MATHEMATICS Linear Equations by Dr. Eman Saad & Dr. Shorouk Ossama

Linear Equations Recall that in two dimensions a line in a rectangular xy-coordinate system can be represented by an equation of the form ax + by = c (a, b not both 0) And in three dimensions a plane in a rectangular xyz- coordinate system can be represented by an equation of the form ax + by + cz = d (a, b, c not all 0)

Examples of “linear equations”: the first being a linear equation in the variables x and y and the second a linear equation in the variables x, y, and z. More generally, we define a linear equation in the n variables x1, x2, …., xn to be one that can be expressed in the form: a1x1 + a2x2 + ….. + anxn = b In the special case where b=0, that called a homogeneous linear equation a1x1 + a2x2 + ….. + anxn = 0

The following are linear equations: x + 3y = 7 x1 – 2x2 – 3x3 + x4 = 0 (½) x – y + 3z = -1 x1 + x2 + ….. + xn = 1 The following are not linear equations: X + 3y2 = 4 3x + 2y – xy = 5 Sinx + y = 0 + 2 x2 + x3 = 1 x1

Linear Systems With Two And Three Unknowns: Linear systems in two unknowns arise in connection with intersections of lines. For example, consider the linear system. a1x + b1y = c1 a2x + b2y = c2

Inverse Method: If we have any system: x1 + x2 = 4 9x1 – x2 = 14 So: Multiply from left by A-1 A-1 A X = A-1 B X = A-1 B Get A-1 Multiply A-1 by B These are x and y A X = B

X = A-1 d X1 X2

Cramer’s Rule: a11 x1 + a12x2 = d1 a21 x1 + a22x2 = d2 The eliminated of x2 and x1 by this process leads to the solution This formula is known as Cramer’s Rule. If the denominator is zero then the two equations can have no solutions or an infinity of solutions.

Example: Solve The Equations x1 + 2x2 = 4 x1 - x2 = 1 Δ det A = = - 1 – 2 = -3 Δ1 = = -4 -2 = -6 Δ2 = = 1 - 4 = -3

+ - + Δ det A = = 1 x (2+3) +2 x (4+1) + 1 x (6-1) = 5 + 10 + 5 = 20 Δ1 = = -4 x (2+3) +2 x (-2+7) + 1 x (-3-7) = -20 +10 -10 = -20

Δ2 = = 1 x (-2+7) + 4 x (4+1) + 1 x ( 14+1) = 5 + 20 + 15 = 40 Δ3 = = Δ2 = = 1 x (-2+7) + 4 x (4+1) + 1 x ( 14+1) = 5 + 20 + 15 = 40 Δ3 = = 1 x (7+3) + 2 x (14+1) - 4 x ( 6-1) = 10 + 30 - 20 = 20 20 3 1 3

Augmented Matrices and Elementary Row Operations: As the number of equations and unknowns in a linear system increases, so does the complexity of the algebra involved in finding solutions. The required computations can be made more manageable by simplifying notation and standardizing procedures. we can abbreviate the system by writing only the rectangular array of numbers.

This is called the augmented matrix for the system This is called the augmented matrix for the system. For example, the augmented matrix for the system of equations: Since the rows (horizontal lines) of an augmented matrix correspond to the equations in the associated system,

these three operations correspond to the following operations on the rows of the augmented matrix: 1. Multiply a row through by a nonzero constant. 2. Interchange two rows. Add a constant times one row to another. These are called elementary row operations on a matrix.

Example: x + y + 2z = 9 2x + 4y – 3z = 1 3x + 6y -5z = 0 -2 R1 + R2 = R`2 -3 R2 + R3 = R`3 -3 R1 + R3 = R`3 (-2) R3 = R`3 (1/2) R2 = R`2 - R1 + R2 = R`1 - (11/2) R3 + R1 = R`1 (7/2) R3 + R2 = R`2 x = 1 y = 2 z = 3 1 4 2 5 3 6 7

Gauss Elimination: We need write the equations in form of augmented matrix for the system. The fourth column consists of the constant on the right- hand sides.

The elementary operations referred to previously become elementary row operations on the matrix. The final matrix is said to be in echelon form: that is, it has zeros below the diagonal elements starting from the top left. We can solve the equations by back substation.

Example: Find the solution by using gauss elimination - R1 + R2 = R`2 -3 R1 + R3 = R`3 R2 - R3 = R`3 1 2 3

back substation Three Equations: 0 + 0 + 3Z = 19 0 + 3Y + Z = 5 X Y Z = back substation Three Equations: 0 + 0 + 3Z = 19 0 + 3Y + Z = 5 X + Y + Z = 5

Example: Find the solution by using gauss elimination -3 R1 + R2 = R`2 -4 R1 + R3 = R`3 R2 ÷ -5 = R`2 11 R2 + R3 = R`3 1 2 3

back substation Three Equations: 0 + 0 + 6Z = 6 0 + Y + Z = 1 X Y Z = back substation Three Equations: 0 + 0 + 6Z = 6 0 + Y + Z = 1 X +2Y + Z = 2

SUMMARY Pages From 51 To 52: Cramer’s Rule Pages From 53 : Cancelled Pages From 54 To 57: Gauss Elimination Sheet (4)

Thanks