Solution:
e)
d)
c)
b)
a)
then , the third row and second column of C equals
If C = AB where
10.2.1. (001-Final-5 \$10.2)
Section 10.2
Thus,
Solve the system for a, b and c.
Note: We only need to compute the first row of the product and equate it to the first row of the identity matrix.
Solution:
e)
d)
c)
b)
a)
The sum x + y +z equals:
The matrix M and its inverse are given by
10.3.1 (001-Final-18 \$10.3)
Section 10.3
is obtained by multiplying the 3rd row of A by the second column of B.
and
or
==>
and
Case III (No solution)
==>
and
or
Substitute the given point into the equation of the parabola to get a system of 3 equations in 3 unknowns.
Solution:
e)
d)
c)
b)
a)
10.1.6 (002-Final-17 \$10.1)
==>
Solution:
e)
d)
c)
b)
a)
10.4.2 (002-Final-18 \$10.4)
==>
==>
2R1+R4
----------->
Solution:
e)
Solution:
e)
d)
c)
b)
a)
10.4.3 (002-Final-25 \$10.4)
Solve for x.
Compute the determinant.
d)
c)
b)
a)
10.3.2 002-Final-16 \$10.3
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==>
d)
c)
b)
a)
What is the value of the following determinant?
10.4.1 (001-Final-8 \$10.4)
Section 10.4
Thus,
Compute the reduce row echelon form of the augmented matrix to find
Augment the matrix A with the identity matrix
Solution:
e)
Use the rref command to compute the reduced row echelon form.
and
Thus,
This is the reduced row echelon form.
5R3+R1
==>
-3R3+R2
==>
This is the echelon form. We can now use back substitution to find x, y and z.
However, it is better to complete the reduction to find the reduced row echelon form and then read the solution from the reduce matrix as shown below.
-R2+R1
==>
-4R2+R3
then the system becomes
and
Let
Solution:
e)
d)
c)
b)
a)
10.1.3 (001-Final-26 \$10.1)
==>
-2R1+R2
Solution:
e) The system is consistent for c > 0.
d) The system is inconsistent for all values of c.
c) The system can be made consistent for a suitable choice of c.
b) The system is consistent for all , with exactly one solution.
a) The system is consistent if c = 0, with infinitely many solutions.
Which one of the following statements is TRUE about the linear system of equations which has the augmented matrix
10.1.1 (001-Final-12 \$10.1)
Section 10.1
A. M. Farhat
Solution of Old Exams Problems

Chapter 10
Math002
R1 <==> R3
-R1+R2
==>
-4R1+R3
Solution:
e)
d)
c)
b)
a)
10.1.2 (001-Final-22 \$10.1)
The system has a leading entry in the last column ==> the system has no solution regardless of the choice of c.
-R1+R3
==>
d) and
c) and
b)
a) and
10.1.5 (002-Final-15 \$10.1)
Not equal
(c)
Not equal
(b)
==>
and
Case II (Infinite number of solutions)
or
==>
Case I (Unique solution)
Reduce the augmented matrix.
-2R1+R2
Solution:
e)
==>
-2R3+R2
==>
-4R3+R1
==>
R2+R1
==>
-5R2+R3
R2<==>R3
-3R1+R2 ==>
Augmented matrix of the new system
(a)
Examples:
10.1.4 (002-Final-7 \$10.2 )
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