Mere, Sam and Nathan go to a bookshop to buy supplies for their school. Mere bought two erasers and four pencils and paid $2.40 for it. Sam bought six erasers, one pencil and two pens and paid $3.95 while Nathan paid $4.15 for an eraser, a pencil and four pens. Write a system of equations that can be used to find the cost of each item.Use the inverse matrix method to solve the system of equations.
[Hint: use Gauss Jordan elimination]
If there are e erasers, p pencils, and q pens, we have
2e+4p = 240
6e+p+2q = 395
e+p+4q = 415
If A is the matrix of coefficients, then A^-1 =
1/84
(-2 16 -8)
(22 -8 4)
(-5 -2 22)
Then you have
(e p q) = (30 45 85)
See here:
http://www.wolframalpha.com/input/?i=inverse+%7B%7B2,4,0%7D,%7B6,1,2%7D,%7B1,1,4%7D%7D
http://www.wolframalpha.com/input/?i=%7B%7B2,4,0%7D,%7B6,1,2%7D,%7B1,1,4%7D%7D*%7B%7Bx%7D,%7By%7D,%7Bz%7D%7D+%3D+%7B%7B240%7D,%7B395%7D,%7B415%7D%7D
cost of pencil --- p
cost of eraser --- e
cost of pen ----- x
2e + 4p = 2.4 (1)
6e + p + 2x = 3.95 (2)
e + p + 4x = 4.15 (3)
I would not use a matrix method, the equations are much too "nice".
from (1) : e = 1.2 - 2p
sub into (2)
6(1.2 - 2p) + p + 2x = 3.95
-11p + 2x = -3.25
2x = 11p - 3.25
4x = 22p - 6.5
now sub those into (3)
e + p + 4x = 4.15
1.2 - 2p + p + 22p - 6.5 = 4.15
21p = 9.45
p = .45
e = 1.2 - 2p = .30
x = 13.60
You can confirm my answer with this Gauss-Jordan applet
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
Looks like I messed up for my x
I multiplied 4x by 4 instead of dividing by 4
4x = 3.4
x = .85
see Steve's solution
DEAR SIR I NEED THE ANSWER OF THIS QUESTION PLEASE.Mere, Sam and Nathan go to a bookshop to buy supplies for their school. Mere bought
two erasers and four pencils and paid $2.40 for it. Sam bought six erasers, one pencil and
two pens and paid $3.95 while Nathan paid $4.15 for an eraser, a pencil and four pens.
a) Write a system of equations that can be used to find the cost of each item.
b) Use an appropriate method to solve the system of equations
To find the cost of each item, we can create a system of equations based on the given information. Let's define the variables as follows:
Let e be the cost of an eraser.
Let p be the cost of a pencil.
Let n be the cost of a pen.
Based on Mere's purchase:
2e + 4p = 2.40
Based on Sam's purchase:
6e + p + 2n = 3.95
Based on Nathan's purchase:
e + p + 4n = 4.15
To solve this system of equations using the inverse matrix method and Gauss Jordan elimination, we need to represent the system in matrix form.
First, rewrite the equations:
2e + 4p + 0n = 2.40
6e + 1p + 2n = 3.95
1e + 1p + 4n = 4.15
Now, write the coefficients of the variables into a matrix A:
[ 2 4 0 ]
[ 6 1 2 ]
[ 1 1 4 ]
Write the constants on the right-hand side into a matrix B:
[ 2.40 ]
[ 3.95 ]
[ 4.15 ]
To use the inverse matrix method, we need to find the inverse of matrix A. If the inverse exists, we can solve for the variables x by multiplying the inverse of A with B.
The inverse of A can be found by performing Gauss Jordan elimination with the augmented matrix [A | I], where I represents the identity matrix.
[A | I] = [ 2 4 0 | 1 0 0 ]
[ 6 1 2 | 0 1 0 ]
[ 1 1 4 | 0 0 1 ]
Now, perform row operations to convert matrix A to the identity matrix.
1. Row 1/2 --> Row 1
[ 1 2 0 | 0.5 0 0 ]
[ 6 1 2 | 0 1 0 ]
[ 1 1 4 | 0 0 1 ]
2. Row 1*(-6) + Row 2 --> Row 2
[ 1 2 0 | 0.5 0 0 ]
[ 0 -11 2 | -3 1 0 ]
[ 1 1 4 | 0 0 1 ]
3. Row 1*(-1) + Row 3 --> Row 3
[ 1 2 0 | 0.5 0 0 ]
[ 0 -11 2 | -3 1 0 ]
[ 0 -1 4 | -0.5 0 1 ]
4. Row 2*(-1/11) --> Row 2
[ 1 2 0 | 0.5 0 0 ]
[ 0 1 -2/11 | 3/11 -1/11 0 ]
[ 0 -1 4 | -0.5 0 1 ]
5. Row 2 + Row 1*(-2) --> Row 1
[ 1 0 4/11 | -1/11 2/11 0 ]
[ 0 1 -2/11 | 3/11 -1/11 0 ]
[ 0 -1 4 | -0.5 0 1 ]
6. Row 3 + Row 2 --> Row 3
[ 1 0 4/11 | -1/11 2/11 0 ]
[ 0 1 -2/11 | 3/11 -1/11 0 ]
[ 0 0 40/11 | 2.5 -1/11 1 ]
7. Row 3*(11/40) --> Row 3
[ 1 0 4/11 | -1/11 2/11 0 ]
[ 0 1 -2/11 | 3/11 -1/11 0 ]
[ 0 0 1 | 0.275 -0.025 0.025 ]
8. Row 1 + Row 3*(-4/11) --> Row 1
[ 1 0 0 | -1.1 0.2 0.1 ]
[ 0 1 -2/11 | 3/11 -1/11 0 ]
[ 0 0 1 | 0.275 -0.025 0.025 ]
9. Row 2 + Row 3*(2/11) --> Row 2
[ 1 0 0 | -1.1 0.2 0.1 ]
[ 0 1 0 | 0.405 -0.055 0.050 ]
[ 0 0 1 | 0.275 -0.025 0.025 ]
The augmented matrix is now in the form [I | A_inv].
The inverse of matrix A (A_inv) is:
[ -1.1 0.2 0.1 ]
[ 0.405 -0.055 0.050 ]
[ 0.275 -0.025 0.025 ]
To find the variables, multiply the inverse of A (A_inv) with matrix B:
[ -1.1 0.2 0.1 ] * [ 2.40 ] [ 0.80 ]
[ 0.405 -0.055 0.050 ] * [ 3.95 ] = [ 1.05 ]
[ 0.275 -0.025 0.025 ] * [ 4.15 ] [ 1.20 ]
Therefore, the cost of each item is:
The cost of an eraser (e) is $0.80.
The cost of a pencil (p) is $1.05.
The cost of a pen (n) is $1.20.