 # Dr. Mark V. Sapir

## Systems of linear equations

A linear equation is an equation of the form

```               a1x1+a2x2+...+anxn=b             (1)
```

where x1,...,xn are unknowns, a1,...,an,b are coefficients.

Example:

```          3x-4y+5z=6                                   (2)
```

This equation has three unknowns and four coefficients (3, -4, 5, 6).

Example:

```       x=2, y=0, z=0
```

is a solution of equation (2).

A linear equation can have infinitely many solutions, exactly one solution or no solutions at all.

Equation (2) has infinitely many solutions. To find them all we can set arbitrary values of x and y and then solve (2) for z.

We get:

```
x = s
y = t
z = (6-3s+4t)/5

```

These formulas give all solutions of our equation meaning that for every choice of values of t and s we get a solution and every solution is obtained this way. Thus this is a (the) general solution of our equation.

There may be many formulas giving all solutions of a given equation. For example Maple gives another formula:
> with(linalg);

This command starts the linear algebra package.
> A:=matrix(1,3,[3,-4,5]):b:=vector():linsolve(A, b);

This command asks Maple to solve the system of equations.

The solution has two parameters t1 and t2;

x=4/3t1-5/3t2+2, y=t1, z=t2.

In order to get this solution "by hand" one can give y and z arbitrary values (t1 and t2) and solve for x.

Examples. 1. Consider the system:

```				          x +  y =  7
2x + 4y = 18
```

This system has just one solution: x=5, y=2. This is a general solution of the system.

2. Consider the system:

```				       x +  y + z =  7
2x + 4y + z = 18.
```

This system has infinitely many solutions given by this formula:

```				        x = 5 - 3s/2
y = 2 + s/2
z = s
```

This is a general solution of our system.

In order to find a general solution of a system of equations, one needs to simplify it as much as possible. The simplest system of linear equations is

```					        x = a
y = b
.....
```

where every equation has only one unknown and all these unknowns are different. It is not possible to reduce every system of linear equations to this form, but we can get very close. There are three operations that one can apply to any system of linear equations:

## Replace an equation by the sum of this equation and another equation multiplied by a number. Swap two equations. Multiply an equation by a non-zero number.

The system obtained after each of these operations is equivalent to the original system, meaning that they have the same solutions.

For example consider the system

```				         x +  y =  7
2x + 4y = 18
```

We can first replace the second equation by the second equation plus the first equation multiplied by -2. We get

```
x + y = 7
2y = 4

```

Now we can use the third operation and multiply the second equation by 1/2:

```
x + y = 7
y = 2

```

Finally we can replace the first equation by the sum of the first equation and the second equation multiplied by -1:

```
x = 5
y = 2

```

Since this system is equivalent to the original system, we get that x=5, y=2 is the general solution of the original system.

## The geometric meaning of systems of linear equations

Consider an (x,y)-plane and the set of points satisfying ax+by=c. This set of points is either a line (if a or b is not 0) or the whole plane (if a=b=c=0), or empty (if a=b=0 but c is not 0).

The set of solutions of the system

```
ax +  by = c
a'x + b'y = c'

```

is the intersection of the sets of solutions of the individual equations. For example if these equations define lines on the plane, the intersection may be a point -- if the lines are not parallel, a line -- if the lines coincide, or empty -- if the lines are parallel.   A system of equations in 3 or more variables has similar geometric meaning.

## Augmented matrices.

Consider the following problem:

Given the system of equations

```	                  x + y + 2z = a
x +      z = b                        (1)
2x + y + 3z = c,
```

show that it has a solution only if a+b=c.

In order to prove that, replace the first equation by the sum of the first two equations:

```
2x + y + 3z = a + b
x +      z = b
2x + y + 3z = c

```

This system is equivalent to the previous one, so it has a solution if and only if the initial system has a solution. But comparing the first and the third equations of this system we notice that it has a solution only if a+b=c. The problem is solved.

Now suppose that we have that a+b=c and we want to find the general solution of this system.

