Products of Matrices

Learning Outcomes

  • Multiply a matrix by a scalar, sum scalar multiples of matrices.
  • Multiply two matrices together.
  • Use a calculator to perform operations on matrices.

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. A scalar is a real number quantity that has magnitude but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication.

Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment from the fall 2013 semester to the fall of 2014. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in the table below.

Lab A Lab B
Computers 15 27
Computer Tables 16 34
Chairs 16 34

Converting the data to a matrix, we have the computer inventory in fall 2013 given by

[latex]{C}_{2013}=\left[\begin{array}{c}15\\ 16\\ 16\end{array}\begin{array}{c}27\\ 34\\ 34\end{array}\right][/latex]

To calculate how much computer equipment will be needed in 2014, we multiply all entries in matrix [latex]C[/latex] by 0.15.

[latex]\left(0.15\right){C}_{2013}=\left[\begin{array}{c}\left(0.15\right)15\\ \left(0.15\right)16\\ \left(0.15\right)16\end{array}\begin{array}{c}\left(0.15\right)27\\ \left(0.15\right)34\\ \left(0.15\right)34\end{array}\right]=\left[\begin{array}{c}2.25\\ 2.4\\ 2.4\end{array}\begin{array}{c}4.05\\ 5.1\\ 5.1\end{array}\right][/latex]

We must round up to the next integer, so the amount of new equipment needed is

[latex]\left[\begin{array}{c}3\\ 3\\ 3\end{array}\begin{array}{c}5\\ 6\\ 6\end{array}\right][/latex]

Adding the two matrices as shown below, we see the new inventory amounts.

[latex]\left[\begin{array}{c}15\\ 16\\ 16\end{array}\begin{array}{c}27\\ 34\\ 34\end{array}\right]+\left[\begin{array}{c}3\\ 3\\ 3\end{array}\begin{array}{c}5\\ 6\\ 6\end{array}\right]=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\begin{array}{c}32\\ 40\\ 40\end{array}\right][/latex]

This means

[latex]{C}_{2014}=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\begin{array}{c}32\\ 40\\ 40\end{array}\right][/latex]

Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs.

A General Note: Scalar Multiplication

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

[latex]A=\left[\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\ {a}_{21}& & & {a}_{22}\end{array}\right][/latex]

the scalar multiple [latex]cA[/latex] is

[latex]\begin{array}{ll}cA & =c\left[\begin{array}{ccc}{a}_{11}& & {a}_{12}\\ {a}_{21}& & {a}_{22}\end{array}\right]\hfill \\ & =\left[\begin{array}{ccc}c{a}_{11}& & c{a}_{12}\\ c{a}_{21}& & c{a}_{22}\end{array}\right]\hfill \end{array}[/latex]

Scalar multiplication is distributive. For the matrices [latex]A,B[/latex], and [latex]C[/latex] with scalars [latex]a[/latex] and [latex]b[/latex],

[latex]\begin{array}{l}\\ \begin{array}{c}a\left(A+B\right)=aA+aB\\ \left(a+b\right)A=aA+bA\end{array}\end{array}[/latex]

Example: Multiplying the Matrix by a Scalar

Multiply matrix [latex]A[/latex] by the scalar 3.

[latex]A=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right][/latex]

Try It

Given matrix [latex]B,\text{}[/latex] find [latex]-2B[/latex] where

[latex]B=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right][/latex]

Example: Finding the Sum of Scalar Multiples

Find the sum [latex]3A+2B[/latex].

[latex]A=\left[\begin{array}{rrr}\hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill 4& \hfill 3& \hfill -6\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 1\\ \hfill 0& \hfill -3& \hfill 2\\ \hfill 0& \hfill 1& \hfill -4\end{array}\right][/latex]

Try it


Finding the Product of Two Matrices

In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If [latex]A[/latex] is an [latex]\text{ }m\text{ }\times \text{ }r\text{ }[/latex] matrix and [latex]B[/latex] is an [latex]\text{ }r\text{ }\times \text{ }n\text{ }[/latex] matrix, then the product matrix [latex]AB[/latex] is an [latex]\text{ }m\text{ }\times \text{ }n\text{ }[/latex] matrix. For example, the product [latex]AB[/latex] is possible because the number of columns in [latex]A[/latex] is the same as the number of rows in [latex]B[/latex]. If the inner dimensions do not match, the product is not defined.

A has two rows and three columns and B has three rows and three columns. Because the number of columns in A matches the number of rows in B, the product of A and B is defined.

We multiply entries of [latex]A[/latex] with entries of [latex]B[/latex] according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.

tip for success

Work through the example model of matrix multiplication below on paper, then apply the process to the example problem below. It may take more than once to gain familiarity with it. Don’t be discouraged if you don’t understand fully right away. Matrix multiplication is a new skill and it will take time and practice for it to feel comfortable.

To obtain the entries in row [latex]i[/latex] of [latex]AB,\text{}[/latex] we multiply the entries in row [latex]i[/latex] of [latex]A[/latex] by column [latex]j[/latex] in [latex]B[/latex] and add. For example, given matrices [latex]A[/latex] and [latex]B,\text{}[/latex] where the dimensions of [latex]A[/latex] are [latex]2\text{ }\times \text{ }3[/latex] and the dimensions of [latex]B[/latex] are [latex]3\text{ }\times \text{ }3,\text{}[/latex] the product of [latex]AB[/latex] will be a [latex]2\text{ }\times \text{ }3[/latex] matrix.

