Upcoming SlideShare
×

# 0.3.e,ine,det.

244 views

Published on

Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
244
On SlideShare
0
From Embeds
0
Number of Embeds
4
Actions
Shares
0
3
0
Likes
0
Embeds 0
No embeds

No notes for slide

### 0.3.e,ine,det.

1. 1. Equations, Inequations and Determinants
2. 2. Equations  An equation is a mathematical statement that has two expressions separated by an equal sign. The expression on the left side of the equal sign has the same value as the expression on the right side.  One or both of the expressions may contain variables. Solving an equation means manipulating the expressions and finding the value of the variables.  An example might be :  x=4+8 to solve this equation we would add 4 and 8 and find that x = 12.  A mathematical expression can have a variable as part of the expression. If x=3, the expression 7x + 4 becomes 7 * 3 + 4 which is equal to 21 + 4 or 25. To evaluate an expression with a variable, simply substitute the value of the variable into the expression and simplify.  A mathematical expression can have variables as part of the expression. If x=3, and y=5, the expression 7x + y - 4 becomes 7 * 3 +5 - 4 which is equal to 21 + 5 - 4 or 22. To evaluate an expression with two or more variables, substitute the value of the variables into the expression and simplify.
3. 3. Examples 1. Solve: 7x - 7 = 42 Solution: 7x - 7 = 42 7x - 7 + 7 = 42 + 7 7x = 49 (7x) / 7 = 49 / 7 x = 7 The variable needs to be isolated. To undo subtracting 7, add 7 to both sides. Adding 7 hasn't isolated the variable, so we need to continue. To undo multiplying by 7, divide both sides by 7. 2. Solve: 5(x + 2) = 25 Solution: 5(x + 2) = 25 [5(x + 2]/5 = 25/5 x + 2 = 5 x + 2 -2 = 5 -2 x = 3 The variable needs to be isolated. To undo multiplying by 5, divide by 5 on both sides. Dividing by 5 hasn't isolated the variable, so we need to continue. To undo adding 2, subtract 2, from both sides.
4. 4. Equations with Multi-variables Solve for 3x + 2y = 3 and x = 3y -10 . Solution: Replace x in the first equation with its equivalent, (3y - 10) from the second equation. 3x + 2y = 3 (Top equation. 3(3y - 10) + 2y = 3 Replaced x with (3y - 10). 9y - 30 + 2y = 3 Multiplied out. 11y = 33 Simplified. y = 3 Divide each side by 11 to get answer.) Now that y has a value, you can plug that value in either equation and find a value for x. Because the second equation has already been solved for x, it will be easier to plug 3 in for y in that equation. x = 3(3) - 10 x = 9 - 10 x = -1 . The solution is the ordered pair (- 1,3).
5. 5. Inequalities Math problems containing <, >, <=, and >= are called inequalities. A solution to any inequality is any number that makes the inequality true. On many occasions, you will be asked to show the solution to an inequality by graphing it on a number line. 1. Addition Principle for Inequalities - If a > b then a + c > b + c. Example: 1. Solve: x + 3 > 6 Solution: Using the Addition Principle, add -3 to each side of the inequality. x + 3 - 3 > 6 - 3 After simplification, x > 3. 2. Multiplication Principle for Inequalities - If a >b and c is positive, then ac > bc. If a > b and c is negative, then ac < bc (notice the sign was reversed). Example: 2. Solve: -4x < .8 Solution: Using the Multiplication Principle, multiply both sides of the inequality by -.25. Then reverse the signs. -.25(-4x) > -.25(.8) x > -.2 Absolute value becomes even more complicated when dealing with equations. However, there is a theorem that tells us how to deal with equations with absolute value and complicated inequalities. 1. If X is any expression, and b any positive number, and |X| = b it is the same as |X| = b or |X|=-b. 2. If X is any expression, and b any positive number, and |X| < b it is the same as -b < X < b. 3. If X is any expression, and b any positive number, and |X| > b it is the same as X < -b, X > b. Example: 3. Solve: |5x - 4| = 11 Solution: Use the theorem stated above to rewrite the equation. |X| = b X = 5x - 4 and b = 11 5x - 4 = 11, 5x - 4 = -11 Solve each equation using the Addition Principle and the Multiplication Principle. 5x = 15, 5x = -7 x = 3, x = -(7/5)