Then we need to simplify the system by using three operations (adding, swapping, multiplying). It is more convenient to work not with the system but with its augmented matrix, the array (table, matrix) consisting of the coefficients of the left sides of the equations and the right sides. For example the system (1) from the problem that we just solved has the following augmented matrix:

 [ 1 1 2 a ] [ 1 0 1 b ] [ 2 1 3 c ]

The number of equations in a system of linear equations is equal to the number of rows in the augmented matrix, the number of unknowns is equal to the number of columns minus 1, the last column consists of the right sides of the equations.

When we execute the operations on the systems of equations, the augmented matrix changes. If we add equation i to equation j, then row i will be added to row j, if we swap equations, the corresponding rows get swapped, if we multiply an equation by a (non-zero) number, the corresponding row is multiplied by this number.

Thus, in order to simplify a system of equations it is enough to simplify its augmented matrix by using the following row operations:

## Replace a row by this row plus another row multiplied by a number. Swap two rows. Multiply a row by a non-zero number.

For example let us simplify the augmented matrix of the system (1) from the problem that we just solved.

First we replace the first row by the sum of the first and the second rows:

 [ 2 1 3 a+b ] [ 1 0 1 b ] [ 2 1 3 c ]

Then we subtract the first row from the third row (remember that a+b=c):

 [ 2 1 3 a+b ] [ 1 0 1 b ] [ 0 0 0 0 ]

Then we subtract the second row multiplied by 2 from the first row. :

 [ 0 1 1 a-b ] [ 1 0 1 b ] [ 0 0 0 0 ]

Then we swap the first two rows and obtain the following matrix

 [ 1 0 1 b ] [ 0 1 1 a-b ] [ 0 0 0 0 ]

The last matrix has several important features:

## All zero rows (rows consisting of zeroes) are at the buttom. Every non-zero row starts with several zeroes followed by 1. This 1 is called the leading 1 of the row. The position of the leading 1 in the row with bigger number is further to the right than for the row with smaller number. Every number below the leading 1 is zero. Every number above the leading 1 is zero.

A matrix which satisfies all five conditions is called a matrix in the reduced row echelon form or a reduced row echelon matrix.

It is very easy to find the general solution of a system of linear equations whose augmented matrix has the reduced row echelon form.

Consider the system of equations corresponding to the last matrix that we got:

```
x +     z = b
y + z = a - b

```

The unknowns corresponding to the leading 1's in the row echelon augmented matrix are called leading unknowns. In our case the leading 1's are in the first and the second positions, so the leading unknowns are x and y. Other unknowns are called free.

In our case we have only one free unknown, z. If we move it to the right and denote it by t, we get the following formulas:

```
x = b - t
y = a - b - t
z = t

```

This system gives us the general solution of the original system with parameter t. Indeed, giving t arbitrary values, we can compute x, y and z and obtain all solutions of the original system of equations.

Similarly, we can get a general solution of every system of equations whose matrix is in the reduced row echelon form:

One just has to move all free variables to the right side of the equations and consider them as parameters.

Example Consider the system of equations:

```                         x1 + 2x2 +         x4   =  6
x3 +  6x4  =  7
x5=1
```

Its augmented matrix is

 [ 1 2 0 1 0 6 ] [ 0 0 1 6 0 7 ] [ 0 0 0 0 1 1 ]

The matrix has the reduced row echelon form. The leading unknowns are x1, x3 and x5; the free unknowns are x2 and x4. So the general solution is:

```
x1= 6-2t-s
x2= s
x3= 7-6t
x4= t
x5= 1

```

If the augmented matrix does not have the reduced row echelon form but has the (ordinary) row echelon form then the general solution also can be easily found.

The method of finding the solution is called the back-substitution.

First we solve each of the equations for the leading unknowns The last non-zero equation gives us the expression for the last leading unknown in terms of the free unknowns. Then we substitute this leading unknown in all other equations by this expression. After that we are able to find an expression for the next to the last leading unknown, replace this unknown everywhere by this expression, etc. until we get expressions for all leading unknowns. The expressions for leading unknowns that we find in this process form the general solution of our system of equations.

Example. Consider the following system of equations.

```                      x1-3x2+ x3-x4 = 2
x2+2x3-x4 = 3
x3+x4 = 1
```

Its augmented matrix

 [ 1 -3 1 -1 2 ] [ 0 1 2 -1 3 ] [ 0 0 1 1 1 ]

is in the row echelon form.

