Augmented Matrix - Applied Linear Algebra - Exam Key, Exams for Linear Algebra. Amity Business School
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alaknanda18 February 2013

Augmented Matrix - Applied Linear Algebra - Exam Key, Exams for Linear Algebra. Amity Business School

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This is the Exam Key of Applied Linear Algebra which includes Homogeneous Coordinates, Linear Equations, Precise Description, Linearly Independent etc. Key important points are: Augmented Matrix, System, Corresponding, ...
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Math 232 First Midterm Solution February 5, 2007

1. We consider the following system of equations: x1 +x2 +2x3 −4x4 = 1 x1 +2x2 +x3 +x4 = 2 2x1 +4x2 +2x3 −x4 = 1

(a) (1 point) Write down the augmented matrix corresponding to this system.

Answer  1 1 2 −4 11 2 1 1 2 2 4 2 −1 1



(b) (3 points) Determine the reduced row echelon form of this matrix. Show your work. (use the back of the previous page if you need more room)

Answer  1 1 2 −4 11 2 1 1 2 2 4 2 −1 1

 ∼  1 1 2 −4 10 1 −1 5 1

0 2 −2 7 −1

 ∼  1 1 2 −4 10 1 −1 5 1

0 0 0 −3 −3

 ∼  1 1 2 −4 10 1 −1 5 1

0 0 0 1 1

 ∼  1 1 2 0 50 1 −1 0 −4

0 0 0 1 1

 ∼  1 0 3 0 90 1 −1 0 −4

0 0 0 1 1



First Midterm Solution, Math 232, Spring 2007 Page 2 of 5

(c) (2 points) Write down the full solution set to the system in parametric form.

Answer

From part (b) we see that the given system is equivalent to x1 = 9− 3x3 x2 = −4 + x3 x3 free variable x4 = 1

Hence, in parametric form, the set of solution vectors

 x1 x2 x3 x4

 is:  

9 −4 0 1

+ x3  −3 1 1 0

 : x3 ∈ R 

(d) (2 points) Let b =

12 1

 and a1 = 11 2

 , a2 = 12 4

 , a3 = 21 2

 , a4 = −41 −1

. Can you write b as a linear combination of a1, . . . , a4? Motivate your answer and, if it is “yes”, give such a linear combination. [HINT: Relate the given vectors to the system given in (a)]

Answer

In vector notation, the system in this question is

x1a1 + x2a2 + x3a3 + x4a4 = b

hence, any solution (x1, x2, x3, x4 to this system gives a way of expressing b as a linear combination of the ai. We have determined all solutions in (c), so we can for instance write:

9a1 − 4a2 + 0a3 + 1a4 = b

First Midterm Solution, Math 232, Spring 2007 Page 3 of 5

2. (a) (2 points) Give the definition of linear independence for a set {v1, . . . ,vr} of vectors in Rn. Your answer should start with: “A set of vectors {v1, . . . ,vr} is called linearly independent if. . . ”

Answer

A set of vectors {v1, . . . ,vr} is called linearly independent if the only way to express the zero-vector 0 as linear combination of v1, . . . ,vr is by taking all weights 0, i.e., the only solution to the vector equation

x1v1 + · · ·+ xrvr = 0

is x1 = x2 = · · · = xr = 0.

(b) (3 points) Can a set of r vectors in Rn ever be linearly independent if n < r? Prove your statement.

Answer

No. As stated above, the vector equation x1v1 + · · · + xrvr = 0 would have to have a unique solution, i.e., no free variables.

However, the matrix corresponding to this system would have n rows and r columns. The row echelon form has at most one pivot per row. Since r > n, there are more columns than rows, so there is a column without a pivot. This corresponds to a free variable.

(c) (4 points) Determine if the set

 13 2

 ,  1−1

0

 , 62 4

 is linearly independent. Show that your answer is correct.

Answer

We test if the equation

x1

13 2

+ x2  1−1

0

+ x3 62 4

 = 00 0

 has a unique solution:1 1 63 −1 2

2 0 4

 ∼ 1 1 60 −4 −16 0 −2 −8

 ∼ 1 1 60 1 2 0 0 0

 We see that the corresponding echelon form does not have a pivot in every column, the system does not have a unique solution and therefore the set is not linearly independent.

First Midterm Solution, Math 232, Spring 2007 Page 4 of 5

3. (a) (4 points) Determine the inverse of the matrix. Show your work.

M =

2 0 10 1 0 1 0 1



Answer  2 0 1 1 0 00 1 0 0 1 0 1 0 1 0 0 1

 ∼  1 0 1 0 0 10 1 0 0 1 0

2 0 1 1 0 0

 ∼  1 0 1 0 0 10 1 0 0 1 0

0 0 −1 1 0 −2

 ∼  1 0 1 0 0 10 1 0 0 1 0

0 0 1 −1 0 2

 ∼  1 0 0 1 0 −10 1 0 0 1 0

0 0 1 −1 0 2

 Hence,

M−1 =

 1 0 −10 1 0 −1 0 2



(b) (3 points) Prove that, for an invertible n× n matrix A, the linear transformation T : Rn → Rn defined by T (x) = Ax, is onto.

Answer

A transformation is called onto if for any vector b ∈ Rn (the codomain), we can find a vector x in the domain such that T (x) = b. Therefore, in order to show that T is onto, we need to show that for any b ∈ Rn, the equation Ax = b has a solution. If A is invertible, we can take x = A−1b. Then Ax = A(A−1b) = AA−1b = Inb = b, so indeed T is onto if A is invertible.

First Midterm Solution, Math 232, Spring 2007 Page 5 of 5

4. Mark true or false and give a reason.

(a) (2 points) A linear transformation T : Rn → Rm can never be one-to-one if n > m.

Answer

True. Let A be the standard matrix of T , so that T (x) = Ax. The matrix A will be m × n. If n > m, this means that the row echelon form of A will have columns without pivot. Hence, the system Ax = 0 has free variables and therefore a non-trivial solution, say v 6= 0. It follows that T (v) = Av = 0 and T (0) = A0 = 0. It follows that two distinct vectors have the same image under T and therefore T is not one-to-one.

(b) (2 points) The following map is linear:

R2 → R2( x1 x2

) 7→

( x1 + x2 x1 − x2

)

Answer

True. We call the transformation T . We need to show that for all v,w ∈ R2 and c ∈ R we have T (v + bw) = T (v) + T (w) and T (cv) = cT (v):

T (v+w) = T

( v1 + w1 v2 + w2

) =

( (v1 + w1) + (v2 + w2) (v1 + w1)− (v2 + w2)

) =

( v1 + v2 v1 − v2

) +

( w1 + w2 w1 − w2

) = T (v)+T (w)

T (cv) = T

( cv1 cv2

) =

( cv1 + cv2 cv1 − cv2

) = c

( v1 + v2 v1 − v2

) = cT (v)

(c) (2 points) A matrix transformation is always an linear transformation.

Answer

True. A matrix transformation is of the form T (x) = Ax. The fact that A(v + w) = Av + Aw and A(cv) = cAv are properties of matrix multiplication.

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