Many employees screen gpa in addition to a variety of skills. Suppose a student is in the last semester of college and has a 2.6
1 answer:
Answer:
- Overall GPA=2.81
- It is not possible to get his GPA to 3.0 for graduation.
Step-by-step explanation:
The Student already has a GPA of 2.68 after 108 credit hours.
If he is taking 12 credit hours in his last semester and gets a perfect 4.0 GPA
Total Credits Earned Before = 2.68 X 108=2894.4
Projected Credit to be earned = 12 X 4= 48 Credits
Total credit Hour= 108+12=120 Hours
His Cumulative GPA = Total credits earned ÷ Total Credit Hour
Since 2.81 is less than 3.00, it is not possible to get his GPA to 3.0 for graduation.
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Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.
Step-by-step explanation:
To determine if solution exist or not, you test the equation for consistency.
A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.
Consider the following scenarios:
(1) For example:Given the matrix A=, to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:
Let A transpose be B.
∵B=
the system Bx=0 can be represented in matrix form as:
=
................................eq(1)
Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,
|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).
Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have :
=
. The rank of
can be found by using the second column and third column pair as follows:
||=(3*0)-(0*2)=0 i.e,
is a singular matrix with rank of order 1.
Note: a matrix is singular if its determinant is = 0 and non-singular if it is 0.
Comparing the rank of both B and , it is obvious that
Rank of BRank of
since (-2)<1.
Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution. </em>
(2) If B= is the transpose of matrix A=
, then
Then the equation Bx=0 is represented as:
=
..................................eq(2)
|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.
=
,
||=(5*0)-(0*10)=0-0=0 i.e
has a rank of order 1.
we can therefor conclude that since
rank B=rank =1, equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (
and
).
<u>Summary:</u>
- Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
- Determine the rank of both the coefficients matrix B and
which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
- If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.