Linear Equations Dan Sloughter - Furmanmath.furman.edu/~dcs/courses/math13/lectures/lecture-6.pdf · 2017. 3. 7. · Systems of linear equations I We call an equation of the form

Mathematics 13: Lecture 6Linear Equations

Dan Sloughter

Furman University

January 15, 2008

Dan Sloughter (Furman University) Mathematics 13: Lecture 6 January 15, 2008 1 / 18

Systems of linear equations

I We call an equation of the form

a1x1 + a2x2 + · · · + anxn = b

a linear equation.

I Note:

I If n = 2, the set of all solutions is a line in R2.I If n = 3, the set of all solutions is a plane in R3.I More generally,the set of all solutions is a hyperplane in Rn.

I We may think of the variables as the components of a vector ~x :

~x =

x1

x2...

xn

.




a1x1 + a2x2 + · · · + anxn = b

a linear equation.I Note:



~x =

x1

x2...

xn

.




a1x1 + a2x2 + · · · + anxn = b


I If n = 2, the set of all solutions is a line in R2.

I If n = 3, the set of all solutions is a plane in R3.I More generally,the set of all solutions is a hyperplane in Rn.


~x =

x1

x2...

xn

.




a1x1 + a2x2 + · · · + anxn = b


I If n = 2, the set of all solutions is a line in R2.I If n = 3, the set of all solutions is a plane in R3.

I More generally,the set of all solutions is a hyperplane in Rn.


~x =

x1

x2...

xn

.




a1x1 + a2x2 + · · · + anxn = b




~x =

x1

x2...

xn

.




a1x1 + a2x2 + · · · + anxn = b




~x =

x1

x2...

xn

.


Systems of linear equations (cont’d)

I In particular, we may denote a solution x1 = s1, x2 = s2, . . . , xn = sn

as a vector:

~x =

s1

s2...

sn

.


Example

I If 2x + y = 5, then y = 5 − 2x .

I It follows that for any real number t, x = t and y = 5 − 2t satisfiesthe equation.

I That is, for any value of t,

~x =

[t

5 − 2t

]= t

[1

−2

]+

[05

]is a solution of 2x + y = 5.

I We call t a parameter for the set of solutions to the equation.

I Note: the set of solutions is infinite. Geometrically, the set ofsolutions is a line in the plane.


Example

I If 2x + y = 5, then y = 5 − 2x .



~x =

[t

5 − 2t

]= t

[1

−2

]+

[05





Example

I If 2x + y = 5, then y = 5 − 2x .



~x =

[t

5 − 2t

]= t

[1

−2

]+

[05





Example

I If 2x + y = 5, then y = 5 − 2x .



~x =

[t

5 − 2t

]= t

[1

−2

]+

[05





Example

I If 2x + y = 5, then y = 5 − 2x .



~x =

[t

5 − 2t

]= t

[1

−2

]+

[05





SystemsI We often want to find simultaneous solutions to two or more linear

equations.

I Example: To solve the system

2x − y = 5

x + 3y = 2,

we note that it is equivalent to the the system (obtained bymultiplying the second equation by −2 and adding it to the first)

−7y = 1

x + 3y = 2.

I This gives us the solution y = −17 and x = 17

7 , that is,

~x =

[177

−17

].



equations.I Example: To solve the system

2x − y = 5

x + 3y = 2,


−7y = 1

x + 3y = 2.


7 , that is,

~x =

[177

−17

].



equations.I Example: To solve the system

2x − y = 5

x + 3y = 2,


−7y = 1

x + 3y = 2.


7 , that is,

~x =

[177

−17

].


Example

I The system

4x + y − z = 6

x − 3y + z = 5

is equivalent to the system

13y − 5z = −14

x − 3y + z = 5.

I If we let z = t, where t can be any real number, then

y =5

13t − 14

13and x =

2

13t +

23

13.


Example

I The system

4x + y − z = 6

x − 3y + z = 5

is equivalent to the system

13y − 5z = −14

x − 3y + z = 5.

I If we let z = t, where t can be any real number, then

y =5

13t − 14

13and x =

2

13t +

23

13.


Example (cont’d)

I Hence the solutions are

~x =

213 t + 23

13513 t − 14

13

t

= t

213513

1

+

2313

−1413

0

.

I Note: the solution set is infinite. Geometrically, this is a line inthree-space.


Example (cont’d)

I Hence the solutions are

~x =

213 t + 23

13513 t − 14

13

t

= t

213513

1

+

2313

−1413

0

.

I Note: the solution set is infinite. Geometrically, this is a line inthree-space.


Example

I The previous examples are examples of consistent systems, that is,systems which have solutions.

I The system of equations

3x − 4y = 12

−6x + 8y = 13

is inconsistent because there are no solutions: if 3x − 4y = 12, then,multiplying by −2, we have to have −6x + 8y = −24.

I We can also see this by multiplying the first equation by 2 and addingit to the second to obtain the equivalent system

3x − 4y = 12

0 = 37,

which clearly has no solutions.