The leading unknowns are x1, x2, x3; the free unknown is x4.

Solving each equation for the leading unknown we get:

```	                    x1=2+3x2-x3+x4
x2=3-2x3+x4
x3=1-x4
```

The last equation gives us an expression for x3: x3=1-x4. Substituting this into the first and the second equations gives:

```                    x1=2+3x2-1+x4+x4=1+3x2+2x4
x2=3-2(1-x4)+x4=1+3x4
x3=1-x4
```

Now substituting x2=1+3x4 into the first equation, we get

```                    x1=1+3(1+3x4)+2x4=4+11 x4
x2=1+3x4
x3=1-x4
```

Now we can write the general solution:

```
x1=4+11 s
x2=1+ 3 s
x3=1-   s
x4=     s

```

Let us check if we made any arithmetic mistakes. Take x4=1 and compute x1=15, x2=4, x3=0, x4=1. Substitute it into the original system of equations:

```
15 - 3 * 4 + 0 - 1 = 2
4 + 2 * 0 - 1 = 3
0 + 1 = 1

```

OK, it seems that our solution is correct.

## The Gauss-Jordan elimination procedure

There exists a standard procedure to obtain a reduced row echelon matrix from a given matrix by using the row operations.

This procedure consists of the following steps.

1. Locate the leftmost column which does not consist of zeroes.
2. If necessary swap the first row with the row which contains a non-zero number a in the column found on step 1.
3. If this number a is not 0, multiply the first row by 1/a, to get a leading 1 in the first row.
4. Use the first row to make zeroes below the leading 1 in the first row (by using the adding operation).
5. Cover the first row and apply the first 4 steps to the remaining sub-matrix. Continue until the whole matrix is in the row echelon form.
6. Use the last non-zero row to make zeroes above the leading 1 in this row. Use the second to last non-zero row to make zeroes above the leading 1 in this row. Continue until the matrix is in the reduced row echelon form.

Example  The Gauss-Jordan elimination procedure allows us to prove the following important theorem. Theorem. A system of linear equations either has no solutions or has exactly one solution or has infinitely many solutions. A system of linear equations has infinitely many solutions if and only if its reduced row echelon form has free unknowns and the last column of the reduced row echelon form has no leading 1's. It has exactly one solution if and only if the reduced row echelon form has no free unknowns and the last column of the reduced row echelon form has no leading 1. It has no solutions if and only if the last column of the reduced row echelon form has a leading 1. ## Homogeneous systems

Example:

```
2x + 3y - 4z = 0
x -  y +  z = 0
x -  y      = 0

```

A homogeneous system of equations always has a solution (0,0,...,0). Therefore the theorem about solutions of systems of linear equations implies the first part of the following result. Theorem. Every homogeneous system has either exactly one solution or infinitely many solutions. If a homogeneous system has more unknowns than equations, then it has infinitely many solutions. ## Matrices and matrix operations

Examples. Here are three matrices:

 [ 2 3 4 ] [ 1 0 5 ]
,
 [ 0 0 0 0 ]
,
 [ 2 ] [ 3 ]

The size of a matrix is the pair of numbers: the number of rows and the number of columns. The matrices above have sizes (2,3), (1,4), (2,1), respectively.

A matrix with one row is called a row-vector. A matrix with one column is called a column-vector. In the example above the second matrix is a row-vector, the third one is a column-vector. The entry of a matrix A which stays in the i-th row and j-th column will be usually denoted by Aij or A(i,j).

Discussing matrices, we shall call numbers scalars. In some cases one can view scalars as 1x1-matrices.

Matrices were introduced first in the middle of 19-th century by W. Hamilton and A. Cayley. Following Cayley, we are going to describe an arithmetic where the role of numbers is played by matrices.

## Motivation.

In order to solve an equation

ax=b

with a not equal 0 we just divide b by a and get x. We want to solve systems of linear equations in a similar manner. Instead of the scalar a we shall have a matrix of coefficients of the system of equations, that is the array of the coefficients of the unknowns (i.e. the augmented matrix without the last column). Instead of x we shall have a vector of unknowns and instead of b we shall have the vector of right sides of the system.

In order to do that we must learn how to multiply and divide matrices.

But first we need to learn when two matrices are equal, how to add two matrices and how to multiply a matrix by a scalar.

