Dr. Neal, WKU

MATH 307 Systems of Equations Let AX = B represent a system of

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m linear equations and

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n unknowns. Exactly one of the following is true:

1. The system AX = B has no solutions. 2. The system AX = B has exactly one solution.

3. The system AX = B has infinitely many solutions. We can determine which of the above cases is true by observing the reduced row echelon form rref (A B) of the augmented matrix A|B . 1. When there are no solutions (i.e., an inconsistent system), then there will be a row in which every element becomes 0 except for a non-zero entry in the augmented column. 2. When there is exactly one solution, then some extraneous rows may become all 0 (when

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m >

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n). But the remainder of A will become the

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n × n identity matrix and the augmented column will become the unique solution for x1 , x2 , . . . , xn . 3. When there are infinitely many solutions, then one or more rows at the bottom of rref (A B) may become entirely 0. However, the last non-zero row will have non-zero coefficients for two or more of the unknowns.

1 0 . . .0 1 . . .. .. .0 0 . . 00 0 . . 0

cd..10

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No Solutions An inconsistent row says

0x1+ . . . + 0xn =1

1 0 . . . 00 1 . . . 0. . . . . .. . . . . .. . . . . .0 0 . . . 10 0 . . . 0

d1d2...dn0

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Exactly One Solution ( x1 = d1 , . . . , xn = dn )

1 0 . . . . .0 1 . . . . .. . . . . . .. . . . . . .0 0 c1 c2 . . cn0 0 . . . . 0

d1d2..dk0

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Infinitely Many Solutions

n equations, n unknowns When we have

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n equations and

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n unknowns, then the matrix of coefficients A is

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n × n . Thus A−1 may exist. When A−1 exists, then the system AX = B has exactly one solution which can be found by the matrix product X = A−1 B (or from rref (A B) ). When A is singular (i.e., A−1 does not exist), then AX = B will have either no solutions or infinitely many solutions. To determine which of these two cases is true, use rref (A B) and look at the last rows as described above. Recall: An

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n × n matrix A has an inverse if and only if det (A) ≠ 0. So if det (A) ≠ 0, then AX = B has exactly one solution. If det (A) = 0, then AX = B has infinite solutions or no solutions.

Dr. Neal, WKU

The Homogeneous System The system of equations AX = 0 is called the homogeneous system:

a11 x1 + a12 x2 + . . .+ a1n xn = 0

a21 x1 + a2 2 x2 + . . .+ a2n xn = 0

. . .

am1 x1 + am 2 x2 + . . .+ am n xn = 0

The homogeneous system always has the trivial solution of X = 0. That is, x1 = 0, x2 = 0 , . . . , xn = 0 is always a solution. Thus, there are only two possibilities:

1. AX = 0 has only the trivial solution of X = 0; or 2. AX = 0 has infinitely many solutions.

(It is not possible for there to be no solution since X = 0 is always a solution.)

To determine which of the two cases for the homogeneous system is true, use rref (A 0) on the augmented matrix A|0. (The inconsistent case cannot happen.) When there are n equations and n unknowns, then we still have two cases for the homogeneous system: 1. If A is invertible (i.e., A−1 exists), then AX = 0 has only the trivial solution of X = 0. 2. If A is singular (i.e., A−1 does not exist), then AX = 0 has infinitely many solutions.

Equivalently,

1. If det (A) ≠ 0, then AX = 0 has only the trivial solution of X = 0.

2. If det (A) = 0, then AX = 0 has infinitely many solutions.

Writing the Form of Infinite Solutions

Example 1. Consider the following rref form of a system of equations:

1 0 0 −90 1 0 80 0 1 20 0 0 0

8−5−30

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This form has infinitely many solutions. For every column that did not result in a leading 1 in the rref form, we need an independent variable. (We generally let the independent variables be r , s , t , u , v as needed.) In this case, the fourth column did not result in a leading 1, so we can let x4 = t , where t can be any real number. Now consider the last row that is not all 0’s: 0 0 1 2 | –3

