Recall that the n×n
having ones on its main diagonal and zeros elsewhere. It is not difficult to deduce directly from the definition of the matrix product that I acts like an identity for matrix multiplication:
if the sizes of A and B are such that the products AI and IB are defined. It is, nevertheless, instructive to derive the identities in (2) formally from the two basic facts about matrix multiplication that we state below. First, recall that the notation
expresses the m×n
Fact 1 Ax in terms of columns of A
If A=[a1a2⋯an]
The reason is that when each row vector of A is multiplied by the column vector x, its jth element is multiplied by xj
Fact 2 AB in terms of columns of B
If A is an m×n
That is, the j th column of AB is the product of A and the j th column of B. The reason is that the elements of the j th column of AB are obtained by multiplying the individual rows of A by the j th column of B.
The third column of the product AB of the matrices
is
To prove that AI=A,
where the jth column vector of I is the jth basic unit vector
If A=[a1a2⋯an],
Hence Fact 2 gives
that is, AI=A.
If a≠0,
If
then
BA=I
Let
If the matrix B had the property that AB=BA=I,
But upon equating corresponding elements of AB and the 2×2
It is clear that these equations are inconsistent. Thus there can exist no 2×2
Thus the matrix A of Example 2 is invertible, whereas the matrix A of Example 3 is not invertible.
A matrix B such that AB=BA=I
The unique inverse of an invertible matrix A is denoted by A−1.
In the case of a 2×2
Equation (9) gives us the following prescription for writing the inverse of an invertible 2×2
First, interchange the two main diagonal entries.
Then, change the signs of the two off-diagonal elements.
Finally, divide each element of the resulting matrix by ad−bc
You might check that this is how B=A−1
If
then ad−bc=36−30=6≠0,
Arbitrary integral powers of a square matrix A are defined as follows, though in the case of a negative exponent we must assume that A is also invertible. If n is a positive integer, we define
In Problem 28 of Section 3.4, we asked you to verify the laws of exponents
in the case of positive integral exponents, and Problem 31 of this section deals with the case of negative integral exponents. In Problem 29 we ask you to establish parts (a) and (b) of the following theorem.
In mathematics it is frequently important to note the surprises. The surprise in Eq. (11) is the reversal of the natural order of the factors in the right-hand side. You should now be able to show that
In general, any product of invertible matrices of the same size is again invertible, and the inverse of a product of invertible matrices is the product in reverse order of their inverses.
To solve the system
we use the inverse of the coefficient matrix
that we found in Example 4. Then Eq. (13) yields
Thus x1=−9, x2=7
Theorem 2 tells us only how to invert 2×2
We obtain some typical elementary matrices as follows.
The three elementary matrices E1, E2,
Now, suppose that the m×m
Elementary row operations are reversible. That is, to every elementary row operation there corresponds an inverse elementary row operation that cancels its effects (see Figure 3.5.1). It follows that every elementary matrix is invertible. To see why, let E be a given elementary matrix and let E1
Elementary Row Operation | Inverse Operation |
---|---|
(c)Ri |
1cRi |
SWAP(Ri,Rj) |
SWAP(Ri,Rj) |
(c)Ri+Rj |
(−c)Ri+Rj |
Elementary matrices are not ordinarily used for computational purposes; it is simpler to carry out row operations directly than to multiply by elementary matrices. Instead, their principal role is in the proof of the following theorem, which leads in turn to a practical method for inverting matrices.
The proof of Theorem 6 actually tells us how to find the inverse matrix of A. If we invert each side in Eq. (15) (remembering to reverse the order on the right), we get
Because each left multiplication by an elementary matrix is equivalent to performing the corresponding row operation, we see by comparison of Eqs. (14) and (16) that the same sequence of elementary row operations that transforms A into I also transforms I into A−1
As a practical matter, it generally is more convenient to carry out the two reductions—from A to I and from I to A−1
Find the inverse of the 3×3
We want to reduce A to the 3×3
We now apply the following sequence of elementary row operations to this 3×6
Now that we have reduced the left half of the 3×6
Ordinarily, we do not know in advance whether a given square matrix is invertible or not. To find out, we attempt to carry out the reduction process illustrated in Example 7. If we succeed in reducing A to I, then A is invertible and thereby we find A−1.