Example



3x − 4y = 12

−6x + 8y = 13



3x − 4y = 12

0 = 37,



Example



3x − 4y = 12

−6x + 8y = 13



3x − 4y = 12

0 = 37,



Number of solutions

I Note: we have seen an example of a system of linear equations havingexactly one solution, an example with an infinite number of solutions,and an example with no solutions.

I We shall see that these are the only possibilities for the number ofsolutions to a system of linear equations.


Number of solutions

I Note: we have seen an example of a system of linear equations havingexactly one solution, an example with an infinite number of solutions,and an example with no solutions.

I We shall see that these are the only possibilities for the number ofsolutions to a system of linear equations.


Coefficient matrix

I The coefficient matrix of the system of linear equations

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

...

am1x1 + am2x2 + · · · + amnxn = bm

is the m × n matrix a11 a12 · · · a1n

a21 a22 · · · a2n...

.... . .

...am1 am2 · · · amn

.


Augmented matrix

I The augmented matrix of the system of linear equations

a11x1 + a12x2 + · · · + a1nxn = b1

a21x1 + a22x2 + · · · + a2nxn = b2

...

am1x1 + am2x2 + · · · + amnxn = bm

is the m × (n + 1) matrixa11 a12 · · · a1n b1

a21 a22 · · · a2n b2...

.... . .

......

am1 am2 · · · amn bm

.


Row operations

I Idea: the operations we used to solve a system of linear equationsaffected only the coefficients and the constants. Hence we can solvethe system more easily by working directly with the augmented matrix.

I The elementary row operations are

I interchange rows,I multiply a row by nonzero scalar,I add a multiple of one row to another row.

I Important: applying a row operation produces an equivalent system(that is, a system with exactly the same solution set).


Row operations


I The elementary row operations are

I interchange rows,I multiply a row by nonzero scalar,I add a multiple of one row to another row.



Row operations


I The elementary row operations areI interchange rows,

I multiply a row by nonzero scalar,I add a multiple of one row to another row.



Row operations


I The elementary row operations areI interchange rows,I multiply a row by nonzero scalar,

I add a multiple of one row to another row.



Row operations


I The elementary row operations areI interchange rows,I multiply a row by nonzero scalar,I add a multiple of one row to another row.



Row operations


I The elementary row operations areI interchange rows,I multiply a row by nonzero scalar,I add a multiple of one row to another row.



Example

I Consider the system of linear equations:

x − y − z = 2

3x − 3y + 2z = 16

2x − y + z = 9.

I The augmented matrix for this system is1 −1 −1 23 −3 2 162 −1 1 9

.


Example

I Consider the system of linear equations:

x − y − z = 2

3x − 3y + 2z = 16

2x − y + z = 9.

I The augmented matrix for this system is1 −1 −1 23 −3 2 162 −1 1 9

.


Example (cont’d)

I Multiplying the first row by −3 and adding to the second row gives us:1 −1 −1 20 0 5 102 −1 1 9

.

I Multiplying the first row by −2 and adding to the third row gives us:1 −1 −1 20 0 5 100 1 3 5

.


Example (cont’d)

I Multiplying the first row by −3 and adding to the second row gives us:1 −1 −1 20 0 5 102 −1 1 9

.

I Multiplying the first row by −2 and adding to the third row gives us:1 −1 −1 20 0 5 100 1 3 5

.


Example (cont’d)

I Interchanging the second and third rows, we have:1 −1 −1 20 1 3 50 0 5 10

.

I Dividing the third row by 5, we have:1 −1 −1 20 1 3 50 0 1 2

.


Example (cont’d)

I Interchanging the second and third rows, we have:1 −1 −1 20 1 3 50 0 5 10

.

I Dividing the third row by 5, we have:1 −1 −1 20 1 3 50 0 1 2

.


Example (cont’d)

I The equivalent system of equations is:

x − y − z = 2

y + 3z = 5

z = 2,

from which we obtain (by back-substitution)

z = 2

y = 5 − 6 = −1

x = 2 − 1 + 2 = 3.


Example (cont’d)

I That is, our system of equations has the unique solution

~x =

xyz

=

3−1

2

.


Example (cont’d)

I Note: we could have continued simplifying the augmented matrix,obtaining first 1 0 2 7

0 1 3 50 0 1 2

,

and then 1 0 0 30 1 0 −10 0 1 2

.

I From this we have immediately

x = 3

y = −1

z = 2.


Example (cont’d)

I Note: we could have continued simplifying the augmented matrix,obtaining first 1 0 2 7

0 1 3 50 0 1 2

,

and then 1 0 0 30 1 0 −10 0 1 2

.

I From this we have immediately

x = 3

y = −1

z = 2.


Documents

Linear Equations Dan Sloughter - Furmanmath.furman.edu/~dcs/courses/math13/lectures/lecture-6.pdf · 2017. 3. 7. · Systems of linear equations I We call an equation of the form