midterm 1 solutions - ucsd mathematicsjinovak/d.math18/math18_solutions_1.pdfmidterm 1 solutions....

Midterm 1 Solutions

Remarks

. The exam is graded out of

30points .

. The Max . score is 34points ( up to 4 bonus points ).

. These solutions arefor Version A

(Solutions for Versions B. Care nearly identical ).

Free Plain Graph Paper from http://incompetech.com/graphpaper/plain/

Problem I.(a) Find all real numbers b.

,b.

,b such that the following

system is consistent :

X ,-3×2 + 2×3 - Xy + 2 × ,

= b,

3×,

- 9×2+7×3 - Xy + 3 xs = bz

Zx,

- 6×2+7×3 + 4×4 - 5 xs= b3

.


Solution :

. The augmented matrix of the system is [ 1 point ]

1 -3 2 - 1 2 b,k:3::b;t

. Use a sequenceof elementary row operations to bring this matrix into

row echelon form [ 3 points ]

I -3 2 - I 2 b,

| }If}a}bb;) ~ ( 01821-2'

-23k¥row 2- 3. rowi.

0 0 3 6 - 9 b , - 2b, row 3 - 2. row

I -3 2 - 1 2 b

foo;to2083215:#rows . 3. rowz


. The matrix

go-32,13!:b . )0 0 0 0 0 7b

,- Zbztb

,

is the augmented matrix of a consistent system if and only if b, ,b< ,b,

satisfy

7b, -34+4=0 .

[ I point ]

. We conclude that our original system is consistent if and only if

7b, -34+4=0 .


Problem 1. (b ) Writeyour answer to part (a) in the form Span { Iii } .

Solution :

. The consistency set from part (a) is

{ ¥,

) e R3 : 7b,-3 b. + b. =o ) = { (bhnb;] :b

" bi e R) [ I point I

= §, (g) + bz|°g| :b " b. c- R} [ 1 point I

= Span {µ,

µ).

H point I


Problem I.

(c) Solve the system

X ,-3×2 + 2×3 - Xy t 2 × ,

= 2

3×,

- 9×2+7×3 - Xy + 3 xs = 7

Zx,

- 6×2+7×3 + 4×4 - 5 xs= 7

.

Solution :

. The augmented matrix of the system is [ I point]

I - 3 2 - 1 2 2

k:3:3't. By our computation in part (a) , [ I point]

I -3 2 - 1 2 2 Z

k:::S t.fm::p0 0 0 0 0 0.


. This matrix is in row echelon form - run backward elimination to getreduced row echelon form

,[ 1 point ]

1 -3 2 - 1 2 2 1 -3 0 -5 8 0 row 1- 2. row 2Kgbfoot - fool2-31 )0 0 0 0 0 0

. From the RREF,

we see that our original system has the same

solution set as the system [ 1 point 1

X ,-3×2-5×4+8×5=0

At 2×4-3×5 = I.

i Free variables are Xz,Xu

,xs ; solution set in parametric form is [ 1 point ]

X , =3 s + St -

8u, x. =s

,xs=t2t+3u ,

xy=t, xs=u ,

s,t.uc.IR .


Problem 2.

(a) Show that the vectors

HTHTH

are linearly independent .

Solution :

. We need to show that the only solution of the homogeneous vector

equation

1 2 2 0

xfzfaffxfitt:/4 1 -1

is the trivial solution,

×,

= x. =b= 0 .

[ I point ]


. The augmented matrix is [ I point ]

I 2 2 0

| 2 -3 IO|4 I - I 0

. Use row operations to find row echelon form:[ 2 points ]

f; } }f) ft:::o) nwz . zrowi

0 - 7 - 9 row 3 - 4. row I

I 2 2 0

foot}:| rows . row

. Sinceevery

column is pivotal , the system has a unique solution , which mustbe the trivial solution

.

[ 1 point]


Problem 2. ( b ) How manysolutions does the matrix equation

kiHl¥H:Dhave ?

Solution :

. By part (a) ,the columns of the coefficient matrix are linearly

independent .

Therefore,if the equation is consistent

,it has

a unique solution .

[ 2 points ]

. By inspection,

eat .is a solution

,so the equation has

a unique solution .

[ 2 points ]


Problem 3.Consider the linear transformation T:R3→R3 defined by

the,H¥t(a) Show that T is a linear transformation

.

Solution :

. Forany

vector E = x. E + x. E' + xsej ,we have

The ) = TC x. E'

+ xiez + a E) = BE + xzei + x.

E,

by the definition of T.

[ I point ]

. But be + xzei + x.

E= x. TIE ) + xz TIE

' ) +x.

TCE' ).

[1 point]

. Therefore T( x. E + x. Etx , E's )= x. TIE ) + x. TIE ) + x. Ties ,

so T is a linear transformation of R3.

[ I point ]


(b)Find a 3×3 matrix A such that This = Ax for all EER ?

Solution :. The matrix A is given by

A=[ HE ) TIEI TIE ) ] [ 1 point ]

= [ E E E' ] [ 0.5 pH

= fq0g to ].

[ 0.5 pH


(c) Is A an invertible matrix ? Explain .

Solution :

. Note that T is an invertible transformation - indeed,

To T = I,the identity transformation :

T( T ( ¥;)) )=T((4) = ( ¥,

) [ 2 points ]

. Since the matrix of T is A,the matrix of IT is AA

.

[ I point]

. Since T°T=I,the matrix of IT is the identity matrix I [1 pointI

. Combining these statements,

we get AA=I.

Thus A is invertible,

and At = A.

[ 1 point7

midterm 1 solutions - ucsd mathematicsjinovak/d.math18/math18_solutions_1.pdfmidterm 1 solutions....

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