Standard Matrix - Linear Algebra - Quiz Solution, Exercises of Linear Algebra

This is the Quiz Solution of Linear Algebra. Mainly this Course includes Zero Vector, Linearly Dependent, Statement, Vector, Linear Combination, Expressed, Trivial Solution, Inspection, Dependent, Theorem etc. Key important points of his quiz are: Standard Matrix, Linear, Transformation, Defined, Vectors, Find, Explain, One to One, Columns, System

Typology: Exercises

2012/2013

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MATH 205A,B - LINEAR ALGEBRA
WINTER 2013
QUIZ 4
NAME: Section:(Circle one) A(1 : 10) B(2 : 40)
Show ALL your work CAREFULLY.
Let T:R3R2be a linear transformation defined by
T(x1, x2, x3) = (2x13x2+x3,4x3x1).
(a) Find the standard matrix Aof Tso that T(~x) = A~x.
The matrix Ahas columns T(~e1), T (~e2), T (~e3). Since T(~e1) = T(1,0,0) = (2,1),T(~e2) =
T(0,1,0) = (3,0),T(~e3) = T(0,0,1) = (1,4), it follows that the matrix Ais given by
A=23 1
104.
(b) Find all vectors ~x such that T(~x) = ~
0.
Since T(~x) = A~x, the vectors ~x such that T(~x) = ~
0are precisely the solutions to the
homogeneous equation A~x =~
0. Now,
A=23 1
10413 5
10413 5
03 91 0 4
03 9 1 0 4
0 1 3.
The solutions to A~x =~
0are
~x =
x1
x2
x3
=
4x3
3x3
x3
=x3
4
3
1
where x3is the parameter.
(c) Is Tone to one? Explain.
No. From (b), the homogeneous equation A~x =~
0has non-trivial solutions so Tcannot
be one-to-one.
Date: February 1, 2013.
1

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MATH 205A,B - LINEAR ALGEBRA

WINTER 2013

QUIZ 4

NAME: Section:(Circle one) A(1 : 10) B(2 : 40)

Show ALL your work CAREFULLY.

Let T : R^3 → R^2 be a linear transformation defined by

T (x 1 , x 2 , x 3 ) = (2x 1 − 3 x 2 + x 3 , 4 x 3 − x 1 ).

(a) Find the standard matrix A of T so that T (~x) = A~x.

The matrix A has columns T ( e~ 1 ), T ( e~ 2 ), T ( e~ 3 ). Since T ( e~ 1 ) = T (1, 0 , 0) = (2, −1), T ( e~ 2 ) = T (0, 1 , 0) = (− 3 , 0), T ( e~ 3 ) = T (0, 0 , 1) = (1, 4), it follows that the matrix A is given by

A =

[

]

(b) Find all vectors ~x such that T (~x) = ~0.

Since T (~x) = A~x, the vectors ~x such that T (~x) = ~ 0 are precisely the solutions to the homogeneous equation A~x = ~ 0. Now,

A =

[

]

[

]

[

]

[

]

[

]

The solutions to A~x = ~ 0 are

~x =

x 1 x 2 x 3

4 x 3 3 x 3 x 3

 (^) = x 3

where x 3 is the parameter.

(c) Is T one to one? Explain.

No. From (b), the homogeneous equation A~x = ~ 0 has non-trivial solutions so T cannot be one-to-one.

Date: February 1, 2013. 1