Math 225 Exam #1 questions with |\ |\ |\ |\ |\ |\
answers
Augmented matrix - CORRECT ANSWERS ✔✔a+b=c |\ |\ |\ |\ |\
d+e=f
[ a b c ; d e f]
|\ |\ |\ |\ |\ |\ |\
Row echelon form (REF) - CORRECT ANSWERS ✔✔1. All nonzero
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\
rows are above any rows of all zeros, called 0-rows
|\ |\ |\ |\ |\ |\ |\ |\ |\
2. Each leading entry (leftmost nonzero entry) of a row is in a
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
column to the right of the leading entry of the row above it
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
3. All entires in a column below a leading entry are zero
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
Reduced row echelon form - CORRECT ANSWERS ✔✔All the rules
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\
for row echelon for plus...
|\ |\ |\ |\
1. The leading entry in each nonzero row is 1
|\ |\ |\ |\ |\ |\ |\ |\ |\
2. Each leading 1 is the only nonzero entry in its column
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General solution of a SLE in parametric vector form - CORRECT
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ANSWERS ✔✔-set free variables equal to s and t |\ |\ |\ |\ |\ |\ |\ |\
-set up constant matrix, s matrix, and t matrix
|\ |\ |\ |\ |\ |\ |\ |\
ex.
x1-x4=-1
x2+x3+2x4=2
x3=s
, x4=t
x1=t-1
x2=2-s-2t
[x1 ; x2 ; x3 ; x4]= [-1 ; 2 ; 0 ; 0] + s*[0 ; -1 ; 1 ; 0] + t*[1 ; -2 ; 0
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
; 1]
|\ |\
Span of vectors in Rn - CORRECT ANSWERS ✔✔The Span{v1, v2,
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\
..., vp} is the set of all linear combinations of vectors v1, v2, ...,
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
vp. In other words, Span{v1, v2, ..., vp} is the collection of all
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
vectors that can be written in the form x1v1 + x2v2 + ... + xpvp
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
where x1, x2, ..., xp are all scalars |\ |\ |\ |\ |\ |\ |\
Linearly independent - CORRECT ANSWERS ✔✔a set of vectors|\ |\ |\ |\ |\ |\ |\ |\ |\
{v1, v2, ..., vp} in Rn where the vector equation x1v1 + x2v2 +
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
... + xpvp = 0 has only the trivial solution (x1=0, x2=0, & xp=0)
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
Linearly dependent - CORRECT ANSWERS ✔✔a set of vectors {v1,
|\ |\ |\ |\ |\ |\ |\ |\ |\
v2, ..., vp} in Rn where there are weights c1, ..., cp not all zero
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
such that the vector equation c1v1 + c2v2 + ... + cpvp = 0
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
How to determine linear dependence - CORRECT ANSWERS ✔✔-
|\ |\ |\ |\ |\ |\ |\ |\
Augment vectors with zero matrix |\ |\ |\ |\ |\
-If there are free variables, the set is linearly dependent
|\ |\ |\ |\ |\ |\ |\ |\ |\
-One vector is linearly independent
|\ |\ |\ |\
-For two vectors: a set of two vectors is linearly dependent if at
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
least one of the vectors is a multiple of the other. The set is
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
linearly independent if and only if neither of the vectors is a
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
multiple of the other |\ |\ |\
answers
Augmented matrix - CORRECT ANSWERS ✔✔a+b=c |\ |\ |\ |\ |\
d+e=f
[ a b c ; d e f]
|\ |\ |\ |\ |\ |\ |\
Row echelon form (REF) - CORRECT ANSWERS ✔✔1. All nonzero
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\
rows are above any rows of all zeros, called 0-rows
|\ |\ |\ |\ |\ |\ |\ |\ |\
2. Each leading entry (leftmost nonzero entry) of a row is in a
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
column to the right of the leading entry of the row above it
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
3. All entires in a column below a leading entry are zero
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
Reduced row echelon form - CORRECT ANSWERS ✔✔All the rules
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\
for row echelon for plus...
|\ |\ |\ |\
1. The leading entry in each nonzero row is 1
|\ |\ |\ |\ |\ |\ |\ |\ |\
2. Each leading 1 is the only nonzero entry in its column
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
General solution of a SLE in parametric vector form - CORRECT
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
ANSWERS ✔✔-set free variables equal to s and t |\ |\ |\ |\ |\ |\ |\ |\
-set up constant matrix, s matrix, and t matrix
|\ |\ |\ |\ |\ |\ |\ |\
ex.
x1-x4=-1
x2+x3+2x4=2
x3=s
, x4=t
x1=t-1
x2=2-s-2t
[x1 ; x2 ; x3 ; x4]= [-1 ; 2 ; 0 ; 0] + s*[0 ; -1 ; 1 ; 0] + t*[1 ; -2 ; 0
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
; 1]
|\ |\
Span of vectors in Rn - CORRECT ANSWERS ✔✔The Span{v1, v2,
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\
..., vp} is the set of all linear combinations of vectors v1, v2, ...,
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
vp. In other words, Span{v1, v2, ..., vp} is the collection of all
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
vectors that can be written in the form x1v1 + x2v2 + ... + xpvp
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
where x1, x2, ..., xp are all scalars |\ |\ |\ |\ |\ |\ |\
Linearly independent - CORRECT ANSWERS ✔✔a set of vectors|\ |\ |\ |\ |\ |\ |\ |\ |\
{v1, v2, ..., vp} in Rn where the vector equation x1v1 + x2v2 +
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
... + xpvp = 0 has only the trivial solution (x1=0, x2=0, & xp=0)
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
Linearly dependent - CORRECT ANSWERS ✔✔a set of vectors {v1,
|\ |\ |\ |\ |\ |\ |\ |\ |\
v2, ..., vp} in Rn where there are weights c1, ..., cp not all zero
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
such that the vector equation c1v1 + c2v2 + ... + cpvp = 0
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
How to determine linear dependence - CORRECT ANSWERS ✔✔-
|\ |\ |\ |\ |\ |\ |\ |\
Augment vectors with zero matrix |\ |\ |\ |\ |\
-If there are free variables, the set is linearly dependent
|\ |\ |\ |\ |\ |\ |\ |\ |\
-One vector is linearly independent
|\ |\ |\ |\
-For two vectors: a set of two vectors is linearly dependent if at
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
least one of the vectors is a multiple of the other. The set is
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
linearly independent if and only if neither of the vectors is a
|\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\ |\
multiple of the other |\ |\ |\
