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Guias de estudo > College Algebra

Products of Matrices

Learning Objectives

  • 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. Recall that 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. 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

C2013=[151616273434]{C}_{2013}=\left[\begin{array}{c}15\\ 16\\ 16\end{array}\begin{array}{c}27\\ 34\\ 34\end{array}\right]

To calculate how much computer equipment will be needed, we multiply all entries in matrix CC by 0.15.

(0.15)C2013=[(0.15)15(0.15)16(0.15)16(0.15)27(0.15)34(0.15)34]=[2.252.42.44.055.15.1]\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]

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

[333566]\left[\begin{array}{c}3\\ 3\\ 3\end{array}\begin{array}{c}5\\ 6\\ 6\end{array}\right]

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

[151616273434]+[333566]=[181919324040]\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]

This means

C2014=[181919324040]{C}_{2014}=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\begin{array}{c}32\\ 40\\ 40\end{array}\right]

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

A=[a11a12a21a22]A=\left[\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\ {a}_{21}& & & {a}_{22}\end{array}\right]

the scalar multiple cAcA is

cA=c[a11a12a21a22] =[ca11ca12ca21ca22]\begin{array}{l}cA=c\left[\begin{array}{ccc}{a}_{11}& & {a}_{12}\\ {a}_{21}& & {a}_{22}\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{ccc}c{a}_{11}& & c{a}_{12}\\ c{a}_{21}& & c{a}_{22}\end{array}\right]\hfill \end{array}

Scalar multiplication is distributive. For the matrices A,BA,B, and CC with scalars aa and bb,

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

Example: Multiplying the Matrix by a Scalar

Multiply matrix AA by the scalar 3.

A=[8154]A=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right]

Answer: Multiply each entry in AA by the scalar 3.

3A=3[8154]=[3⋅83⋅13⋅53⋅4]=[2431512]\begin{array}{l}3A=3\left[\begin{array}{rr}\hfill 8& \hfill 1\\ \hfill 5& \hfill 4\end{array}\right]\hfill \\ = \left[\begin{array}{rr}\hfill 3\cdot 8& \hfill 3\cdot 1\\ \hfill 3\cdot 5& \hfill 3\cdot 4\end{array}\right]\hfill \\ = \left[\begin{array}{rr}\hfill 24& \hfill 3\\ \hfill 15& \hfill 12\end{array}\right]\hfill \end{array}

Try It

Given matrix B,B,\text{} find −2B-2B where

B=[4132]B=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right]

Answer: −2B=[−8−2−6−4]-2B=\left[\begin{array}{cc}-8& -2\\ -6& -4\end{array}\right]

Example: Finding the Sum of Scalar Multiples

Find the sum 3A+2B3A+2B.

A=[1−200−1243−6] and B=[−1210−3201−4]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]

Answer: First, find 3A,3A,\text{} then 2B2B.

3A=[3⋅13(−2)3⋅03⋅03(−1)3⋅23⋅43⋅33(−6)]=[3−600−36129−18]\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ 3A=\left[\begin{array}{lll}3\cdot 1\hfill & 3\left(-2\right)\hfill & 3\cdot 0\hfill \\ 3\cdot 0\hfill & 3\left(-1\right)\hfill & 3\cdot 2\hfill \\ 3\cdot 4\hfill & 3\cdot 3\hfill & 3\left(-6\right)\hfill \end{array}\right]\hfill \end{array}\hfill \\ =\left[\begin{array}{rrr}\hfill 3& \hfill -6& \hfill 0\\ \hfill 0& \hfill -3& \hfill 6\\ \hfill 12& \hfill 9& \hfill -18\end{array}\right]\hfill \end{array} 2B=[2(−1)2⋅22⋅12⋅02(−3)2⋅22⋅02⋅12(−4)]=[−2420−6402−8]\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ 2B=\left[\begin{array}{lll}2\left(-1\right)\hfill & 2\cdot 2\hfill & 2\cdot 1\hfill \\ 2\cdot 0\hfill & 2\left(-3\right)\hfill & 2\cdot 2\hfill \\ 2\cdot 0\hfill & 2\cdot 1\hfill & 2\left(-4\right)\hfill \end{array}\right]\hfill \end{array}\hfill \\ =\left[\begin{array}{rrr}\hfill -2& \hfill 4& \hfill 2\\ \hfill 0& \hfill -6& \hfill 4\\ \hfill 0& \hfill 2& \hfill -8\end{array}\right]\hfill \end{array}

Now, add 3A+2B3A+2B.