Example.

 [ 2 3 4 ] [ 1 2 4 ] [ 0 0 1 ]
+
 [ 1 2 1 ] [ 2 2 3 ] [ 0 0 2 ]
=
 [ 3 5 5 ] [ 3 4 7 ] [ 0 0 3 ]

In order to multiply a matrix by a scalar, one has to multiply all entries of the matrix by this scalar.

Example:

3 *
 [ 2 3 1 2 ] [ 3 1 0 1 ] [ 1 2 2 3 ]
=
 [ 6 9 3 6 ] [ 9 3 0 3 ] [ 3 6 6 9 ]

The product of a row-vector v of size (1, n) and a column vector u of size (n,1) is the sum of products of corresponding entries: uv=u(1)v(1)+u(2)v(2)+...+u(n)v(n)

Example:

```                        [ 3 ]
[1, 2, 3] * [ 4 ] =1*3 + 2*4 + 3*1 = 3+8+3=14
[ 1 ]
```

Example:

```                        [ x ]
[2, 4, 3] * [ y ] = 2x + 4y + 3z
[ z ]
```

As you see, we can represent the left side of a linear equation as a product of two matrices. The product of arbitrary two matrices which we shall define next will allow us to represent the left side of any system of equations as a product of two matrices.

Let A be a matrix of size (m,n), let B be a matrix of size (n,k) (that is the number of columns in A is equal to the number of rows in B. We can subdivide A into a column of m row-vectors of size (1,n). We can also subdivide B into a row of k column-vectors of size (n,1):

```          r1
r2
A =...      B=[c1 c2 ... ck]
rm
```
Then the product of A and B is the matrix C of size (m,k) such that

C(i,j)=ricj

(C(i,j) is the product of the row-vector ri and the column-vector cj).

Matrices A and B such that the number of columns of A is not equal to the number of rows of B cannot be multiplied.

Example:

 [ 2 3 4 ] [ 1 2 3 ]
*
 [ 1 2 ] [ 3 4 ] [ 5 6 ]
=
 [ 31 40 ] [ 22 28 ]

Example:

 [ 2 3 4 ] [ 1 2 3 ]
*
 [ x ] [ y ] [ z ]
=
 [ 2x + 3y + 4z ] [ x + 2y + 3z ]

You see: we can represent the left part of a system of linear equations as a product of a matrix and a column-vector. The whole system of linear equations can thus be written in the following form:

A*v = b

where A is the matrix of coefficients of the system -- the array of coefficients of the left side (do not mix with the augmented matrix), v is the column-vector of unknowns, b is the column vector of the right sides (constants).

## Properties of matrix operations.

1. The addition is commutative and associative: A+B=B+A, A+(B+C)=(A+B)+C
2. The product is associative: A(BC)=(AB)C
3. The product is distributive with respect to the addition: A(B+C)=AB+AC, (A+B)C=AC+BC
4. The multiplication by scalar is distributive with respect to the addition of matrices: a(B+C)=aB+aC
5. The product by a scalar is distributive with respect to the addition of scalars: (a+b)C=aC+bC
6. a(bC)=(ab)C; a(BC)=(aB)C; (Ab)C=A(bC)
7. 1 A=A; 0 A=0 (here the 0 on the left is the number zero, the 0 on the right is the zero matrix of the same size as A, that is a matrix all of whose entries are 0
8. 0+A=A+0=A (here 0 is the zero matrix of the same size as A).
9. 0A=0 (these two zeroes are matrices of appropriate sizes).
10. Let In denote the identity matrix of order n that is a square matrix of order n with 1s on the main diagonal and zeroes everywhere else. Then for every m by n matrix A the product of A and In is A and the product of Im and A is A.
Example  The following properties of matrix operations do not hold:

1. For every square matrices A and B we have AB=BA:

Example:

A =
 [ 1 2 ] [ 0 1 ]
, B =
 [ 1 0 ] [ 2 1 ]

Indeed,

AB =
 [ 5 2 ] [ 2 1 ]
, BA =
 [ 1 2 ] [ 2 5 ]
2. For every matrices A, B, C, if AB=AC and A is not a zero matrix then B=C.

