scalar and matrix multiplication assignment

Matrix Scalar Multiplication

Matrix scalar multiplication is multiplying a matrix by a scalar. A scalar is a real number whereas a matrix is a rectangular array of numbers. When we deal with matrices, we come across two types of multiplications:

• Multiplying a matrix by another matrix and is called "matrix multiplication"
• Multiplying a matrix by a scalar (a number) and is called "matrix scalar multiplication"

Let us learn how to do matrix scalar multiplication and its properties along with examples.

What is Matrix Scalar Multiplication?

The matrix scalar multiplication is the process of multiplying a matrix by a scalar. Let 'A' be a matrix and 'k' be a scalar (real number). Then kA is the result of the matrix scalar multiplication. To find kA, we just multiply every element of A by 'k'. Here are some examples.

Example: If A = \(\left[\begin{array}{ll} -1 & 2 \\ 0 & 3 \end{array}\right]\) then

• 2A = 2 \(\left[\begin{array}{ll} -1 & 2 \\ \\ 0 & 3 \end{array}\right]\) = \(\left[\begin{array}{ll} 2(-1) & 2(2) \\ \\ 2(0) & 2(3) \end{array}\right]\) = \(\left[\begin{array}{ll} -2 & 4 \\ \\ 0 & 6 \end{array}\right]\)
• (1/2) A = (1/2) \(\left[\begin{array}{ll} -1 & 2 \\ \\ 0 & 3 \end{array}\right]\) = (1/2)\(\left[\begin{array}{ll} (1/2)-1 & (1/2)2 \\ \\ (1/2)0 & (1/2)3 \end{array}\right]\) = \(\left[\begin{array}{ll} -1/2 & 1 \\ \\ 0 & 3/2 \end{array}\right]\)
• -A = -1 \(\left[\begin{array}{ll} -1 & 2 \\ \\ 0 & 3 \end{array}\right]\) = \(\left[\begin{array}{ll} 1 & -2 \\ \\ 0 & -3 \end{array}\right]\)
• 0A = 0 \(\left[\begin{array}{ll} -1 & 2 \\ \\ 0 & 3 \end{array}\right]\) = \(\left[\begin{array}{ll} 0 & 0 \\ \\ 0 & 0 \end{array}\right]\)

Thus, matrix scalar multiplication is mathematically defined as follows:

"If A = [aᵢⱼ] ₘ ₓ ₙ and k is a scalar then kA = k [aᵢⱼ] ₘ ₓ ₙ = [kaᵢⱼ] ₘ ₓ ₙ"

i.e., the element in i th row and j th column of kA is obtained by multiplying the corresponding element of A by 'k'. We can visualize this in the figure below.

Properties of Matrix Scalar Multiplication

If A and B are matrices of the same order ; and k, a, and b are scalars then:

• A and kA have the same order. For example, if A is a matrix of order 2 x 3 then any of its scalar multiple, say 2A, is also of order 2 x 3.
• Matrix scalar multiplication is commutative . i.e., k A = A k.
• Scalar multiplication of matrices is associative . i.e., (ab) A = a (bA).
• The distributive property works for the matrix scalar multiplication as follows: k (A + B) = kA + k B A (a + b) = Aa + Ab (or) aA + bA
• The product of any scalar and a zero matrix is the zero matrix itself. For example: k \(\left[\begin{array}{ll} 0 & 0 \\ \\ 0 & 0 \end{array}\right]\) = \(\left[\begin{array}{ll} 0 & 0 \\ \\ 0 & 0 \end{array}\right]\)
• The product of -1 and A gives -A which is the additive inverse of A. For example, the additive inverse of \(\left[\begin{array}{ll} -1 & 2 \\ \\ 0 & 3 \end{array}\right]\) is (-1) \(\left[\begin{array}{ll} -1 & 2 \\ \\ 0 & 3 \end{array}\right]\) = \(\left[\begin{array}{ll} 1 & -2 \\ \\ 0 & -3 \end{array}\right]\).

☛ Related Topics:

• Addition of Matrices
• Matrix Addition Calculator
• Types of Matrices
• Properties of Matrices

Matrix Scalar Multiplication Examples

Example 1: If the matrix A = \(\left[\begin{array}{c} -18 \\ -15\\ 21 \end{array}\right]\) then what is the scalar multiple (-1/3)A?

To find (-1/3) A, we have to multiply every element of A by (-1/3). Then

(-1/3) A = \(\left[\begin{array}{c} -1/3(-18) \\ -1/3(-15)\\ -1/3(21) \end{array}\right]\)

= \(\left[\begin{array}{c} 6 \\ 5\\ -7 \end{array}\right]\)

Answer: (-1/3) A = \(\left[\begin{array}{c} 6 \\ 5\\ -7 \end{array}\right]\).