3A+2B=[3−600−36129−18]+[−2420−6402−8] =[3−2−6+40+20+0−3−66+412+09+2−18−8] =[1−220−9101211−26]\begin{array}{l}\hfill \\ \hfill \\ 3A+2B=\left[\begin{array}{rrr}\hfill 3& \hfill -6& \hfill 0\\ \hfill 0& \hfill -3& \hfill 6\\ \hfill 12& \hfill 9& \hfill -18\end{array}\right]+\left[\begin{array}{rrr}\hfill -2& \hfill 4& \hfill 2\\ \hfill 0& \hfill -6& \hfill 4\\ \hfill 0& \hfill 2& \hfill -8\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 3 - 2& \hfill -6+4& \hfill 0+2\\ \hfill 0+0& \hfill -3 - 6& \hfill 6+4\\ \hfill 12+0& \hfill 9+2& \hfill -18 - 8\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 1& \hfill -2& \hfill 2\\ \hfill 0& \hfill -9& \hfill 10\\ \hfill 12& \hfill 11& \hfill -26\end{array}\right]\hfill \end{array}

 

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 AA is an  m × r \text{ }m\text{ }\times \text{ }r\text{ } matrix and BB is an  r × n \text{ }r\text{ }\times \text{ }n\text{ } matrix, then the product matrix ABAB is an  m × n \text{ }m\text{ }\times \text{ }n\text{ } matrix. For example, the product ABAB is possible because the number of columns in AA is the same as the number of rows in BB. 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 AA with entries of BB according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers. To obtain the entries in row ii of AB,AB,\text{} we multiply the entries in row ii of AA by column jj in BB and add. For example, given matrices AA and B,B,\text{} where the dimensions of AA are 2 × 32\text{ }\times \text{ }3 and the dimensions of BB are 3 × 3,3\text{ }\times \text{ }3,\text{} the product of ABAB will be a 2 × 32\text{ }\times \text{ }3 matrix.

A=[a11a12a13a21a22a23] and B=[b11b12b13b21b22b23b31b32b33]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]

Multiply and add as follows to obtain the first entry of the product matrix ABAB.
  1. To obtain the entry in row 1, column 1 of AB,AB,\text{} multiply the first row in AA by the first column in BB, and add.
    [a11a12a13]⋅[b11b21b31]=a11⋅b11+a12⋅b21+a13⋅b31\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}
  2. To obtain the entry in row 1, column 2 of AB,AB,\text{} multiply the first row of AA by the second column in BB, and add.
    [a11a12a13]⋅[b12b22b32]=a11⋅b12+a12⋅b22+a13⋅b32\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}
  3. To obtain the entry in row 1, column 3 of AB,AB,\text{} multiply the first row of AA by the third column in BB, and add.
    [a11a12a13]⋅[b13b23b33]=a11⋅b13+a12⋅b23+a13⋅b33\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}
We proceed the same way to obtain the second row of ABAB. In other words, row 2 of AA times column 1 of BB; row 2 of AA times column 2 of BB; row 2 of AA times column 3 of BB. When complete, the product matrix will be

AB=[a11⋅b11+a12⋅b21+a13⋅b31a21⋅b11+a22⋅b21+a23⋅b31a11⋅b12+a12⋅b22+a13⋅b32a21⋅b12+a22⋅b22+a23⋅b32a11⋅b13+a12⋅b23+a13⋅b33a21⋅b13+a22⋅b23+a23⋅b33]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]

A General Note: Properties of Matrix Multiplication

For the matrices A,B,A,B,\text{} and CC the following properties hold.
  • Matrix multiplication is associative:
    (AB)C=A(BC)\left(AB\right)C=A\left(BC\right)
  • Matrix multiplication is distributive:
    C(A+B)=CA+CB,(A+B)C=AC+BC.\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}
Note that matrix multiplication is not commutative.

Example: Multiplying Two Matrices

Multiply matrix AA and matrix BB.

A=[1234] and B=[5678]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]

Answer: First, we check the dimensions of the matrices. Matrix AA has dimensions 2×22\times 2 and matrix BB has dimensions 2×22\times 2. The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions 2×22\times 2. We perform the operations outlined previously. The first column of the product of A and B is defined as the result of matrix -vector multiplication of A and the first column of B. Column two of the product of A and B is defined as the result of the matrix-vector multiplication of A and the second column of B.

 

Example: Multiplying Two Matrices

Given AA and B:B:
  1. Find ABAB.
  2. Find BABA.

