1 Matrix multiplication (Section 6.2 , page 610)
2 Equality of matrices (Section 6.2 , page 606)
3 Gauss–Jordan elimination procedure (Section 6.1 , page 597)
4 Leontieff input–output model (Section 6.1 , page 599)
Imagine you are a secret agent who is looking for spies in different parts of the world. Periodically, you communicate with your headquarters via your laptop or by satellite. You expect that your communications will be intercepted. So before you broadcast any message, you must write it using a secret code. A message written with a secret code is called a cryptogram.
First, you might replace each letter in the message with a number and send the message as a sequence of numbers. The problem with this approach is that it is rather easy to crack such a code. For example, if letters such as e are always represented by the same symbol, then a code breaker (hacker) can guess which symbol represents each letter and eventually translate your message. In Example 8, we show you a better way to encode and decode a message using matrix multiplication.
1 Verify the multiplicative inverse of a matrix.
In arithmetic, you know that if a is a nonzero real number, then a has a unique reciprocal, denoted by 1a, or a−1. The number a−1 is called the multiplicative inverse of a, and
Similarly, certain square matrices have multiplicative inverses, with an identity matrix acting as “1.”
Notice that if B is the inverse of A, then A is the inverse of B.
Recall that in general (even for square matrices), AB≠BA. However, if A and B are square matrices with AB=I, then it can be shown that BA=I. Therefore, to determine whether A and B are inverses, you need only check that either AB=I or BA=I.
Show that B is the inverse of A.
You need to verify that AB=I or BA=I.
Because AB=I, it follows that B=A−1.
Show that B is the inverse of A.
In working with real numbers, you know that if a=0, then a−1 does not exist. So it should not come as a surprise that if A is a square zero matrix, then A−1 does not exist. Surprisingly, there are also nonzero matrices that do not have inverses.
Show that the matrix A does not have an inverse.
Suppose A has an inverse B, where
Then you have
Because these two matrices are equal, you must have
Because 0=1 is a false statement, the matrix A does not have an inverse.
Show that the matrix A=[3131] does not have an inverse.
2 Find the inverse of a matrix.
It can be shown that if a square matrix A has an inverse, then the inverse is unique. This means that a square matrix cannot have more than one inverse. If a matrix A has an inverse, then A is called invertible or nonsingular. A square matrix that does not have an inverse is called a singular matrix. The following discussion derives a technique for finding the inverse of an invertible matrix.
Suppose we want to find the inverse of the matrix A=[1235]. We first let B=[xyzw] be the inverse of the matrix A. Then
Equating corresponding entries in the last equation yields two systems of linear equations.
Let’s solve the two systems by Gauss–Jordan elimination.
System of Equations | Matrix Form |
---|---|
{x+2z=13x+5z=0 | [121350] |
{y+2w=03y+5w=1 | [120351] |
Because the coefficient matrix for both systems of equations is the original matrix
the calculations involved in the Gauss–Jordan elimination on the coefficient matrix A will be the same for both systems. Therefore, you can solve both systems at once by considering the “doubly augmented” matrix
Note that the matrix C is formed by adjoining the identity matrix I to matrix A:
Now apply Gauss–Jordan elimination to the matrix C.
Converting the last matrix back into two systems of equations, we have
Similarly,
We conclude that x=−5, y=2, z=3, and w=−1.
Because B=[xyzw], it follows that
You should check that AB=I.
We have shown that to find the inverse of an invertible matrix A, we transform the matrix [A|I] by a sequence of elementary row operations into [I|B], where B=A−1.
Find the inverse (if it exists) of the matrix A=[123257246].
Step 1 Start by adjoining the identity matrix I=[100010001] to the matrix A to form the augmented matrix [A|I].
Step 2 Perform row operations on [A|I] to transform it to a matrix of the form [I|B].
Because the third row of matrix C consists of zeros, it is impossible to transform A into I by row operations. Hence, A is not invertible.
Find the inverse (if it exists) of the matrix A=[14−2−11237−6].
Find the inverse (if it exists) of the matrix A=[110031233].
Step 1 Start with the matrix [A|I]:
Step 2 Use row operations on [A|I]. Then [A|I]→
Hence, A is invertible and
Step 3 You should verify that AA−1=I.
Find the inverse (if it exists) of the matrix A=[123−23145−2].
3 Find the inverse of a 2×2 matrix.
You can quickly determine whether any 2×2 matrix is invertible and, if so, what its inverse is.
Find the inverse (if it exists) of each matrix.