Dr. Neal, WKU

This last (non-zero) equation says x3 + 2x4 = −3 ; hence, x3 = −3 − 2x4 , or x3 = −3 − 2t , where x4 = t . From the second row, we have x2 + 8x4 = −5 , or . Finally, from the first row we have x1 − 9 x4 = 8 , or . So the infinite solutions are given by

x1 = 8 + 9t x2 = −5 − 8t x3 = −3 − 2t x4 = t , where t is any real number. One specific solution, when t = 0 , is (8, –5, –3, 0).

Example 2. Consider the following reduced row echelon form of a system of equations:

1 0 12 −3 20 1 4 −2 00 0 0 0 00 0 0 0 0

−4−200

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Here we need 3 independent variables, because the second, third, and fourth columns do not result in leading 1’s. Let x3 = r , x4 = s , x5 = t . Then the second row says x2 + 4 x3 − 2 x4 = −2 , or x2 = −2 − 4x3 + 2x4. Thus, x2 = −2 − 4r + 2s . The first row says x1 +12 x3 − 3x4 + 2x5 = − 4 , or x1 = −4 − 12 x3 + 3x4 − 2 x5 . Thus, x1 = −4 − 12r + 3s − 2t . Hence, x1 = −4 −12r + 3s − 2t , x2 = −2 − 4r + 2s , x3 = r , x4 = s , x5 = t , where r , s , and t are any real numbers. Letting (r , s , t ) be (1, 0, 0), then (0, 1, 0), then (0, 0, 1), we have three possible solutions being (–16, –6, 1, 0, 0), then (–1, 0, 0, 1, 0), then (–6, –2, 0, 0, 1). Example 3. Consider the rref of a homogeneous system of 5 equations, 5 unknowns:

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1 3 0 0 00 0 1 4 00 0 0 0 10 0 0 0 00 0 0 0 0

00000

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Here the second and fourth columns do not result in leading 1’s. So let

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x2 = s and

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x4 = t . The first row says

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x1 + 3x2 = 0; thus,

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x1 = −3s. The second row says

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x3 + 4x4 = 0 , so

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x3 = −4t . The third row says

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x5 = 0. Thus the infinite solutions are

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x1 = −3s

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x2 = s

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x3 = −4t

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x4 = t

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x5 = 0, where

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s, t ∈ ℜ . Letting

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(s, t) be (1, 0) , then (0, 1), we have two possible solutions being (–3, 1, 0 , 0, 0) and (0, 0, –4, 1, 0).

Dr. Neal, WKU

We now give some theorems regarding when a system of equations AX = B has a unique solution. Theorem 1. Suppose the homogeneous system AX = 0 has only the trivial solution of X = 0 . If AX = B has a solution for some other B , then it can only have one solution for this B . Proof. Suppose we have two solutions AX1 = B and AX2 = B . Then we have A(X1 − X2) = AX1 − AX2 = 0 . Because AX = 0 has only the trivial solution of X = 0 , we must have X1 − X2 = 0 , or X1 = X2 . Thus, there can only be one solution to AX = B .

Theorem 2. Suppose AX = B1 has a unique solution for some B1 . If AX = B has a solution for some other B , then it can only have one solution for this B . Proof. From Theorem 1, it suffices to show that AX = 0 has only the trivial solution. So suppose that AX2 = 0 and let AX1 = B1 where X1 is the unique solution. Then, we have

B1 = B1+ 0 = AX1+ AX2 = A(X1+ X2) .

By the uniqueness assumption, we must have X1+ X2 = X1 , which makes X2 = 0 . Theorem 3. Suppose A is an invertible n×n matrix. Then every system of equations AX = B has a unique solution. Proof. Given AX = B , we always have a solution given by X = A−1B . And if AX1 = B too, then again multiplying on the left by A−1 gives X1 = A

−1B . So there is a solution, and there can only be one solution.