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Material Type: Exam; Professor: Binegar; Class: LINEAR ALGEBRA; Subject: Mathematics ; University: Oklahoma State University - Stillwater; Term: Summer 2007;
Typology: Exams
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Math 3013
SAMPLE FIRST EXAM
(with solutions)
R 4 .
of the triangle correspond to the vectors
Since
the first two sides must be perpendicular. Hence the points A, B and C form the vertices of a
right triangle in R 4
x 1 − x 2 + 2x 3 = 1
2 x 1 + x 2 + x 3 = 5
−x 1 + x 2 + 2x 3 = − 1
x 2 + x 3 = 1
(a) (5 pts) Write down the corresponding augmented matrix and reduce it to row-echelon form.
(b) (5 pts) Reduce the augmented matrix further to reduced row-echelon form.
1 3
1 4
(c) (5 pts) Write down the solution of the original linear system.
1
, inverse: and verify that you have the correct inverse by showing AA
− 1 = I.
(^) and AA − 1 =
3in
to solve
x 1
x 2
x 3
x =
x 1
x 2
x 3
(^) = A−^1 b =
contain pivots. The linear system corresponding to the matrix equation Ax = 0 is
x 1 + x 2 + 2x 4 = 0
x 3 + 3x 4 = 0
0 = 0
0 = 0
x 1 = −x 2 − 2 x 4
x 3 = − 3 x 4
x =
−x 2 − 2 x 4
x 2
− 3 x 4
x 4
= x 2
Therefore, {[− 1 , 1 , 0 , 0] , [− 2 , 0 , − 3 , 1]} will be a basis for the null space of A
F (a) If A and B are invertible n × n matrices, then AB = BA implies B = A
− 1 .
T (b) A square linear system Ax = b has a unique solution if and only if A is row-equivalent to the
identity matrix.
T (c) If A, B and C are invertible n × n matrices, then AC = BC implies A = B.
T (d) If a linear system Ax = b has infinitely many solutions, then it’s possible that another linear
Ax = b
′ (with b
′ 6 = b) will has no solution.
F (e) If a consistent linear system has more equations than unknowns, then there will be a unique
solution.