6. 6. Determinants With each square matrix corresponds just one number. This number is called the determinant of the matrix. The determinant of a matrix A is denoted det(A) or |A|. Now we'll define this correspondence. Determinant of a 1 x 1 matrix The determinant of the matrix is the element itself. Ex: det([-7]) = -7 Row and columns of the determinant If we say the i-th row of a determinant we mean the i-th row of the matrix corresponding with this determinant. If we say the i-th column of a determinant we mean the i-th column of the matrix corresponding with this determinant.
7. 7.  Quick References:  Determinant of a 1x1 matrix  The determinant of a 1x1 matrix is the element itself.  Determinant of a 2x2 matrix  |a b| |c d| = ad – cb  Determinant of a 3x3 matrix  The Sarrus rule : |a b c| |d e f| |g h i| = aei + bfg + cdh - ceg - afh – bdi  Cofactor of ai,j  The cofactor Ai,j is independent of the elements of the i-th row and the elements of the j-th column. The value of Ai,j = (-1)i+j.(the determinant of the sub-matrix of A, obtained from A by crossing out the i-th row and the j-th column.  Example : Take the 3 x 3 matrix A = [4 5 7] [1 2 3] [2 5 6] We calculate the cofactor corresponding with the element a1,3 = 7. We delete the first row and the third column. The cofactor A1,3 = (-1)4.(5 - 4) = 1  Determinant of a nxn matrix : Choose a fixed row value i. The determinant can be calculated emanating from the i-th row. |A| = Ai,1 . ai,1 + Ai,2 . ai,2 + Ai,3 . ai,3 + ... Ai,n . ai,n  Ai, j is the cofactor of ai,j. We say we have unfold the determinant following row i.  Example : Take the 3 x 3 matrix A = [4 5 7] [1 2 3] [2 5 6] We unfold the determinant following row 1. The three cofactors are -3 , 0 , 1. |A| = 4.(-3) + 5.0 + 7.1 = -5. We unfold the determinant following column 3. The three cofactors are 1, -10, 3. |A| = 7.1 - 3.10 + 6.3 = -5
8. 8. Properties A matrix A and its transpose have the same determinant. If we swap two columns in A, |A| changes sign. If we swap two rows in A, |A| changes sign. If we multiply a row in A by a real number r, |A| changes in r.|A| If we multiply a column in A by a real number r, |A| changes in r.|A| If a row of a determinant only consists of zeros, the determinant is 0. If a column of a determinant only consists of zeros, the determinant is 0. If a determinant has two equal rows or two equal columns, the determinant is 0. If a determinant has two proportional rows or two proportional columns, the determinant is 0. |a b c| |a b' c| |a b+b' c| |d e f|+|d e' f| = |d e+e' f| |g h i| |g h' i| |g h+h' i| A determinant does not change if we add a multiple of a row to another row. The same rule holds for columns The determinant of the identity matrix is 1. The determinant of a diagonal matrix is the product of the diagonal elements. |A|.|B| = |A.B|
9. 9. Cramer’s Rule A system of n linear equations in n unknowns is called a Cramer system if and only if the matrix formed by the coefficients is regular. There is a special method to solve such a system. This method is called Cramer's rule. We'll prove the rule for a system of 3 equations in 3 unknowns, but the rule is universal. Take, / ax + by + cz = d | a'x+ b'y + c'z = d‘ (1) a"x+ b"y + c"z = d" |a b c | with |N| = |a' b' c'| |a" b" c"| Then we have x.|N| = |xa b c | |xa' b' c'| |xa" b" c"| and using the properties of determinants |xa +by +cz b c | x.|N| = |xa'+b'y+c'z b' c'| |xa"+b"y+c"z b" c"| and appealing to (1) |d b c | x.|N| = |d' b' c'| |d" b" c"| Thus, x =|d b c | |d' b' c'| / |N| (2) |d" b" c"| Similarly, y =|a d c | |a' d' c‘|/|N| (3) |a" d" c"| z = |a b d | |a' b' d'| / |N| (4) |a" b" d"| The formulas (2), (3), (4) constitute Cramer's rule. It can be proved that this solution is the only solution of (1). Example: Solve the system in x, y and z . This system has parameter p. / p x - y + z = 0 | 6 x + y - 2z = 2 px - 2y - z = 1 We find : |N| = -4p - 18 We assume that p is not -4.5. Men find : x.|N| = -5 y.|N| = -2p + 6 z.|N| = 3p + 6 So, x = 5/(4p+18) ; y = (p- 3)/(2p+9) ; z = (3p+6)/(-4p-18)