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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 ).
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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
.
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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
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. 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 .
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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
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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.
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. 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 .
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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 ]
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. 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]
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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 ]
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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 ]
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(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
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(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
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