In certain applications, one needs to solve a system Ax=b
By Fact 2 at the beginning of this section,
So the k equations in (17) are equivalent to the single matrix equation
where
If A is invertible and we know A−1,
Note that this equation is a generalization of Eq. (13) in Theorem 4. If k=1,
Find a 3×4
The coefficient matrix is the matrix A whose inverse we found in Example 7, so Eq. (19) yields
and hence
By looking at the third columns of B and X, for instance, we see that the solution of
is x1=14, x2=8, x3=−39
Theorem 6 tells us that the square matrix A is invertible if and only if it is row equivalent to the identity matrix I, and Theorem 4 in Section 3.3 implies that the latter is true if and only if the system Ax=0
The proof of Theorem 7 is a bit long, but it summarizes most of the basic theory of Chapter 1 and is therefore well worth the effort. Indeed, this theorem is one of the central theorems of elementary linear algebra, and we will need to refer to it repeatedly in subsequent chapters.
In Problems 1–8, first apply the formulas in (9) to find A−1.
A=[3243],b=[56]
A=[3725],b=[−13]
A=[6756],b=[2−3]
A=[512717],b=[55]
A=[3254],b=[56]
A=[4736],b=[105]
A=[7957],b=[32]
A=[815510],b=[73]
In Problems 9–22, use the method of Example 7 to find the inverse A−1
[5645]
[5746]
[151250271]
[1322833106]
[273132379]
[356243235]
[11514133212]
[1−3−3−1122−3−3]
[1−30−12−10−22]
[1−223011−12]
[143145251]
[20−1103111]
[0010100001203001]
[4011313101203241]
In Problems 23–28, use the method of Example 8 to find a matrix X such that AX=B
A=[4354],B=[13−5−1−25]
A=[7687],B=[20405−3]
A=[141283274],B=[103022−110]
A=[15121−2172],B=[201030102]
A=[1−23217227],B=[001101011010]
A=[653532342],B=[2102−13501105]
Verify parts (a) and (b) of Theorem 3.
Problems 30 through 37 explore the properties of matrix inverses.
Suppose that A, B, and C are invertible matrices of the same size. Show that the product ABC is invertible and that (ABC)−1=C−1B−1A−1
Suppose that A is an invertible matrix and that r and s are negative integers. Verify that ArAs=Ar+s
Prove that if A is an invertible matrix and AB=AC,
Let A be an n×n
Show that a diagonal matrix is invertible if and only if each diagonal element is nonzero. In this case, state concisely how the inverse matrix is obtained.
Let A be an n×n
Show that A=[abcd]
Suppose that ad−bc≠0
Problems 38 through 40 explore the effect of multiplying by an elementary matrix.
Let E be the elementary matrix E1
Let E be the elementary matrix E2
Let E be the elementary matrix E3
Problems 41 and 42 complete the proof of Eq. (2).
Show that the ith row of the product AB is AiB,
Apply the result of Problem 41 to show that if B is an m×n
Suppose that the matrices A and B are row equivalent. Use Theorem 5 to prove that B=GA,
Show that every invertible matrix is a product of elementary matrices.
Extract from the proof of Theorem 7 a self-contained proof of the following fact: If A and B are square matrices such that AB=I,
Deduce from the result of Problem 45 that if A and B are square matrices whose product AB is invertible, then A and B are themselves invertible.
Linear systems with more than two or three equations are most frequently solved with the aid of calculators or computers. If an n×nwith(linalg): inverse(A)
, the Mathematica command Inverse[A]
, or the Matlab command inv(A)
. Consequently, the solution vector x is calculated by the Maple command
with(linalg): x := multiply(inverse(A),b);
or the Mathematica command
x = Inverse[A].b
or the Matlab command
x = inv(A)*b
Figure 3.5.2 illustrates a similar calculator solution of the linear system
for the solution x1=59, x2=13, x3=17, x4=47.
A = ((3, −2, 7, 5), (2, 4, −1, 6), (5, 1, 7, −3),
(4, −6, −8, 9)),
b = (505, 435, 286, 445),
inv(A).b
Whereas the preceding commands illustrate the handy use of conveniently available inverse matrices to solve linear systems, it might be mentioned that modern computer systems employ direct methods—involving Gaussian elimination and still more sophisticated techniques—that are more efficient and numerically reliable to solve a linear system Ax=b
Use an available calculator or computer system to solve the linear systems in Problems 1–6 of the 3.3 Application. The applied problems below are elementary in character—resembling the “word problems” of high school algebra—but might illustrate the practical advantages of automated solutions.