Example 2: If A = \(\left[\begin{array}{cc} a & -2 \\ \\ -3 & -2b \end{array}\right]\) and B = \(\left[\begin{array}{cc} 5a & 2 \\ \\ 3 & 4b \end{array}\right]\), and A + B = 2I, where I is the identity matrix of order 2x2. Then find the values of a and b.

2I is the scalar multiple of the identity matrix I = \(\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\). i.e., 2I = \(\left[\begin{array}{ll} 2 & 0 \\ 0 & 2 \end{array}\right]\).

It is given that A + B = 2I

\(\left[\begin{array}{cc} a & -2 \\ \\ -3 & -2b \end{array}\right]\) + \(\left[\begin{array}{cc} 5a & 2 \\ \\ 3 & 4b \end{array}\right]\) = \(\left[\begin{array}{ll} 2 & 0 \\ \\ 0 & 2 \end{array}\right]\)

\(\left[\begin{array}{ll} 6a & 0 \\ \\ 0 & 2b \end{array}\right]\) = \(\left[\begin{array}{ll} 2 & 0 \\ \\ 0 & 2 \end{array}\right]\)

We will set the corresponding elements equal.

6a = 2 ⇒ a = 1/3

2b = 2 ⇒ b = 1

Answer: a = 1/3 and b = 1.

Example 3: If A = \(\left[\begin{array}{ll} -5 & 1 & 3\\ -4 & -2 & -1 \end{array}\right]\) and B = \(\left[\begin{array}{ll} 6 & -7 & 2\\ 0 & -8 & 3 \end{array}\right]\), then find -2A + 3B.

= -2 \(\left[\begin{array}{ll} -5 & 1 & 3\\ -4 & -2 & -1 \end{array}\right]\) + 3 \(\left[\begin{array}{ll} 6 & -7 & 2\\ 0 & -8 & 3 \end{array}\right]\)

= \(\left[\begin{array}{ll} 10 & -2 & -6\\ 8 & 4 & 2 \end{array}\right]\) + \(\left[\begin{array}{ll} 18 & -21 & 6\\ 0 & -24 & 9 \end{array}\right]\)

= \(\left[\begin{array}{ll} 28 & -23 & 0\\ 8 & -20 & 11 \end{array}\right]\)

Answer: -2A + 3B = \(\left[\begin{array}{ll} 28 & -23 & 0\\ 8 & -20 & 11 \end{array}\right]\).

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Practice Questions on Matrix Scalar Multiplication

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FAQs on Matrix Scalar Multiplication

What is the difference between matrix scalar multiplication and matrix multiplication.

Matrix scalar multiplication is multiplying a matrix by a scalar whereas matrix multiplication is multiplying two matrices. For any two matrices A and B, and for a scalar 'k', kA and kB represent the scalar multiplications of A and B respectively by k whereas AB represents the multiplication of matrices A and B.

How Do You Solve Matrix Scalar Multiplication?

The result of multiplying a matrix by a scalar is again a matrix of the same order where each of its elements is obtained by multiplying the corresponding elements of the original matrix by the scalar. For example, if P = \(\left[\begin{array}{ccc} 2 & -1 & 3 \\ 0 & 5 & 2 \\ 1 & -1 & -2 \end{array}\right]\) then

= (-3) \(\left[\begin{array}{ccc} 2 & -1 & 3 \\ 0 & 5 & 2 \\ 1 & -1 & -2 \end{array}\right]\)

= \(\left[\begin{array}{ccc} -3(2) & -3(-1) & -3(3) \\ -3(0) & -3(5) & -3(2) \\ -3(1) & -3(-1) & -3(-2) \end{array}\right]\)

= \(\left[\begin{array}{ccc} -6 & 3 & -9 \\ 0 & -15 & -6 \\ -3 & 3 & 6 \end{array}\right]\)

Can We Multiply a Matrix by a Scalar?

Yes, we can multiply a matrix by a scalar. For doing this, we just need to multiply every element of the matrix by the scalar. For example, if A = \(\left[\begin{array}{ccc} 2 & -1 & 3 \\ \\ 0 & 5 & 2 \\ \end{array}\right]\) then 2A = \(\left[\begin{array}{ccc} 4 & -2 & 6\\\\ 0 & 10 & 4 \\ \end{array}\right]\).

Can You Multiply a Matrix by 3?

A matrix can be multiplied by any scalar and hence it can be multiplied by 3 as well. For example, if A = \(\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 1 \end{array}\right]\) then 3A = \(\left[\begin{array}{ll} 3 & 6 \\ 9 & 12 \\ 15 & 3 \end{array}\right]\).