Example:

A =
 [ 1 0 ] [ 0 0 ]
, B =
 [ 0 0 ] [ 1 0 ]
, C =
 [ 0 0 ] [ 0 1 ]

Then AB=AC=0 but B and C are not equal. Notice that this example shows also that a product of two non-zero matrices can be zero.

## Transpose, trace, inverse

There are three other important operations on matrices.

If A is any m by n matrix then the transpose of A, denoted by AT, is defined to be the n by m matrix obtained by interchanging the rows and columns of A, that is the first column of AT is the first row of A, the second column of AT is the second row of A, etc.

Example.
The transpose of

 [ 2 3 ] [ 4 5 ] [ 6 7 ]

is
 [ 2 4 6 ] [ 3 5 7 ]

If A is a square matrix of size n then the sum of the entries on the main diagonal of A is called the trace of A and is denoted by tr(A).

Example.
The trace of the matrix

 [ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ]
is equal to 1+5+9=15

A square matrix of size n A is called invertible if there exists a square matrix B of the same size such that AB = BA = In, the identity matrix of size n. In this case B is called the inverse of A.

Examples. 1. The matrix In is invertible. The inverse matrix is In: In times In is In because In is the identity matrix.

2. The matrix A

 [ 1 3 ] [ 0 1 ]

is invertible. Indeed the following matrix B:

 [ 1 -3 ] [ 0 1 ]

is the inverse of A since A*B=I2=B*A.

3. The zero matrix O is not invertible. Indeed, if O*B=In then O=O*B=In which is impossible.

4. A matrix A with a zero row cannot be invertible because in this case for every matrix B the product A*B will have a zero row but In does not have zero rows.

5. The following matrix A:

 [ 1 2 3 ] [ 3 4 5 ] [ 4 6 8 ]

is not invertible. Indeed, suppose that there exists a matrix B:

 [ a b c ] [ d e f ] [ g h i ]

such that A*B=I3. The corresponding entries of A*B and I3 must be equal, so we get the following system of nine linear equations with nine unknowns:

 a+2d+3g = 1 ; the (1,1)-entry b+2e+3h = 0 ; the (1,2)-entry c+2f+3i = 0 ; the (1,3)-entry 3a+4d+5g = 0 3b+4e+5h = 1 3c+4f+5i = 0 4a+6d+8g = 0 4b+6e+8h = 0 4c+6f+8i = 1

This system does not have a solution which can be shown with the help of Maple.

Now we are going to prove some theorems about transposes, traces and inverses. Theorem. The following properties hold:

1. (AT)T=A, that is the transpose of the transpose of A is A (the operation of taking the transpose is an involution).
2. (A+B)T=AT+BT, the transpose of a sum is the sum of transposes.
3. (kA)T=kAT.
4. (AB)T=BTAT, the transpose of a product is the product of the transposes in the reverse order.  Theorem. The following properties of traces hold:

1. tr(A+B)=tr(A)+tr(B)
2. tr(kA)=k tr(A)
3. tr(AT)=tr(A)
4. tr(AB)=tr(BA)  Theorem. The following properties hold:

1. If B and C are inverses of A then B=C. Thus we can speak about the inverse of a matrix A, A-1.
2. If A is invertible and k is a non-zero scalar then kA is invertible and (kA)-1=1/k A-1.
3. If A and B are invertible then AB is invertible and

(AB)-1=B-1 A-1

that is the inverse of the product is the product of inverses in the opposite order. In particular

(An)-1=(A-1)n.

4. (AT)-1=(A-1)T, the inverse of the transpose is the transpose of the inverse.
5. If A is invertible then (A-1)-1=A. The proofs of 2, 4, 5 are left as exercises. Notice that using inverses we can solve some systems of linear equations just in the same way we solve the equation ax=b where a and b are numbers. Suppose that we have a system of linear equations with n equations and n unknowns. Then as we know, this system can be represented in the form Av=b where A is the matrix of the system, v is the column-vector of unknowns, b is the column-vector of the right sides of the equations. The matrix A is a square matrix. Suppose that it has an inverse A-1. Then we can multiply both sides of the equation A v = b by A-1 on the left. Using the associativity, the fact that A-1 A=I and the fact that Iv=v, we get: v=A-1b. This is the solution of our system. 