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Super Augmented - Linear Algebra - Quiz, Exercises of Linear Algebra

This is the Quiz of Linear Algebra which includes Zero Vector, Linearly Dependent, Statement, Vector, Linear Combination, Expressed, Trivial Solution, Inspection, Dependent, Theorem etc. Key important points are: Super Augmented, Matrix Technique, Conditions, Guarantee, Ensure, Value, Columns, Specific Linear Combination, Linear Combination, Actual Values

Typology: Exercises

2012/2013

Uploaded on 02/27/2013

senapathy_101
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Math 205A Quiz 02 page 1 September 19, 2008 NAME
1. Let A=
52512
6113 13
20 6 4
and let b=
b1
b2
b3
.
1A. Use the “super-augmented” matrix technique discussed in class to find what conditions (if any)
there are on b1,b2, and b3which guarantee that Ax=bhas a solution.
1B. What value must b1have to ensure that w=
b1
13
12
can be written as a linear combination of the
columns of A?
1C. Express wfrom (1B) as a specific linear combination of the columns a1,a2,etc. of A. Your answer
will have the form w=x1a1+x2a2+...
00 (etc) where you’ll supply the actual values of x1,x2,etc..
1D. Let z=
43
11
6
. Express all solutions of Ax=zin the form p+vhwhere pis a particular solution
of Ax=zand vhrepresents all solutions of Ax=0.
1E. Explicitly give the “trivial solution” of Ax=0.

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Math 205A Quiz 02 page 1 September 19, 2008 NAME

  1. Let A =

 (^) and let b =

b 1 b 2 b 3

1A. Use the “super-augmented” matrix technique discussed in class to find what conditions (if any) there are on b 1 , b 2 , and b 3 which guarantee that Ax = b has a solution.

1B. What value must b 1 have to ensure that w =

b 1 − 13 12

 (^) can be written as a linear combination of the

columns of A?

1C. Express w from (1B) as a specific linear combination of the columns a 1 , a 2 , etc. of A. Your answer will have the form “w = x 1 a 1 + x 2 a 2 +.. .′′^ (etc) where you’ll supply the actual values of x 1 , x 2 , etc..

1D. Let z =

. Express all solutions of Ax = z in the form p + vh where p is a particular solution

of Ax = z and vh represents all solutions of Ax = 0.

1E. Explicitly give the “trivial solution” of Ax = 0.