A=[5243]
B=[4623]
For the matrix A, a=5, b=2, c=4, and d=3. Here ad−bc=(5)(3)−(2)(4)=15−8=7≠0; so A is invertible and
You should verify that AA−1=I.
For the matrix B, a=4, b=6, c=2, and d=3; so
Therefore, the matrix B does not have an inverse.
Find the inverse (if it exists) of each matrix.
A=[8241]
B=[8−231]
4 Use matrix inverses to solve systems of linear equations.
Matrix multiplication can be used to write a system of linear equations in matrix form.
System of Equations | Matrix Form |
---|---|
{3x−2y=44x−3y=5 | [3−24−3][xy]=[45] |
{2x1+4x2−x3=93x1+x2+2x3=7x1+3x2−3x3=4 | [24−131213−3][x1x2x3]=[974] |
Solving a system of linear equations amounts to solving a corresponding matrix equation of the form
where A is the coefficient matrix of the system, X is the column matrix containing the variables, and B is the column matrix of the system’s constants.
If A is a square matrix and A is invertible, then the matrix equation AX=B has a unique solution, X=A−1B, as the following derivation shows.
Use a matrix inverse to solve the linear system: {x+y=43y+z=72x+3y+3z=21.
Write the linear system in matrix form.
We need to solve the matrix equation AX=B for X. Because the matrix A is invertible (see Example 4), the system has a unique solution X=A−1B. Use the following:
The solution set is {(3, 1, 4)}, which you can check in the original system.
Use a matrix inverse to solve the linear system.
5 Use matrix inverses in applied problems.
We can use the inverse of a matrix to analyze input–output models that we studied in Section 6.1.
Suppose a simplified economy depends on two products: energy (E) and food (F).
To produce 1 unit of E requires 14 unit of E and 12 unit of F. To produce 1 unit of F requires 13 unit of E and 14 unit of F. Then the interindustry consumption is given by the following matrix.
The matrix A is the interindustry technology input–output matrix, or simply the technology matrix, of the system. If the system is producing x1 units of energy and x2 units of food, then the column matrix X=[x1x2] is called the gross production matrix.
Consequently,
represents the interindustry consumptions. If the column matrix D=[d1d2] represents consumer demand, then
So if (I−A) is invertible, then the gross production matrix is X=(I−A)−1D.
In the preceding discussion, suppose the consumer demand for energy is 1000 units and for food is 3000 units. Find the level of production (X) that will meet interindustry and consumer demand.
For the matrix A, we have
Use the formula for the inverse of a 2×2 matrix.
The matrix D=[10003000]. Therefore, the gross production matrix X is
To meet the consumer demand for 1000 units of energy and 3000 units of food, the energy produced must be 108,00019 units and food production must be 124,00019 units.
In Example 7 , find the level of production (X) that will meet both the interindustry and consumer demand if consumer demand for energy is 800 units and consumer demand for food is 3400 units.
To encode or decode a message, first associate each letter in the message with a number, in the obvious manner. The blank corresponds to 0.
For example, the message
JOHNSON IS IN DANGER
becomes
This message is easy to decode, so we will make the coding more complicated. First, partition these 20 numbers into groups of three, enclosed in brackets, and insert zeros at the end of the last bracket if necessary.
Let’s write these seven groups of three numbers as the seven columns of a 3×7 matrix.
Now choose any invertible 3×3 matrix A, say,
We can use the matrix A to convert the message into code and then use A−1 to decode the message, as in the next example. The matrix A is called the coding matrix.
Encode and decode the message
JOHNSON IS IN DANGER
by using the matrix A given above.
Step 1 Express the message numerically and partition the numbers into groups of three.
Step 2 Write each group of three numbers in Step 1 as a column of a matrix M.
Step 3 Select an invertible 3×3 matrix A, such as
Step 4 Multiply the matrices in Steps 2 and 3 to form AM.
The coded message sent will be the numbers from column 1 of AM, followed by the numbers from column 2, and so on.
Step 5 To decode the message in Step 4, write the numbers received in the message in groups of three and as columns of the matrix AM. Find the decoding matrix A−1 of the coding matrix A. You can verify that
Step 6 Compute the message matrix M.
Step 7 Finally, convert the numerical message from the matrix M in Step 6 back into English.
JOHNSON IS IN DANGER
Rework Example 8 , replacing the message JOHNSON IS IN DANGER with the message JACK IS NOW SAFE.
For n×n matrices A and B, if AB=I=BA, then B is called the of A.
