402135.

Recall that two matrices are equal if they have the same order and each pair of entries from the same row and column are equal. We can see that the matrices have the same order since they both have 2 rows and 2 columns. We must therefore match the pairs of entries in the same row and column from the two matrices.

This produces the system of linear equations 2𝑥=14,0=0,3𝑥=21,5𝑥=35.

Apart from the trivial second equation, which is clearly true, we observe that every equation is solved by setting 𝑥=7.

Often, when working with a matrix of interest, we would choose to factor out a scalar multiple from each entry, if possible. For instance, if we were given the matrix 𝐴=362404168444, it is easily observed that this can be written as 𝐴=4×(9)4×(6)4×(0)4×(1)4×(4)4×(2)4×(1)4×(1)4×(1).

And therefore each entry has a factor of 4 that can be removed, thus obtaining 𝐴=4960142111.

In some situations, it may actually be preferable to define a new matrix 𝐵=960142111, thereby allowing us to write 𝐴=4𝐵.

Let us consider an example that involves factoring a scalar from a matrix.

Example 3: Scalar Multiplication

Given the matrix 𝐴=1165243174, what is the greatest number 𝑘 for which no entry of 𝑘𝐴 is greater than 1?

Answer

Recall that we can multiply a number (a scalar) by a matrix by multiplying the number by each entry in the matrix.

Given the matrix 𝐴 as defined above, this means 𝑘𝐴=𝑘16𝑘5𝑘2𝑘4𝑘3𝑘𝑘7𝑘4𝑘.

If 𝑘=0, every entry would be zero (which is the zero matrix), which would satisfy the condition that no entry is greater than 1. Since we are looking for the greatest number 𝑘 for which this condition is met, we can focus on finding a positive number 𝑘>0.

Examining the entries that are positive, we have 𝑘, 2𝑘, 4𝑘, 5𝑘, and 7𝑘. Of these, 7𝑘 should be the greatest entry, which means that the given condition will be met if 7𝑘1. This leads to 𝑘17. Since this value of 𝑘 is positive, the remaining values 𝑘, 3𝑘, 4𝑘, and 16𝑘 would all be negative and hence less than 1.

Since we are searching for an upper limit on the value of 𝑘, the answer is 𝑘=17.

Scalar multiplication is an operation that features regularly in linear algebra. Along with addition, it is perhaps the most simple algebraic operation to understand. However, this does not mean that a problem in linear algebra can be thought of as simple, purely because it involves scalar multiplication. The following two questions will illustrate how scalar multiplication of matrices can provide rich and interesting examples that allow for a better level of comprehension to be developed.

Example 4: Solving Equations Involving Scalar Multiplication

Consider the matrix equation 8193=𝑚3021+1130.

Find the value of 𝑚 which solves this equation.

Answer

Recall that we can multiply a number (a scalar) by a matrix by multiplying the number by each entry in the matrix. Similarly, we know that we can add two matrices of the same order by adding the pairs of entries corresponding to the same row and column from the two matrices.

First, we rewrite the equation after incorporating the scalar multiplication by 𝑚: 8193=3𝑚02𝑚𝑚+1130.

We complete the addition on the right-hand side of this equation, working entry by entry to find 8193=3𝑚112𝑚+3𝑚.

Recall that two matrices are equal if they have the same order and each pair of entries from the same row and column are equal. We can see that the matrices have the same order since they both have 2 rows and 2 columns.

Matching each pair of entries in the same row and column from the two matrices, we obtain 8=3𝑚1,1=1,9=2𝑚+3,3=𝑚.

The final equation gives 𝑚=3 and it can be checked that all of the given equations are also true if 𝑚=3, which means that this must be the answer. This can be checked by substituting 𝑚=3 back into the original matrix equation and then observing that both sides of the equation match.

In the previous example, we found an unknown constant in a matrix equation. If we examine the process closer, we can note that a matrix equation leads to simultaneous equations. Hence, a matrix equation can be used to find more than one unknown constant. In our final example, we will consider a matrix equation involving three unknown constants.

Example 5: Solving Equations Involving Scalar Multiplication

Find the numbers 𝑎, 𝑏, and 𝑐 so that 𝑎1101+𝑏1001+𝑐0110=1013.

Answer

Recall that we can multiply a number (a scalar) by a matrix by multiplying the number by each entry in the matrix. Similarly, we know that we can add two matrices of the same order by adding the pairs of entries corresponding to the same row and column from the two matrices.

We begin by incorporating the scalar terms into the matrices, giving 𝑎𝑎0𝑎+𝑏00𝑏+0𝑐𝑐0=1013.

Since matrix addition is completed entry by entry, we have 𝑎+𝑏𝑎𝑐𝑐𝑎+𝑏=1013.

Recall that two matrices are equal if they have the same order and each pair of entries from the same row and column are equal. We can see that the matrices have the same order since they both have 2 rows and 2 columns.

Matching each pair of entries in the same row and column from the two matrices, we obtain the simultaneous equations 𝑎+𝑏=1,𝑎𝑐=0,𝑐=1,𝑎+𝑏=3.

The third equation gives 𝑐=1, which can be substituted into the second equation to show that 𝑎=1. Then, substituting the value of 𝑎 into either the first or the fourth equation gives that 𝑏=2.

Scalar multiplication has many attractive properties when combined with matrix addition. If we were working with conventional algebra then we know that the quantities 𝑎, 𝑏, and 𝑐 will always obey the rule 𝑎×(𝑏+𝑐)=𝑎×𝑏+𝑎×𝑐, which is known as the “distributive property.” It transpires that the same property holds for matrix addition and scalar multiplication.

Theorem: Distributive Property

Scalar multiplication is “distributive” when combined with matrix addition. In other words, assuming that 𝑎 is a scalar constant and that 𝐵 and 𝐶 are matrices with the same order, then 𝑎(𝐵+𝐶)=𝑎𝐵+𝑎𝐶.

We demonstrate this result by way of example. Let us set 𝑎=3 and define the two matrices 𝐵=460872,𝐶=375661.

Then, 𝐵+𝐶=460872+375661=113514131; and hence 𝑎(𝐵+𝐶)=3113514131=3391542393.

Equally, we could have chosen a different route and first calculated 𝑎𝐵=1218024216,𝑎𝐶=9211518183.

From this we could then have calculated that 𝑎𝐵+𝑎𝐶=1218024216+9211518183=3391542393.

Thus, we have shown in this example that 𝑎(𝐵+𝐶)=𝑎𝐵+𝑎𝐶. Naturally, it is possible to prove the above theorem rigorously and without reference to any particular example, although here we have simply given a demonstration.

Although it may seem as though scalar multiplication is a trivial matrix operation, being able to work fluently with this concept is often the difference between the solution to a problem being short and simple and the solution being long and complicated. When examining any particular matrix, it is usually a good idea to check whether any scalar multiple can be factored from all entries.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Multiplying a matrix 𝐴 by a scalar 𝑘 means that every entry of the matrix 𝐴 will be multiplied by 𝑘.
  • When each entry of a matrix has a common factor, we can factor the scalar from the matrix by reversing the scalar multiplication.
  • Scalar multiplication is distributive. In other words, given a scalar 𝑎 and matrices 𝐵 and 𝐶 of the same order, 𝑎(𝐵+𝐶)=𝑎𝐵+𝑎𝐶.
  • For any matrix 𝐴, 1𝐴=𝐴 and 0𝐴=𝑂, where 𝑂 is the zero matrix of the same order as 𝐴.

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