[latex]A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill {b}_{11}& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill {b}_{21}& \hfill {b}_{22}& \hfill {b}_{23}\\ \hfill {b}_{31}& \hfill {b}_{32}& \hfill {b}_{33}\end{array}\right][/latex]

Multiply and add as follows to obtain the first entry of the product matrix [latex]AB[/latex].

  1. To obtain the entry in row 1, column 1 of [latex]AB,\text{}[/latex] multiply the first row in [latex]A[/latex] by the first column in [latex]B[/latex] and add.
    [latex]\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}[/latex]
  2. To obtain the entry in row 1, column 2 of [latex]AB,\text{}[/latex] multiply the first row of [latex]A[/latex] by the second column in [latex]B[/latex] and add.
    [latex]\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}[/latex]
  3. To obtain the entry in row 1, column 3 of [latex]AB,\text{}[/latex] multiply the first row of [latex]A[/latex] by the third column in [latex]B[/latex] and add.
    [latex]\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}[/latex]

We proceed the same way to obtain the second row of [latex]AB[/latex]. In other words, row 2 of [latex]A[/latex] times column 1 of [latex]B[/latex]; row 2 of [latex]A[/latex] times column 2 of [latex]B[/latex]; row 2 of [latex]A[/latex] times column 3 of [latex]B[/latex]. When complete, the product matrix will be

[latex]AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdot {b}_{32}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}\\ \end{array}\\ {a}_{21}\cdot {b}_{13}+{a}_{22}\cdot {b}_{23}+{a}_{23}\cdot {b}_{33}\end{array}\right][/latex]

A General Note: Properties of Matrix Multiplication

For the matrices [latex]A,B,\text{}[/latex] and [latex]C[/latex] the following properties hold.

  • Matrix multiplication is associative:
  • Matrix multiplication is distributive:
    [latex]\begin{array}{l}\begin{array}{l}\\ C\left(A+B\right)=CA+CB,\end{array}\hfill \\ \left(A+B\right)C=AC+BC.\hfill \end{array}[/latex]

Note that matrix multiplication is not commutative.

Example: Multiplying Two Matrices

Multiply matrix [latex]A[/latex] and matrix [latex]B[/latex].

[latex]A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right][/latex]

Try It

Example: Multiplying Two Matrices

Given [latex]A[/latex] and [latex]B:[/latex]

  1. Find [latex]AB[/latex].
  2. Find [latex]BA[/latex].

[latex]A=\left[\begin{array}{ccc}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\text{ and }B=\left[\begin{array}{cc}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right][/latex]

Q & A

Is it possible for AB to be defined but not BA?

Yes, consider a matrix A with dimension [latex]3\times 4[/latex] and matrix B with dimension [latex]4\times 2[/latex]. For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and 3 so the product is undefined.

Example: Using Matrices in Real-World Problems

Let’s return to the problem presented at the opening of this section. We have the table below, representing the equipment needs of two soccer teams.

Wildcats Mud Cats
Goals 6 10
Balls 30 24
Jerseys 14 20

We are also given the prices of the equipment, as shown in the table below.

Goal $300
Ball $10
Jersey $30

We will convert the data to matrices. Thus, the equipment need matrix is written as

[latex]E=\left[\begin{array}{c}6\\ 30\\ 14\end{array}\begin{array}{c}10\\ 24\\ 20\end{array}\right][/latex]

The cost matrix is written as

[latex]C=\left[\begin{array}{ccc}300& 10& 30\end{array}\right][/latex]
We perform matrix multiplication to obtain costs for the equipment.
[latex]\begin{array}{l}\hfill \\ \hfill \\ CE & =\left[\begin{array}{rrr}\hfill 300& \hfill 10& \hfill 30\end{array}\right]\cdot \left[\begin{array}{rr}\hfill 6& \hfill 10\\ \hfill 30& \hfill 24\\ \hfill 14& \hfill 20\end{array}\right]\hfill \\ & =\left[\begin{array}{rr}\hfill 300\left(6\right)+10\left(30\right)+30\left(14\right)& \hfill 300\left(10\right)+10\left(24\right)+30\left(20\right)\end{array}\right]\hfill \\ & =\left[\begin{array}{rr}\hfill 2,520& \hfill 3,840\end{array}\right]\hfill \end{array}[/latex]

The total cost for equipment for the Wildcats is $2,520, and the total cost for equipment for the Mud Cats is $3,840.

How To: Given a matrix operation, evaluate using a calculator

  1. Save each matrix as a matrix variable
  2. Enter the operation into the calculator, calling up each matrix variable as needed.
  3. If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message.

Example: Using a Calculator to Perform Matrix Operations

Find [latex]AB-C[/latex] given

[latex]A=\left[\begin{array}{rrr}\hfill -15& \hfill 25& \hfill 32\\ \hfill 41& \hfill -7& \hfill -28\\ \hfill 10& \hfill 34& \hfill -2\end{array}\right],B=\left[\begin{array}{rrr}\hfill 45& \hfill 21& \hfill -37\\ \hfill -24& \hfill 52& \hfill 19\\ \hfill 6& \hfill -48& \hfill -31\end{array}\right],\text{and }C=\left[\begin{array}{rrr}\hfill -100& \hfill -89& \hfill -98\\ \hfill 25& \hfill -56& \hfill 74\\ \hfill -67& \hfill 42& \hfill -75\end{array}\right][/latex].