Linear Algebra Study Questions for Exam 1

[warwick]

7. It seems like there was some mathematical trickery here - as my Cal 2 professor would say. Instead of writing the solution in terms of the free variable, x3, they wrote it in terms of the basic variable, x2. Is this correct?

10. It's independent because there is a pivot in every column of the coefficient matrix of Ax=0 meaning that the only solution is the trivial one.

a)

2 3 0
1 0 -1
0 1 2

~

1 0 -1
0 1 -2
0 0 1

b) It's dependent because there is not a pivot in every column which yields a free variable in the coefficient matrix of Ax=0 meaning that there is a infinite number of non-trivial solutions.

1 -1 0
0 1 1
1 0 1

~

1 0 1
0 1 1
0 0 0

15. I'm also not sure how to approach this.

 

[pka]

Here is help on #15.
\(\displaystyle \left[ \begin{gathered} 1 \hfill \\ - 1 \hfill \\ \end{gathered} \right] = \frac{{ - 2}}{3}\left[ \begin{gathered} - 1 \hfill \\ 2 \hfill \\ \end{gathered} \right] + \frac{1}{3}\left[ \begin{gathered} 1 \hfill \\ 1 \hfill \\ \end{gathered} \right]\)
Now apply the linear transformation.

 

[stapel]

For those others in whose browsers the wide graphics do not display, the text is as follows:

\(\displaystyle \mbox{[7] Let } T:R^3\, \rightarrow\, R^2\, \mbox{ be defined by }\, T\left( \overset{\rightarrow}{x} \right)\, =\, A \overset{\rightarrow}{x},\)

. . . . .\(\displaystyle \mbox{where }\, A \, =\, \left[\begin{array}{rrr}-1&2&0\\1&1&-1\end{array}\right]\)

. . . . .\(\displaystyle \mbox{(a) Is the vector } \left[\begin{array}{rr}1\\-1\end{array}\right]\, \mbox{in the range of }\, T\, \mbox{?}\)

. . . . .\(\displaystyle \mbox{(b) Find all } \overset{\rightarrow}{x}\, \mbox{in } R^3\, \mbox{such that } T\overset{\rightarrow}{x}\, =\, \overset{\rightarrow}{0}.\)

\(\displaystyle \mbox{[10] Suppose that the vectors } \overset{\rightarrow}{u},\, \overset{\rightarrow}{v},\, \overset{\rightarrow}{w}\, \mbox{are linearly independent.}\)

. . . . .\(\displaystyle \mbox{Determine if each of the following sets of vectors is linearly }\)
. . . . .\(\displaystyle \mbox{independent or dependent.}\)

. . . . .\(\displaystyle \mbox{(a) } 2\overset{\rightarrow}{u}\, + 3\overset{\rightarrow}{v}, \, \overset{\rightarrow}{u}\, -\, \overset{\rightarrow}{w},\, \overset{\rightarrow}{v}\, +\, 2\overset{\rightarrow}{w}\)

. . . . .\(\displaystyle \mbox{(b) } \overset{\rightarrow}{u}\, -\, \overset{\rightarrow}{v},\, \overset{\rightarrow}{v}\,+\, \overset{\rightarrow}{w}, \, \overset{\rightarrow}{u}\, +\, \overset{\rightarrow}{w}\)

\(\displaystyle \mbox{[15] Suppose that } T:R^2\, \rightarrow\, R^3\, \mbox{is a linear transformation such that:}\)

. . . . .\(\displaystyle T\left({\left[\begin{array}{r}-1\\2\end{array}\right]}\right)\, =\, \left[\begin{array}{r}-1\\2\\1\end{array}\right]\, \mbox{and } T\left({\left[\begin{array}{r}1\\1\end{array}\right]}\right)\, =\, \left[\begin{array}{r}0\\-1\\3\end{array}\right]\)
. . . . .\(\displaystyle \mbox{Calculate } T\left({\left[\begin{array}{r}1\\-1\end{array}\right]}\right)\)

 

[warwick]

Here is help on #15.
\(\displaystyle \left[ \begin{gathered} 1 \hfill \\ - 1 \hfill \\ \end{gathered} \right] = \frac{{ - 2}}{3}\left[ \begin{gathered} - 1 \hfill \\ 2 \hfill \\ \end{gathered} \right] + \frac{1}{3}\left[ \begin{gathered} 1 \hfill \\ 1 \hfill \\ \end{gathered} \right]\)
Now apply the linear transformation.

I was thinking about doing something like that. I just didn't think to setup a matrix. Durrr. lol. Thanks for the hint.

T(1, -1) = ( 2/3,-5/3, 1/3)

Is my observation correct on number seven?