# Solving Systems with Inverse Matrices ```Sec. 7.3c
Let A be the coefficient matrix of a system of n linear
equations in n variables given by AX = B, where X is the
n x 1 matrix of variables and B is the n x 1 matrix of
numbers of the right-hand side of the equations. If A–1
exists, then the system of equations has the unique solution
X=A
–1
B
Write the system of equations as a matrix equation AX = B,
with A as the coefficient matrix of the system.
x  3y  z  9
2x  4z  1
8 x  y  z  5
 1 3 1  x  9






AX = B:  2 0 4   y   1
 8 1 1   z  5
Write the matrix equation as a system of equations
 1 2 3 1   x   2  x  2 y  3 z  w  2
 0 0 2 8   y  3 

   
2z  8w  3
 9 0 1 5  z   9 
9x  z  5w  9

   
 1 1 6 3   w  2  x  y  6 z  3w  2
Solve the given system using inverse matrices
3x  2 y  0
0 
 x
 3 2
X   B 
x  y  5 A  

5 
 y
 1 1 
To solve for X, apply the
inverse of A to both sides
of the matrix equation:
10 
X A B  
15 
1
3 x  2 y 
A X 
B

 x  y 
Solution:
(x, y) = (10, 15)
Solve the given system using inverse matrices
3x  3 y  6 z  20
x  3 y  10 z  40
 x  3 y  5 z  30
Find
1
XA B
Solution:
(x, y, z) = (18, 118/3, 14)
 3 3 6 
x
 20 






A   1 3 10  X   y  B   40 
30 
 1 3 5
 z 
Solve the given system using inverse matrices
x  4 y  2z  0
2x  y  z  6
3x  3 y  5 z  13
Find
1
XA B
Solution:
(x, y, z) = (3, –1/2, 1/2)
 1 4 2
x
 0 






A   2 1 1  X   y B   6 
 3 3 5
 z 
 13
Solve the given system using inverse matrices
2x  y  2z  8
3x  2 y  z  w  10
2 x  3w  y  1
4 x  3 y  2 z  5w  39
Find
1
XA B
Solution:
(x, y, z, w) =
(4, –2, 1, –3)
2 1 2 0
x
8
 3 2 1 1
 y
10 
 X   B 
A
 2 1 0 3
z
 1


 
 
 4 3 2 5
 w
39 
Use a method of your choice to solve the given system.
x yz 6
x  y  2 z  2
Augmented Matrix:
1 1 1 6 
1 1 2 2 


1
0
1.5
2


RREF:
0 1 0.5 4 


Solution:
(x, y, z) = (2 – 1.5z, –4 – 0.5z, z)
Fitting a parabola to three points. Determine a, b, and c so
that the points (–1, 5), (2, –1), and (3, 13) are on the graph of
f  x   ax  bx  c
2
How about a diagram to start???
We need f(–1) = 5, f(2) = –1, and f(3) = 13:
f  1  a  b  c  5
f  2  4a  2b  c  1
f  3  9a  3b  c  13
Now, simply solve
this system!!!
 (a, b, c) = (4, –6, –5)
 f  x   4x  6x  5
2
Double-check with a graph?
Mixing Solutions. Aileen’s Drugstore needs to prepare a 60-L
mixture that is 40% acid using three concentrations of acid. The
first concentration is 15% acid, the second is 35% acid, and the
third is 55% acid. Because of the amounts of acid solution on
hand, they need to use twice as much of the 35% solution as
the 55% solution. How much of each solution should they use?
x = liters of 15% solution
y = liters of 35% solution
z = liters of 55% solution
x  y  z  60
y  2z  0
0.15x  0.35 y  0.55z  0.40  60 
Solve the system!!!
Need 3.75 L of 15% acid, 37.5 L of 35% acid, and
18.75 L of 55% acid to make 60 L of 40% acid solution.
Manufacturing. Stewart’s metals has three silver alloys on hand.
One is 22% silver, another is 30% silver, and the third is 42%
silver. How many grams of each alloy is required to produce 80
grams of a new alloy that is 34% silver if the amount of 30% alloy
used is twice the amount of 22% alloy used?
x = amount of 22% alloy
y = amount of 30% alloy
z = amount of 42% alloy
x  y  z  80
0.22 x  0.30 y  0.42 z  27.2
2x  y  0
Solve the system!!!
Need approximately 14.545g of the 22% alloy,
29.091g of the 30% alloy, and 36.364g of the 42% alloy
to make 80g of the 34% alloy.
Vacation Money. Heather has saved \$177 to take with her on
the family vacation. She has 51 bills consisting of \$1, \$5, and
\$10 bills. If the number of \$5 bills is three times the number of
\$10 bills, find how many of each bill she has.
x = number of \$1 bills
y = number of \$5 bills
z = number of \$10 bills
x  y  z  51
x  5 y  10 z  177
y  3z  0
Solve the system!!!
Heather has 27 one-dollar bills, 18 fives, and 6 tens.
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