An n×n matrix A is invertible if there is a matrix B such that _=I.
To find the inverse of an invertible matrix A, we transform [A | I] by a sequence of row operations into [I | B], where B=_.
A=[abcd] is invertible if and only if
True or False. Every square matrix is invertible.
True or False. A=[81648] is invertible.
True or False. The inverse of A=[abcd] is 1ad−bc[d−b−ca] if ad−bc≠0.
True or False. If A is a square, invertible matrix and AX=B, then X=A−1B.
In Exercises 9–18, determine whether B is the inverse of A by computing AB and BA.
A=[1213], B=[3−2−11]
A=[3243], B=[3−2−43]
A=[3214], B=[25−15−110310]
A=[2−34−3], B=[−1212−2313]
A=[10−1−111], B=[211111]
A=[−2130−11120], B=18[−2641−32152]
A=[1−21−86−25−31], B=12[012123311]
A=[231123312], B=118[1−5771−5−571]
A=[111123−11−1], B=14[5−2−1202−32−1]
18. A=[1−1001−1101], B=12[111−111−1−11]
In Exercises 19–28, find A−1 (if it exists) by forming the matrix [A | I] and using row operations to obtain [I | B], where B=A−1.
A=[2013]
A=[4310]
A=[2436]
A=[9664]
A=[164023012]
A=[20−1510013]
A=[231241372]
A=[11543−5110]
A=[3−342−340−11]
A=[12−2−1300−21]
In Exercises 29–34, find the inverse (if it exists) of A=[abcd] by using the formula A−1=1ad−bc[d−b−ca]. Check that A−1A=I.
A=[1032]
A=[3456]
A=[2−3−35]
A=[3−42−2]
A=[a−bb−a],a2≠b2
A=[2−11−1]
In Exercises 35–38, write the given linear system of equations as a matrix equation AX=B.
{2x+3y=−9x−3y=13
{5x−4y=74x−3y=5
{3x+2y+z=82x+y+3z=7x+3y+2z=9
{x+3y+z=4x−5y+2z=73x+y−4z=−9
In Exercises 39–42, write each matrix equation as a linear system of equations.
[1−221][xy]=[05]
[233−1][x1x2]=[011]
[23157−1430][x1x2x3]=[−155]
[3−23504270][rst]=[43−8]
Let A=[125238−112]. Show that A−1=[2−1−112−7−2−531].
In Exercises 44–46, use the result of Exercise 43 to solve the system of equations.
{x+2y+5z=42x+3y+8z=6−x+y+2z=3
{x1+2x2+5x3=12x1+3x2+8x3=3−x1+x2+2x3=−3
{x+2y+5z=−42x+3y+8z=−6−x+y+2z=−52
Find the inverse of the matrix A=[111123149].
Use the result of part (a) to solve the linear system.
Find the inverse of the matrix A=[24−131213−3].
Use the result of part (a) to solve the linear system.
In Exercises 49–54, solve each linear system by using an inverse matrix.
{3x+7y=11−5x+4y=13
{x−7y=32x+3y=23
{x+y+2z=7x−y−3z=−62x+3y+z=4
{x+y+z=62x−3y+3z=53x−2y−z=−4
{2x+2y+3z=75x+3y+5z=33x+5y+z=−5
{9x+7y+4z=126x+5y+4z=54x+3y+z=7
In Exercises 55–58, use the method presented in this section to solve the problem.
Investment. At the beginning of 2016, Liz decided to invest $90,000, splitting it three different ways. Part of the money she invested in a treasury bill that yields 3% annual interest, she invested another part in bonds with an annual yield 7%, and the rest she invested in a mutual fund that estimated an annual interest rate of 8%. Liz expected her net income for 2016 to be $5920. Unfortunately, the mutual fund lost 11% of its assets during the year so that Liz received only $980 income from all three investments combined. How much money did Liz put in each investment?
Sandwich sales. The Fresh Sandwich specializes in chicken, fish, and ham sandwiches. In April, May, and June 2016, they sold the following quantities.
Sales Volume in Hundreds | |||
---|---|---|---|
April | May | June | |
Chicken | 8 | 7 | 4 |
Fish | 11 | 8 | 6 |
Ham | 6 | 7 | 3 |
Monthly revenue, in hundreds, from the sales of these sandwiches | $164 | $141 | $86 |
Find the selling price of each of the three types of sandwiches.
Euler’s formula. A plane graph connects points called vertices with nonintersecting lines called edges and divides the plane into regions. The accompanying figure has seven vertices, nine edges, and four regions. Euler’s theorem states that in a plane graph,
where
V=the number of vertices,E=the number of edges, andR=the number of regions.
