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# Name: AMAT 220: Linear Algebra Practice Exam March, 2021 Show all work for each problem in the space provided. If you run out of room for an answer, continue on the back of the page. You may NOT use a...

Name:
AMAT 220: Linear Alge
a
Practice Exam
March, 2021
Show all work for each problem in the space provided. If you run out of room for an answer, continue on
the back of the page. You may NOT use a calculato
Question Points Bonus Points Score
1 0 0
2 0 0
3 0 0
4 0 0
5 0 0
6 0 0
7 0 0
8 0 0
9 0 0
Total: 0 0
1. Define what both a linear transformation is and the span of a set of vectors.
2. Find the general solution of the linear system
3x1 âˆ’ 4x2 + 2x3 = 0 (1)
âˆ’9x1 + 12x2 âˆ’ 6x3 = 0 (2)
âˆ’6x1 + 8x2 âˆ’ 4x3 = 0 (3)
OR
x1 âˆ’ 7x2 + 6x4 = 5 (4)
x3 âˆ’ 2x4 = âˆ’3 (5)
âˆ’x1 + 7x2 âˆ’ 4x3 + 2x4 = 7 (6)
y row reducing the co
esponding augmented matrix and interpreting your result in terms of the co
esponding
linear system.
3. Show by computation whether the given vector b âˆˆ Span{a1, a2, . . . , an}, where ai are vectors. Specifically,
show whethe
=
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
âˆ’9
âˆ’7
âˆ’15
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
is contained in the spanning set
Span{a1 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
2
1
4
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» a2 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
1
âˆ’1
3
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» a3 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
3
2
5
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»}
OR
=
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
9
2
7
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
is contained in the spanning set
Span{a1 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
1
2
âˆ’1
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» a2 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
6
4
2
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»}
4. Describe in parametric form all solutions of the system Ax = 0 for the given A.
A =
ï£®ï£°2 1 3
1 2 0
ï£¹ï£»
OR
A =
ï£®ï£°3 1 1 1
5 âˆ’1 1 1
ï£¹ï£»
5. Determine if the given set of vectors are linearly dependent:
S = {a1 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
âˆ’2
0
1
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» , a2 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
3
2
5
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» , a3 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
6
âˆ’1
1
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» , a4 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
7
0
2
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»}
OR
S = {a1 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
8
âˆ’1
3
ï£¹ï£ºï£ºï£ºï£ºï£ºï£» , a2 =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
4
0
1
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»}
6. Compute the standard matrix A associated to the given linear transformation T : Rn â†’ Rm, where
T (x1, x2) = (x1 + 2x2, 3x1 âˆ’ x2)
OR
T (x1, x2, x3) = (2x1 âˆ’ x2 + x3, x2 âˆ’ 4x3)
7. Compute the matrix product AB for the given A and B.
A =
ï£®ï£°1 2 4
2 6 0
ï£¹ï£»
and
B =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
4 1 4 3
0 âˆ’1 3 1
2 7 5 2
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
OR
A =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
6 1 3
âˆ’1 1 2
4 1 3
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
and
B =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
3 0
âˆ’1 2
1 1
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
8. Compute the inverse of the given matrix A by the algorithm for computing a matrixâ€™s inverse; that is, row
educe the matrix of A augmented by I.
A =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
1 2 3
2 5 3
1 0 8
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
OR
A =
ï£®ï£¯ï£¯ï£¯ï£¯ï£¯ï£°
3 4 âˆ’1
1 0 3
2 5 âˆ’4
ï£¹ï£ºï£ºï£ºï£ºï£ºï£»
9. Use your answer from the question 8. to find a solution x to the linear system
x1 + 2x2 + 3x3 = 7 (7)
2x1 + 5x2 + 3x3 = 5 (8)
x1 + 8x3 = 8 (9)
RESPECTIVELY
3x1 + 4x2 âˆ’ x3 = 7 (10)
x1 + 3x3 = 5 (11)
2x1 + 5x2 âˆ’ 4x3 = 8 (12)
Respectively means apply the inverse of the first answer in 8 to solve the first problem in this question and to
apply the second answer in 8 to solve the second problem in this question.
Answered 1 days After Mar 27, 2022

## Solution

Rajeswari answered on Mar 29 2022
102779 assignment
1. A linear transformation is a function from one vector space to another where the variables in the second vector space can be expressed as a linear combination of vectors in the first one.
It is called linear because it preserves all other properties of the variables ,
Span of a set of vectors is the set of all linear combinations of those vectors.
2. a. Write in matrix form as:
* =0
Doing linear trans as R3= R3+2R1 and R2 = R2+3R1 we have
* =0
i.e. 3x-4y+2z =0 is the equation and all points lying on the plane are solution
General solution is vectors of the form (x y )
. Augmented matrix we write first and do linear transformations

1
-7
0
6
5

0
0
1
-2
-3

-1
7
-4
2
7

R3+R1 =R3
1
-7
0
6
5

0
0
1
-2
-3

0
0
-4
8
12

R3+4R2=R3
1
-7
0
6
5

0
0
1
-2
3

0
0
0
0
0
i.e the general solution is all points lying on intersection of x1-7x2+0x3+6x4=5 and x3-2x4 =3
If we give values for x3 =s then x4 = (s-3)/2 . Also give value for x1 = t
Then t-7x2+3s-9 =5
Or x2 =-...
SOLUTION.PDF