A=[−123405] and B=[5−42−103]A=\left[\begin{array}{l}\begin{array}{ccc}-1& 2& 3\end{array}\hfill \\ \begin{array}{ccc}4& 0& 5\end{array}\hfill \end{array}\right]\text{ and }B=\left[\begin{array}{c}5\\ -4\\ 2\end{array}\begin{array}{c}-1\\ 0\\ 3\end{array}\right]

Answer:

  1. As the dimensions of AA are 2×32\text{}\times \text{}3 and the dimensions of BB are 3×2,3\text{}\times \text{}2,\text{} these matrices can be multiplied together because the number of columns in AA matches the number of rows in BB. The resulting product will be a 2×22\text{}\times \text{}2 matrix, the number of rows in AA by the number of columns in BB.
    AB=[−123405] [5−1−4023] =[−1(5)+2(−4)+3(2)−1(−1)+2(0)+3(3)4(5)+0(−4)+5(2)4(−1)+0(0)+5(3)] =[−7103011]\begin{array}{l}\hfill \\ AB=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -1\left(5\right)+2\left(-4\right)+3\left(2\right)& \hfill -1\left(-1\right)+2\left(0\right)+3\left(3\right)\\ \hfill 4\left(5\right)+0\left(-4\right)+5\left(2\right)& \hfill 4\left(-1\right)+0\left(0\right)+5\left(3\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -7& \hfill 10\\ \hfill 30& \hfill 11\end{array}\right]\hfill \end{array}
  2. The dimensions of BB are 3×23\times 2 and the dimensions of AA are 2×32\times 3. The inner dimensions match so the product is defined and will be a 3×33\times 3 matrix.
    BA=[5−1−4023] [−123405] =[5(−1)+−1(4)5(2)+−1(0)5(3)+−1(5)−4(−1)+0(4)−4(2)+0(0)−4(3)+0(5)2(−1)+3(4)2(2)+3(0)2(3)+3(5)] =[−910104−8−1210421]\begin{array}{l}\hfill \\ BA=\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 5\left(-1\right)+-1\left(4\right)& \hfill 5\left(2\right)+-1\left(0\right)& \hfill 5\left(3\right)+-1\left(5\right)\\ \hfill -4\left(-1\right)+0\left(4\right)& \hfill -4\left(2\right)+0\left(0\right)& \hfill -4\left(3\right)+0\left(5\right)\\ \hfill 2\left(-1\right)+3\left(4\right)& \hfill 2\left(2\right)+3\left(0\right)& \hfill 2\left(3\right)+3\left(5\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill -9& \hfill 10& \hfill 10\\ \hfill 4& \hfill -8& \hfill -12\\ \hfill 10& \hfill 4& \hfill 21\end{array}\right]\hfill \end{array}

Analysis of the Solution

Notice that the products ABAB and BABA are not equal.

AB=[−7103011]≠[−910104−8−1210421]=BAAB=\left[\begin{array}{cc}-7& 10\\ 30& 11\end{array}\right]\ne \left[\begin{array}{ccc}-9& 10& 10\\ 4& -8& -12\\ 10& 4& 21\end{array}\right]=BA

This illustrates the fact that matrix multiplication is not commutative.

Q & A

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

Yes, consider a matrix A with dimension 3×43\times 4 and matrix B with dimension 4×24\times 2. 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

E=[63014102420]E=\left[\begin{array}{c}6\\ 30\\ 14\end{array}\begin{array}{c}10\\ 24\\ 20\end{array}\right]

The cost matrix is written as

C=[3001030]C=\left[\begin{array}{ccc}300& 10& 30\end{array}\right] We perform matrix multiplication to obtain costs for the equipment. CE=[3001030]⋅[61030241420] =[300(6)+10(30)+30(14)300(10)+10(24)+30(20)] =[2,5203,840]\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 \\ \text{ }=\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 \\ \text{ }=\left[\begin{array}{rr}\hfill 2,520& \hfill 3,840\end{array}\right]\hfill \end{array}

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
    [A],[B],[C],..\left[A\right],\left[B\right],\left[C\right],..
  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 AB−CAB-C given

A=[−15253241−7−281034−2],B=[4521−37−2452196−48−31],and C=[−100−89−9825−5674−6742−75]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].

Answer: On the matrix page of the calculator, we enter matrix AA above as the matrix variable [A]\left[A\right], matrix BB above as the matrix variable [B]\left[B\right], and matrix CC above as the matrix variable [C]\left[C\right]. On the home screen of the calculator, we type in the problem and call up each matrix variable as needed.

[A]×[B]−[C]\left[A\right]\times \left[B\right]-\left[C\right] The calculator gives us the following matrix. [−983−4621361,8201,897−856−3112,032413]\left[\begin{array}{rrr}\hfill -983& \hfill -462& \hfill 136\\ \hfill 1,820& \hfill 1,897& \hfill -856\\ \hfill -311& \hfill 2,032& \hfill 413\end{array}\right]

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