In a certain plane graph, twice the number of edges is three times the number of vertices and twice the number of regions is one less than the number of edges. Find the number of vertices, edges, and regions for this graph. Sketch a plane graph that satisfies the conditions of this exercise.
Projectile. On a planet, the height h of a projectile fired up from an initial height h0 with initial vertical velocity of v0 ft /sec after t seconds is given by h=12at2+v0t+h0, where a is the gravitational constant. The following measurements were recorded.
Time t in seconds | 1 | 2 | 3 |
Height h in feet | 100 | 92 | 76 |
Find the initial height.
Find the height of the projectile after 2.5 seconds.
In Exercises 59–62, technology matrix A, representing interindustry demand, and matrix D, representing consumer demand for the sectors in a two- or three-sector economy, are given. Use the matrix equation X=(I−A)−1D to find the production level X that will satisfy both demands.
Two-sector economy. In a two-sector economy,
Two-sector economy. In a two-sector economy,
Three-sector economy. In a three-sector economy,
Three-sector economy. In a three-sector economy,
In Exercises 63 and 64, use a 2×2 invertible coding matrix A of your choice. Remember to partition the numerical message into groups of two. (a) Write a cryptogram for each message. (b) Check by decoding.
Coding and decoding. CANNOT FIND COLOMBO.
Coding and decoding. FOUND HEAD OF TERRORIST NETWORK.
In Exercises 65 and 66, use a 3×3 invertible coding matrix A of your choice.
Code and then decode the message of Exercise 63.
Code and then decode the message of Exercise 64.
Suppose A is an invertible square matrix.
Is A−1 invertible? Why or why not?
If your answer to part (a) is yes, then what is (A−1)−1?
Let A−1=[213−4]. Find A.
Let A be a square matrix. Show that if A2 is invertible, then A must be invertible. [Hint: A2B=A(AB).]
Show that if A is invertible, then A2 is invertible.
Show that if A and B are n×n invertible matrices, then AB is invertible and (AB)−1=B−1A−1. (Note the order.)
Verify the result of Exercise 71 for
Let A, B, and C be n×n matrices such that AB=AC. Show that if A is invertible, then B=C.
Is the result of Exercise 73 true if A is not invertible?
Let A and B be n×n matrices such that AB=A. Show that if A is invertible, then B=I.
Give an example to show that the result of Exercise 75 is not true if A is not invertible.
Let A=[3423].
Show that the matrix A satisfies the equation
Use the equation in part (a) to show that A−1=6I−A.
Use the equation in part (b) to find A−1.
Let A=[2−11−12−11−12].
Show that A satisfies the equation A3−6A2+9A−4I=0.
Use the equation in part (a) to show that
[Hint: Consider the product 14 (A2−6A+9I)A.]
Use part (b) to find A−1.
In Exercises 79 and 80, use your graphing utility to find the inverse of the coefficient matrix. Then use the inverse to solve each linear system.
{x−y−z−w=−4x+y+z−w=22x+y+z−w=3x−y+z−w=−2
{x+y+z−w=1x−y−z+w=1x−y+z+3w=1x+y+z+w=3
Suppose A and B are 2×2 matrices and B−1=[1234].
Find A.
If AB=[−1302].
If BA=[−1302].
If BAB−1=[1112].
If B−1AB=[1112].
Is the matrix [x−12x+2] invertible for all real numbers x?
Is the matrix [x−1−13x−5] invertible for all real numbers x?
True or False. Justify your answers.
If A and B are matrices for which AB=BA, then the formula A2B=BA2 must hold.
If A and B are invertible matrices, then A+B must be invertible.
Give an example of a nonzero 2×2 matrix A such that A2=0. Show that the matrix I+A is invertible and that (I+A)−1=I−A.
Is it true that every nonsingular matrix is a square matrix? Why or why not?
In Exercises 86–89, state whether each expression is equal to 1 or −1.
(−1)0
(−1)10
(−1)17
(−1)6+9
In Exercises 90–93, perform the indicated operation.
[−1, 4][23]
[−1−3][5, 0]
3[715−2]
(−1)[−674−1]
In Exercises 94–97, solve each system of equations.
{2x−3y=16x−y=7
{16x−9y=−510x+18y=−11
{x−y+5z=−63x+3y−z=10x+3y+2z=5
{2x−y+2z=32x+2y−z=0−x+2y+2z=−12