Is Matrix Scalar Multiplication Commutative?

Yes, the matrix scalar multiplication is commutative. i.e., for any matrix M and a scalar 'a', we have aM = Ma. For example:

• 2 \(\left[\begin{array}{ll} 1 & -1 \\ 2 & 1 \end{array}\right]\) = \(\left[\begin{array}{ll} 2 & -2 \\ 4 & 2 \end{array}\right]\).
• \(\left[\begin{array}{ll} 1 & -1 \\ 2 & 1 \end{array}\right]\) 2 = \(\left[\begin{array}{ll} 2 & -2 \\ 4 & 2 \end{array}\right]\).

Can a Matrix be a Scalar?

No, a matrix cannot be a scalar. A matrix is a rectangular array of elements where the elements are arranged in rows and columns. A scalar is just a real number . Hence, a matrix cannot be a scalar.

Scalar Matrix Multiplication

We need to consider only one equation

Find the values of x and y.

2 x – 6 = 5 2 x = 11 x = 5.5

4 – y = 3 y = 1

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Scalar & Matrix Multiplication

Scalar & Matrix Multiplication When is Multiplication Defined? Entry-wise / The Identity

There are two types of multiplication for matrices: scalar multiplication and matrix multiplication.

What is scalar multiplication?

Scalar multiplication is the process of multiplying every entry in a matrix by the same number; this number is called the "scalar".

Content Continues Below

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(The word "scalar" comes from the Latin "scalaris", related to the word for "ladder". A ladder can be used to "scale", or climb up and over, a wall. If you take linear algebra — after calculus — you may learn more about this.)

How do you multiply by a scalar?

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To multiply a matrix by a scalar, multiply each entry of the matrix by the scalar's value. For instance, given a matrix M and the scalar −1 , the scalar product −1 M will multiply each entry in M by −1 , so each entry in −1 M will have the opposite sign of each entry in the original matrix M .

What are the properties of scalar multiplication?

The properties of scalar multiplication of a matrix are similar to the properties for multiplying numbers. Given a matrices X and Y and scalars c and d , we have the following:

• c X is a matrix
• c(d X ) = (cd) X
• c( X + Y ) = c X + c Y

What is an example of multiplying by a scalar?

The following is an example of multiplying a matrix by a scalar.

They've given me a matrix and two numbers; namely, the numbers 2 and −1 . These two numbers are the scalars. So they're asking me to do scalar multiplication.

To do the first scalar multiplication to find 2 A , I just multiply a 2 on every entry in the matrix.

Being excruciatingly complete, the process looks like this:

The other scalar multiplication, to find −1 A , works the same way:

I can do the multiplications in my head, of course, but if an instructor wants to see every frickin' step, then I need to write things out like the above. You probably won't need to be this complete in your own work.

Having done the multiplications, my final answer (including which matrix goes with which multiplication) is:

You can use the Mathway widget below to practice multiplying a matrix by a scalar. (Or skip the widget and continue with the lesson.) Try the entered exercise, or type in your own exercise. Then click the button and select "Multiply" to compare your answer to Mathway's.

Please accept "preferences" cookies in order to enable this widget.

(Click "Tap to view steps" to be taken directly to the Mathway site for a paid upgrade.)

Scalar multiplication is easy. Matrix multiplication, however, is quite another story. In fact, it's a royal pain in the hindquarters.

What is matrix multiplication?

Matrix multiplication is the process of multiplying one matrix by another matrix, when such multiplication is well-defined — that is, when the matrices fit the rule that make matrix multiplication work.

(We'll get to what that rule is on the next page .)

What is the formula for multiplying matrices?

Your textbook probably gave you a complex summation formula for the process of matrix multiplication, and that formula probably didn't make any sense at all to you. A typical statement of the formula is as follows:

Let A = α ij be an s × m matrix, and let B  =  β ij be an m × t matrix. Let AB  = C =  γ ij , where C is an s × t matrix.

Then, for 1 ≤ i ≤ s and 1 ≤ j ≤ t , the entries of C are defined as being:

γ ij = Σ m k =1 α ik β kj

...which is about as clear as mud.

That's okay. The process is messy, and that complicated formula is the best they can do for an explanation in a formal setting like a textbook. I'm way less formal, so I'll show you how the process really works:

To calculate the product AB , I first write down A and B next to each other like this:

Now I need to multiply the entries in the ROWS of A by the entries in the COLUMNS of B . By this I mean that I take the first row of A and the first column of B , and I multiply the first entries of A 's row and B 's column, then the second entries of A 's row and B 's column, and then the third entries of A 's row and B 's column; and then I add the three products.

