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PreCalculus
6-3 Solving Linear Systems using Inverses and Cramer’s Rule
Name:___________________
A square system is a system that has the same number of equations as it does variables. If the square
coefficient matrix is invertible, then there is one unique solution to the system.
Invertible Square Linear Systems: Let A be the coefficient matrix of a system of n linear equations in n
variables given by AX=B, where X is the matrix of variables and B is the matrix of constants. If A is
invertible, then the system of equations has one unique solution given by X=A-1B.
To solve a 3x3 system using an inverse matrix, use a graphing calculator.
Ex. 1: Use an inverse matrix to solve each system, if possible.
a. 2x – y = 1
b. -3x + 9y = 36
2x + 3y = 13
7x – 8y = -19
Ex. 2: Solve the system using an inverse matrix.
a.
2x + y = 9
x – 3y + 2z = 12
5y – 3z = -11
b. Marquis has 22 coins that are all nickels, dimes and quarters. The value of the coins is $2.75. He
has three fewer dimes than twice the number of quarters. How many of each type of coin does
Marquis have?
PreCalculus Ch 6B Notes_Page 1
Cramer’s Rule is another method to solve systems of equations. (See page 390)
* Cramer’s Rule: Let A be the coefficient matrix of a system of n linear equations in n variables given by
AX=B. If the det(A) ≠ 0, then the unique solution of the system is given by
x1 
A1
A
, x2 
A2
A
, x3 
A3
A
, ..., xn 
An
A
,
where Ai is obtained by replacing the ith column of A with the column of the constant terms B.
If the det(A) = 0, then AX=B has no unique solution (either no solution or infinitely many solutions).
Ex. 3: Use Cramer’s Rule to find the solution to the system of linear equations, if a unique solution exists.
a.
4 x1  5 x2  49
3x1  2 x2  28
b.
You Try: c.
12 x  9 y  5
4 x  3 y  11
9 x  3 y  8
2 x  y  3
Ex. 4: Use Cramer’s Rule to find the solution to the system of linear equations, if a unique solution
exists.
2 x  4 y  z  3
y  4 z  1
a. 2 x  2 y  z  18
b. 3 x  y  2 z  6
x  3y  1
x  4z  7
PreCalculus Ch 6B Notes_Page 2
PreCalculus
6-4 Partial Fractions
Name:___________________
When a rational function f(x) can be written as the sum of two fractions with denominators that are linear
factors of the original denominator, the fractions in the sum are called partial fractions. The sum of
these partial fractions makes up the Partial fraction decomposition of the original rational function.
Ex. 1: Find the partial fraction decomposition of each rational expression.
x  25
x  11
a. 2
b.
2
x  x  12
2 x  5x  3
f  x
is
d  x
improper (that is, the numerator's degree is greater than or equal to the denominator's degree), then you
first have to use long division to get the "mixed number" form of the rational expression. Then decompose
the remaining fractional part.
* Partial-fraction decomposition only works for "proper" fractions. If a rational expression
Ex. 2: Find the partial fraction decomposition of the rational expression.
2 x 2  9 x  4
5 x 2  30 x  21
a.
b.
x2  x
x2  7 x
PreCalculus Ch 6B Notes_Page 3
If the denominator of a rational expression has a linear factor that is repeated n times, the partial fraction
decomposition must include a partial fraction with its own constant numerator for each power 1 to n of the
5x 1
linear factor. For example, to find the partial fraction decomposition of 3
, you would write
2
x  x  1
5x  1
x  x  1
3
2

A B C
D
E
 2 3

x x
x
x  1  x  12
Ex. 3: Find the partial fraction decomposition of each. (show two different methods)
3x 2  5 x  4
a. 3
x  4 x2  4 x
b.
x2
x  2x2  x
3
PreCalculus Ch 6B Notes_Page 4
Prime Quadratic Factors: If the denominator of a rational expression contains a prime quadratic factor,
the partial fraction decomposition must include a partial fraction with a linear numerator of the form
Bx+C for each power of this factor.
Ex. 4: Find the partial fraction decomposition of each.
x3  x
a.
2
 x 2  3
b.
4 x3  7 x
x
2
 x  1
2
* See the key concept summary on the bottom of page 401.
PreCalculus Ch 6B Notes_Page 5