You are walking down the street minding your own business when you spot a small but heavy leather bag lying on the sidewalk. It turns out to contain U.S. Mint American Eagle gold coins of the following types:
One-half ounce gold coins that sell for $285 each,
One-quarter ounce gold coins that sell for $150 each, and
One-tenth ounce gold coins that sell for $70 each.
A bank receipt found in the bag certifies that it contains 258 such coins with a total weight of 67 ounces and a total value of exactly $40,145. How many coins of each type are there?
Now you really strike it rich! You find a bag containing one-ounce U.S. American Eagle gold coins valued at $550 each, together with half-ounce and quarter-ounce coins valued as in the preceding problem. If this bag contains a total of 365 coins with a total weight of exactly 11 pounds and a total value of $100,130, how many gold coins of each type are there?
A commercial customer orders 81 gallons of paint that contains equal amounts of red paint, green paint, and blue paint—and, hence, could be prepared by mixing 27 gallons of each. However, the store wishes to prepare this order by mixing three types of paint that are already available in large quantity:
a reddish paint that is a mixture of 50% red, 25% green, and 25% blue paint;
a greenish paint that is 12.5% red, 75% green, and 12.5% blue paint; and
a bluish paint that is 20% red, 20% green, and 60% blue paint.
How many gallons of each are needed to prepare the customer’s order?
Now the paint store receives a really big order—for 244 gallons of paint that is 1/2 red paint, 1/4 green paint, and 1/4 blue paint. The store has three already-mixed types of paint available in large quantity—the greenish paint and the bluish paint of the preceding problem, plus a reddish paint that is 2/3 red paint, 1/6 green paint, and 1/6 blue paint. How many gallons of each must be mixed in order to fill this order?
A tour busload of 45 people attended two Florida theme parks on successive days. On Day 1 the entrance fee was $15 per adult, $8 per child, $12 per senior citizen and the total charge was $558. On Day 2 the entrance fee was $20 per adult, $12 per child, $17 per senior citizen and the total charge was $771. How many adults, children, and senior citizens were on this tour bus?
For some crazy reason, the lunches bought at the first theme park were totaled separately for the adults, children, and seniors. The adults ordered 34 hot dogs, 15 French fries, and 24 soft drinks for a total bill of $70.85. The children ordered 20 hot dogs, 14 French fries, and 15 soft drinks for a total bill of $46.65. The senior citizens ordered 11 hot dogs, 10 French fries, and 12 soft drinks for a total bill of $30.05. What were the prices of a hot dog, an order of French fries, and a soft drink?
A fast-food restaurant sells four types of sandwiches—hamburgers, cheeseburgers, roast beef, and chicken—and has four cash registers. At the end of each day, each cash register tallies the number of each type of sandwich sold, and the total sandwich receipts for the day. The four cash register operators work at different speeds, and one day’s totals were as follows:
Hamburgers | Cheeseburgers | Roast Beef | Chicken | Receipts | |
---|---|---|---|---|---|
Register 1 | 37 | 44 | 17 | 23 | $232.99 |
Register 2 | 28 | 35 | 13 | 17 | $178.97 |
Register 3 | 32 | 39 | 19 | 21 | $215.99 |
Register 4 | 47 | 51 | 25 | 29 | $294.38 |
What was the price of each of the four types of sandwiches?
The fast-food restaurant of the preceding problem adds a ham sandwich to its menu and, because of increased business, it also adds a fifth cash register and reduces prices. After this expansion, one day’s totals were as follows:
Hamburgers | Cheeseburgers | Roast Beef | Chicken | Ham | Receipts | |
---|---|---|---|---|---|---|
Register 1 | 41 | 49 | 22 | 26 | 19 | $292.79 |
Register 2 | 34 | 39 | 18 | 20 | 16 | $236.73 |
Register 3 | 36 | 43 | 23 | 24 | 18 | $270.70 |
Register 4 | 49 | 52 | 26 | 31 | 24 | $340.19 |
Register 5 | 52 | 55 | 24 | 28 | 25 | $341.64 |
What were the new prices of the five types of sandwiches?
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