The sum of these three multiplications is one entry in the product matrix AB . In fact, being the product of row 1 and column 1 , the summed result is the 1,1 -entry of the product AB . Then I continue in like manner.

All the entries of the product matrix are found in this way. For instance, the sum of the products from row 2 of A and column 1 of B is the 2,1 -entry of AB .

Which is all well and good, but what does this look like in practice?

What does matrix multiplication look like?

When I multiply matrices, I use my fingers to keep track of what I'm doing. The following animation is my attempt to illustrate this process. (Don't laugh; I'm no artist!)

The above is an animated gif on the live website.

(Now, class; what did I say about laughing?) 😜

My final answer is:

How do you multiply two matrices?

To multiply two matrices, you entry-wise multiply rows of the left-hand matrix by columns of the right-hand matrix. The sum of the products of the entries of the i -th row of the left-hand matrix and the j -th column of the right-hand matrix becomes the i , j -th entry of the product matrix.

This general rule is, in large part, what that complicated formula in your textbook was all about.

So when I, in the above animation, multiplied the first row (of A ) and the second column (of B ), this gave me the first-row - second-column entry in the product matrix AB .

URL: https://www.purplemath.com/modules/mtrxmult.htm

You can use the Mathway widget below to practice multiplying two matrices. (Or skip the widget and continue to the next page .) Try the entered exercise, or type in your own exercise. Then click the button to compare your answer to Mathway's.

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Scalar multiplication of matrices

On this post we explain how to multiply a matrix by a scalar (number). You will see examples of scalar multiplications of matrices that will help you understand it perfectly and solved exercises so that you can practice. You will also find all the properties of the product of a scalar and a matrix.

Table of Contents

How to multiply a matrix by a matrix?

To multiply a matrix by a scalar , multiply each element of the matrix by the scalar.

Note that when we work with matrices we refer to numbers as scalars, therefore, multiplying a matrix by a scalar means multiplying a matrix by a real number.

Example of a multiplication of a matrix by a scalar

Once we have seen how to do a scalar multiplication of matrices, let’s see an example of this type of matrix operation:

As you can see, to solve the product of a scalar and a matrix, you simply have to multiply each entry of the matrix by the scalar.

Having seen this type of matrix operation, you can now learn how to calculate a multiplication of matrices .

Practice scalar multiplication of matrices

Find the following scalar multiplication:

To calculate the multiplication of a scalar by matrix we have to multiply each entry of the matrix by the scalar:

Note that the matrix before performing the scalar operation was a square 2×2 dimension matrix, and so is the result of the operation.

Calculate the following multiplication of a matrix by a scalar:

The result of the multiplication of the 3×3 square matrix by the scalar is:

Note that the dimension of the matrix remains the same after performing the operation.

Multiply the following rectangular matrix by the scalar -2:

The result of the scalar multiplication is:

Perform the following additions and scalar multiplications of matrices:

See how to add matrices .

It is an operation that combines two products of scalar and matrices and a sum of matrices of order 2. Thus, we must first solve the scalar multiplications of matrices:

And then we add the resulting matrices:

Given the following square matrices of order 3:

It is an operation that combines scalar multiplication together with addition and subtraction of 3×3 matrices. Furthermore, matrix I is the identity matrix, which is composed of 1 on the main diagonal and 0 the remainder of the elements:

Therefore, we first do the multiplications:

Secondly, we solve the addition of the first two matrices:

Finally, we perform the matrix subtraction:

Properties of scalar multiplication of matrices

There are many types of matrices : square matrices, triangular matrices, the identity matrix, symmetric and antisymmetric matrices,… But, fortunately, all the properties of the product of numbers and matrices work for all kinds of matrices.

So these are the properties of matrix scalar multiplication:

• Associative property : when two or more scalars are involved in a scalar multiplication of matrices, we can first multiply the scalars and then the result by the matrix.

Look at the following two operations as they give the same result, regardless of how we multiply scalars 2 and 3:

• Distributive property (addition of scalars) : adding two scalars and then multiplying the result by a matrix equals to multiply each scalar by the matrix and then adding the results.

As you can see in the example below, adding 1+2 and then multiplying it by a matrix is the same as multiplying the same matrix separately by 1 and by 2 and then adding the results:

• Distributive property (addition of matrices) : adding two matrices and then multiplying them by a number is equivalent to multiplying the two matrices separately by the same number and then adding the results.

You can check this property of the scalar multiplication of matrices with the following example:

• Neutral element property : the multiplication of any matrix by scalar 1 results in the same matrix.

