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Midterm

MATH240 BOYLE-M SPRING2011 0101 MID SOL 2Exam


Department
Mathematics
Course Code
MATH 240
Professor
All
Study Guide
Midterm

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MATH 240 – Spring 2011 – Exam 3 Solutions
1. Compute the characteristic polynomial of the matrix A=
0 1 2 2
0 1 2 2
0 0 0 2
0 0 0 2
.SOLUTION.
This polynomial p(x) is x2(x1)(x2) = x43x3+ 2x2.
2. Let Abe the matrix
A=
1 2
2 0
11
.
Compute the area of the parallelogram in R3spanned by the columns of A.
SOLUTION.
Atr A= 1 2 1
2 0 1!
1 2
2 0
11
= 6 1
1 5!
Therefore the area of the paralleogram is
qdet(Atr A) = 29 .
3. For the matrix A=
1 2
2 0
11
, find all least squares solutions for Ax =
1
2
3
.
SOLUTION.
The least squares solutions for Ax =
1
2
3
are the solutions to Atr Ax =Atr
1
2
3
.
We solve:
1 2 1
2 0 1!
1 2
2 0
11
x1
x2!=
1 2
2 0
11
1
2
3
6 1
1 5! x1
x2!= 8
1!
x1
x2!= 6 1
1 5!1 8
1!
x1
x2!=1
29 51
1 6 !1 8
1!
=1
29 41
14!.
(The least squares solution is unique in this problem, because Atr Ais invertible – which
we know in advance is guaranteed when Ais n×kwith knand Ahas rank k, as for
this matrix Awith k= 2 and n= 3.)
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