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These are the notes of Solved Exam of Linear Algebra which includes General Solution, Linear Systems, Homogeneous System, Solution Sets, Particular Solution, Nonhomogeneous, Coefficient Matrix etc. Key important points are: Augmented Matrix, Linear System, Values, Consistent, Corresponding, Reduction Algorithm, System is Consistent, General, Solution Set, Linear Combination
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The augmented matrix of a linear system has the form
[ a 1 1 2 a − 1 1
Determine the values of a for which the linear system is consistent.
We apply row-reduction algorithm to the augmented matrix corresponding to the system given above: Assume that a ̸ = 0, then we get
[ a 1 1 2 a − 1 1
( − 2 /a ) R 1 + R (^2) // ÏR 2
a 1 1 0 a − 1 −^2 a 1 −^2 a
By Theorem 2, we know that the system above is consistent if and only if there is no row of the form [0 0 1]. Therefore, we must have either a − 1 −^2 a ̸ = 0 or we must have a − 1 − (^) a^2 = 0 and 1 −^2 a = 0. Let us solve the equation a − 1 −^2 a = 0 or ( a + 1)( a − 2) = 0 or a = − 1 or a = 2. We need to examine the case a = 0. If a = 0, then we have x 2 = 1 and x 1 = 1. So, the system is consistent. Note that the case a = 2 also gives a consistent system. Finally, we conclude that the system above is consistent if and only if a ̸ = − 1.
Write the augmented matrix corresponding the system below:
x 1 − 6 x 2 − 4 x 3 = − 5 2 x 1 − 10 x 2 − 9 x 3 = − 4 −x 1 + 6 x 2 + 5 x 3 = 3_._
Solve the system by applying the row reduction algorithm. If the system is consistent, find the general solution set.
The augmented matrix corresponding to the given system is
We need to reduce the augmented matrix
R 1 + R 3 ÏR 3
R 3 + R 2 ↔R 2
(1 / 2) R 2 ↔R 2
x 1 = − 1 x 2 = 2 x 3 = − 2
A.) Write the given matrix equation below as system of linear equations:
x 1 x 2 x 3
x 1 + x 2 + x 3 = 1 x 1 − x 2 − 2 x 3 = − 5 2 x 1 + − 4 x 3 = 5
B.) Solve the system and write the general solution.
We need to reduce the augmented matrix that represents the given system (I’ll leave the details to you)
x 1 = − 7 / 2 x 2 = 15 / 2 x 3 = − 3
A. Solve the nonhomogeneous system Ax=b and write the solution in parametric vector form where
(^) and b =
− 2 R 2 + R 1 ÏR 1
3 R 3 + R 1 ↔R 1
− 2 R 1 + R 3 ↔R 3 ,−R 1 + R 2 ÏR 2
x 1 = − 2 x 2 = 7 x 3 = 4
B. Using the parametric vector form of the solution set in part A., determine a particular solution p.
We see that p =
(^) is a particular solution.
C. Write the general solution for the system A x = 0 in parametric vector form.
The parametric vector form of homogeneous part of the general solution set is
vh =