Therefore, when multiplying a matrix by 1 we do not modify the matrix:

Other matrix operation related to multiplication, and that is very useful, is the power of a matrix . Here you will find how compute the power of a matrix and what it is for.

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2.2E: Matrix Addition, Scalar Multiplication, and Transposition Exercises

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Exercises for 1

Exercise \(\pageindex{1}\).

Find \(a\), \(b\), \(c\), and \(d\) if

• \(\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} c - 3d & -d \\ 2a + d & a + b \end{array} \right]\)
• \(\left[ \begin{array}{cc} a - b & b - c \\ c - d & d - a \end{array} \right] = 2 \left[ \begin{array}{rr} 1 & 1 \\ -3 & 1 \end{array} \right]\)
• \(3 \left[ \begin{array}{c} a \\ b \end{array} \right] + 2 \left[ \begin{array}{rr} b \\ a \end{array} \right] = \left[ \begin{array}{rr} 1 \\ 2 \end{array} \right]\)
• \(\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = \left[ \begin{array}{cc} b & c \\ d & a \end{array} \right]\)
• \((a\ b\ c\ d) = (-2, -4, -6, 0) + t(1, 1, 1, 1)\),
• \(t\) arbitrary \(a = b = c = d = t\), \(t\) arbitrary

Exercise \(\PageIndex{2}\)

Compute the following:

• \( \left[\begin{array}{lll}3 & 2 & 1 \\5 & 1 & 0\end{array}\right]-5\left[\begin{array}{rrr}3 & 0 & -2 \\1 & -1 & 2\end{array}\right]\)
• \( 3\left[\begin{array}{r}3 \\-1\end{array}\right]-5\left[\begin{array}{l}6 \\2\end{array}\right]+7\left[\begin{array}{r}1 \\-1 \end{array}\right] \)
• \( \left[\begin{array}{rr}-2 & 1 \\3 & 2\end{array}\right]-4\left[\begin{array}{ll}1 & -2 \\0 & -1\end{array}\right]+3\left[\begin{array}{rr}2 & -3 \\-1 & -2\end{array}\right]\)
• \( \left[\begin{array}{lll}3 & -1 & 2\end{array}\right]-2\left[\begin{array}{lll}9 & 3 & 4\end{array}\right]+\left[\begin{array}{lll}3 & 11 & -6\end{array}\right]\)
• \( \left[\begin{array}{rrrr}1 & -5 & 4 & 0 \\2 & 1 & 0 & 6\end{array}\right]^T \)
• \( \left[\begin{array}{rrr}0 & -1 & 2 \\1 & 0 & -4 \\-2 & 4 & 0\end{array}\right]^T \)
• \( \left[\begin{array}{rr}3 & -1 \\2 & 1\end{array}\right]-2\left[\begin{array}{rr}1 & -2 \\1 & 1\end{array}\right]^T \)
• \( 3\left[\begin{array}{rr}2 & 1 \\-1 & 0\end{array}\right]^T-2\left[\begin{array}{rr}1 & -1 \\2 & 3\end{array}\right] \)
• \(\left[ \begin{array}{r} -14 \\ -20 \end{array} \right]\) \((-12, 4, -12)\)
• \(\left[ \begin{array}{rrr} 0 & 1 & -2 \\ -1 & 0 & 4 \\ 2 & -4 & 0 \end{array} \right]\)
• \(\left[ \begin{array}{rr} 4 & -1 \\ -1 & -6 \end{array} \right]\)

Exercise \(\PageIndex{3}\)

Let \(A = \left[ \begin{array}{rr} 2 & 1 \\ 0 & -1 \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 3 & -1 & 2 \\ 0 & 1 & 4 \end{array} \right]\), \(C = \left[ \begin{array}{rr} 3 & -1 \\ 2 & 0 \end{array} \right]\), \(D = \left[ \begin{array}{rr} 1 & 3 \\ -1 & 0 \\ 1 & 4 \end{array} \right]\), and \(E = \left[ \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]\).

Compute the following (where possible).

• \(3A - 2B\)
• \(4A^{T} - 3C\)
• \((A + C)^{T}\)
• \(2B - 3E\)
• ((B - 2E)^{T}\)
• \(\left[ \begin{array}{rr} 15 & -5 \\ 10 & 0 \end{array} \right]\)
• \(\left[ \begin{array}{rr} 5 & 2 \\ 0 & -1 \end{array} \right]\)

Exercise \(\PageIndex{4}\)

Find \(A\) if:

