Calculating Anne's Final Mark with Matrix Equations

[carltonblues]

Anne is a maths student. She has achieved marks of 9 outof 10 for tutorials and 17 out of 25 for her mid semester esam. There are two options for calculation of her final mark. Both options give Annie a final mark of exactly 72%.

F=(t + a + 0.75e)
F=(t + a + m + 0.5e)

a)
Let a be Anne's assignment mark (out of 15) and let e be her final exam mark (out of 100). Write down an equation for her final mark calculation under each option::

For that I get:

72 = 9 + a + 0.75e
63 = a + 0.75e

and...

72 = 9 + a + 17 + 0.5e
46 = a + 0.5e

Is that right?

b) Use matrices to determine Anne's assignment mark and her final exam mark...

For that i get:

|1, 0.75|
|1, 0.5 |

and to find a and e you have to times the inverse and the fraction of ad-bc (-1/4) by 72 and 46. When I do this i get a and e to equal absolutlynothing near what they should be. Have i made a mistake somewhere??

 

[SpaceTiger]

and to find a and e you have to times the inverse and the fraction of ad-bc (-1/4) by 72 and 46.

I think you mean 63 and 46. That gives me 12 and 68 for a and e.

 

[thepatientmental]

a) Yes, your equations for both options are correct. The first equation represents the case where the final mark is calculated using the tutorial mark (t), assignment mark (a), and 75% of the exam mark (e). The second equation represents the case where the final mark is calculated using the tutorial mark (t), assignment mark (a), midterm exam mark (m), and 50% of the final exam mark (e). Both equations give a final mark of 72%, so either option can be used to calculate Anne's final mark.

b) In order to use matrices to determine Anne's assignment mark and final exam mark, we need to set up a system of equations. Using the first equation, we can write:

9 + a + 0.75e = 72

Similarly, using the second equation, we can write:

9 + a + 17 + 0.5e = 72

Simplifying both equations, we get:

a + 0.75e = 63
a + 0.5e = 46

We can then represent this system of equations in matrix form as:

|1, 0.75| |a| = |63|
|1, 0.5 | |e| = |46|

To solve this system, we can multiply both sides by the inverse of the coefficient matrix, which is:

|-2, 3|
|4, -4|

Multiplying both sides by this inverse, we get:

|-2, 3| |-2, 3| |1, 0.75| |a| = |-2, 3| |63|
|4, -4| x |1, 0.5 | x |e| = |4, -4| x |46|

Simplifying, we get:

|1, 0| |a| = |-72|
|0, 1| |e| = |152|

This gives us the solution:

a = -72
e = 152

However, these values are not realistic, as they are negative and greater than 100. This could indicate an error in the calculations or that the system of equations is inconsistent. It is important to double check the calculations to ensure accuracy.