• \(5A - \left[ \begin{array}{rr} 1 & 0 \\ 2 & 3 \end{array} \right] = 3A - \left[ \begin{array}{rr} 5 & 2 \\ 6 & 1 \end{array} \right]\)
• \(3A - \left[ \begin{array}{r} 2 \\ 1 \end{array} \right] = 5A - 2 \left[ \begin{array}{rr} 3 \\ 0 \end{array} \right]\)

b. \(\left[ \begin{array}{r} 2 \\ -\frac{1}{2} \end{array} \right]\)

Exercise \(\PageIndex{5}\)

Find \(A\) in terms of \(B\) if:

• \(A + B = 3A + 2B\)
• \(2A - B = 5(A + 2B)\)

b. \(A = -\frac{11}{3}B\)

Exercise \(\PageIndex{6}\)

If \(X\), \(Y\), \(A\), and \(B\) are matrices of the same size, solve the following systems of equations to obtain \(X\) and \(Y\) in terms of \(A\) and \(B\).

• \(5X + 3Y = A \\ 2X + Y = B\)
• \(4X + 3Y = A \\ 5X + 4Y = B\)

b. \(X = 4A - 3B\), \(Y = 4B - 5A\)

Example \(\PageIndex{7}\)

Find all matrices \(X\) and \(Y\) such that:

• \(3X - 2Y = \left[ \begin{array}{rr} 3 & - 1 \end{array} \right]\)
• \(2X - 5Y = \left[ \begin{array}{rr} 1 & 2 \end{array} \right]\)

\(Y = (s, t)\), \(X = \frac{1}{2}(1 + 5s, 2 + 5t)\); \(s\) and \(t\) arbitrary

Exercise \(\PageIndex{8}\)

Simplify the following expressions where \(A\), \(B\), and \(C\) are matrices.

• \(2 \left[ 9(A - B) + 7(2B - A) \right]\) \(- 2 \left[ 3(2B + A) - 2(A + 3B) - 5(A + B) \right]\)
• \(5 \left[ 3(A - B + 2C) - 2(3C - B) - A \right]\) \(+ 2 \left[ 3(3A - B + C) + 2(B - 2A) - 2C \right]\)

b. \(20A - 7B + 2C\)

Exercise \(\PageIndex{9}\)

If \(A\) is any \(2 \times 2\) matrix, show that:

• \(A = a \left[ \begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right] + b \left[ \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right] + c \left[ \begin{array}{rr} 0 & 0 \\ 1 & 0 \end{array} \right] + d \left[ \begin{array}{rr} 0 & 0 \\ 0 & 1 \end{array} \right]\) for some numbers \(a\), \(b\), \(c\), and \(d\).
• \(A = p \left[ \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right] + q \left[ \begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right] + r \left[ \begin{array}{rr} 1 & 0 \\ 1 & 0 \end{array} \right] + s \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]\) for some numbers \(p\), \(q\), \(r\), and \(s\).

b. If \(A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]\), then \((p, q, r, s) = \frac{1}{2}(2d, a + b - c - d, a - b + c - d, -a + b + c + d)\).

Exercise \(\PageIndex{10}\)

Let \(A = \left[ \begin{array}{rrr} 1 & 1 & -1 \end{array} \right]\), \(B = \left[ \begin{array}{rrr} 0 & 1 & 2 \end{array} \right]\), and \(C = \left[ \begin{array}{rrr} 3 & 0 & 1 \end{array} \right]\). If \(rA + sB + tC = 0\) for some scalars \(r\), \(s\), and \(t\), show that necessarily \(r = s = t = 0\).

Exercise \(\PageIndex{11}\)

• If \(Q + A = A\) holds for every \(m \times n\) matrix \(A\), show that \(Q = 0_{mn}\).
• If \(A\) is an \(m \times n\) matrix and \(A + A^\prime = 0_{mn}\), show that \(A^\prime = -A\

b. If \(A + A^\prime = 0\) then \(-A = -A + 0 = -A + (A + A^\prime) = (-A + A) + A^\prime = 0 + A^\prime = A^\prime\)

Exercise \(\PageIndex{12}\)

If \(A\) denotes an \(m \times n\) matrix, show that \(A = -A\) if and only if \(A = 0\).

Exercise \(\PageIndex{13}\)

A square matrix is called a diagonal matrix if all the entries off the main diagonal are zero. If \(A\) and \(B\) are diagonal matrices, show that the following matrices are also diagonal.

• \(kA\) for any number \(k\)

Write \(A = \diag(a_{1}, \dots, a_{n})\), where \(a_{1}, \dots, a_{n}\) are the main diagonal entries. If \(B = \diag(b_{1}, \dots, b_{n})\) then \(kA = \diag(ka_{1}, \dots, ka_{n})\).

Exercise \(\PageIndex{14}\)

In each case determine all \(s\) and \(t\) such that the given matrix is symmetric:

• \(\left[ \begin{array}{rr} 1 & s \\ -2 & t \end{array} \right]\)
• \(\left[ \begin{array}{cc} s & t \\ st & 1 \end{array} \right]\)
• \(\left[ \begin{array}{crc} s & 2s & st \\ t & -1 & s \\ t & s^{2} & s \end{array} \right]\)
• \(\left[ \begin{array}{ccc} 2 & s & t \\ 2s & 0 & s + t \\ 3 & 3 & t \end{array} \right]\)
• \(s = 1\) or \(t = 0\)
• \(s = 0\), and \(t = 3\)

Exercise \(\PageIndex{15}\)

In each case find the matrix \(A\).

• \(\left(A + 3 \left[ \begin{array}{rrr} 1 & -1 & 0 \\ 1 & 2 & 4 \end{array} \right] \right)^{T} = \left[ \begin{array}{rr} 2 & 1 \\ 0 & 5 \\ 3 & 8 \end{array} \right]\)
• \(\left(3A^{T} + 2 \left[ \begin{array}{rr} 1 & 0 \\ 0 & 2 \end{array} \right] \right)^{T} = \left[ \begin{array}{rr} 8 & 0 \\ 3 & 1 \end{array} \right]\)
• \(\left(2A - 3 \left[ \begin{array}{rrr} 1 & 2 & 0 \end{array} \right] \right)^{T} = 3A^{T} + \left[ \begin{array}{rrr} 2 & 1 & -1 \end{array} \right]^{T}\)
• \(\left(2A^{T} - 5 \left[ \begin{array}{rr} 1 & 0 \\ -1 & 2 \end{array} \right] \right)^{T} = 4A - 9 \left[ \begin{array}{rr} 1 & 1 \\ -1 & 0 \end{array} \right]\)
• \(\left[ \begin{array}{rr} 2 & 0 \\ 1 & -1 \end{array} \right]\)
• \(\left[ \begin{array}{rr} 2 & 7 \\ -\frac{9}{2} & -5 \end{array} \right]\)

Exercise \(\PageIndex{16}\)

Let \(A\) and \(B\) be symmetric (of the same size). Show that each of the following is symmetric.

• \((A - B)\)
• \(kA\) for any scalar \(k\)
• \(A = A^{T}\),
• \((kA)^{T} = kA^{T} = kA\).

Exercise \(\PageIndex{17}\)

Show that \(A + A^{T}\) is symmetric for any square matrix \(A\).

Exercise \(\PageIndex{18}\)

If \(A\) is a square matrix and \(A = kA^{T}\) where \(k \neq \pm 1\), show that \(A = 0\).

Exercise \(\PageIndex{19}\)

In each case either show that the statement is true or give an example showing it is false.

• If \(A + B = A + C\), then \(B\) and \(C\) have the same size.
• If \(A + B = 0\), then \(B = 0\).
• If the \((3, 1)\)-entry of \(A\) is \(5\), then the \((1, 3)\)-entry of \(A^{T}\) is \(-5\).
• \(A\) and \(A^{T}\) have the same main diagonal for every matrix \(A\).
• If \(B\) is symmetric and \(A^{T} = 3B\), then \(A = 3B\).
• If \(A\) and \(B\) are symmetric, then \(kA + mB\) is symmetric for any scalars \(k\) and \(m\).
• False. Take \(B = -A\) for any \(A \neq 0\).
• True. Transposing fixes the main diagonal.
• True. \((kA + mB)^{T} = (kA)^{T} + (mB)^{T} = kA^{T} + mB^{T} = kA + mB\)

Exercise \(\PageIndex{20}\)

A square matrix \(W\) is called skew-symmetric if \(W^{T} = -W\). Let \(A\) be any square matrix.

• Show that \(A - A^{T}\) is skew-symmetric.
• Find a symmetric matrix \(S\) and a skew-symmetric matrix \(W\) such that \(A = S + W\).
• Show that \(S\) and \(W\) in part (b) are uniquely determined by \(A\).

c. Suppose \(A = S + W\), where \(S = S^{T}\) and \(W = -W^{T}\). Then \(A^{T} = S^{T} + W^{T} = S - W\), so \(A + A^{T} = 2S\) and \(A - A^{T} = 2W\). Hence \(S = \frac{1}{2}(A + A^{T})\) and \(W = \frac{1}{2}(A - A^{T})\) are uniquely determined by \(A\).

Exercise \(\PageIndex{21}\)

If \(W\) is skew-symmetric (Exercise \(\PageIndex{20}\) ), show that the entries on the main diagonal are zero.

Exercise \(\PageIndex{22}\)

Prove the following parts of Theorem [thm:002170].

• \((k + p)A = kA + pA\)
• \((kp)A = k(pA)\)

b. If \(A = \left[ a_{ij} \right]\) then \((kp)A = \left[ (kp)a_{ij} \right] = \left[ k(pa_{ij}) \right] = k \left[ pa_{ij} \right] = k(pA)\).

Exercise \(\PageIndex{23}\)

Let \(A, A_{1}, A_{2}, \dots, A_{n}\) denote matrices of the same size. Use induction on \(n\) to verify the following extensions of properties 5 and 6 of Theorem \(\PageIndex{1}\).

• (k(A_{1} + A_{2} + \dots + A_{n}) = kA_{1} + kA_{2} + \dots + kA_{n}\) for any number \(k\)
• \((k_{1} + k_{2} + \dots + k_{n})A = k_{1}A + k_{2}A + \dots + k_{n}A\) for any numbers \(k_{1}, k_{2}, \dots, k_{n}\)

Example \(\PageIndex{1}\)

Let \(A\) be a square matrix. If \(A = pB^{T}\) and \(B = qA^{T}\) for some matrix \(B\) and numbers \(p\) and \(q\), show that either \(A = 0 = B\) or \(pq = 1\).

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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.

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.

Related Topics

• How to Add and Subtract Matrices
• How to Multiply Matrix

A 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 the 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 \(3A\).

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

Scalar Multiplication of Matrices – Example 2:

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

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

Exercises for Scalar Multiplication of Matrices

• \(\color{blue}{A=}\)\(\color{blue}{\begin{bmatrix}0 & 2 \\-2 & -5 \end{bmatrix}}\), \(\color{blue}{6A}\).
• \(\color{blue}{B=}\)\(\color{blue}{\begin{bmatrix}-5 & 0 &2 \\7& -3& 4 \\ -1& 3 & 2 \end{bmatrix}}\), \(\color{blue}{-3B}\).
• \(\color{blue}{D=}\)\(\color{blue}{\begin{bmatrix}6 & -2 \\3 & 7 \end{bmatrix}}\), \(\color{blue}{F=}\)\(\color{blue}{\begin{bmatrix}1 & -2 \\-3 & 4 \end{bmatrix}}\), \(\color{blue}{-2D+5F}\).
• \(\color{blue}{A=}\)\(\color{blue}{\begin{bmatrix}-5 & 2& 0 \\7 & -4 & 3 \\ -1 & 2 & 4 \end{bmatrix}}\), \(\color{blue}{B=}\)\(\color{blue}{\begin{bmatrix}0 & -1 & 7 \\6 & -12 & 2 \\ 9 & 5 & 1 \end{bmatrix}}\), \(\color{blue}{4A – 3B}\).
• \(\color{blue}{\begin{bmatrix}0 & 12 \\-12 & -30 \end{bmatrix}}\)
• \(\color{blue}{\begin{bmatrix}15 & 0 & -6 \\-21 & 9 &-12 \\ 3 & -9 & -6 \end{bmatrix}}\)
• \(\color{blue}{\begin{bmatrix}-7 & -6 \\-21 & 6 \end{bmatrix}}\)
• \(\color{blue}{\begin{bmatrix}-20 & 11 & -21 \\10 & 20 & 6 \\ -31 & -7 & 13 \end{bmatrix}}\)

by: Effortless Math Team about 2 years ago (category: Articles )

Effortless Math Team

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Linear algebra

Course: linear algebra   >   unit 2.

• A more formal understanding of functions
• Vector transformations
• Linear transformations
• Visualizing linear transformations
• Matrix from visual representation of transformation
• Matrix vector products as linear transformations
• Linear transformations as matrix vector products
• Image of a subset under a transformation
• im(T): Image of a transformation
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Video transcript

Scalar Multiplication

Learn how to implement scalar multiplication on matrices using R, Rcpp, Armadillo, and Eigen.

• Scalar multiplication in R
• Scalar multiplication algorithm
• Scalar multiplication in Armadillo
• Scalar multiplication in Eigen

Scalar multiplication is defined over a matrix of any dimension.

𝑅 = 𝑘 ∗ 𝐴 𝑅 = 𝑘∗𝐴 R = k ∗ A

𝑟 𝑖 𝑗 = 𝑘 ∗ 𝑎 𝑖 𝑗 𝑟_{𝑖𝑗} =𝑘∗𝑎_{𝑖𝑗} r ij ​ = k ∗ a ij ​

• 𝐴 𝐴 A is ( 𝑚 × 𝑛 ) (𝑚 \times 𝑛) ( m × n ) matrix
• 𝑘 𝑘 k is a scalar value

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