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Linear Algebra

Exercise 4.1.1

Answers

1.

No. We have

\det (2 I_{2}) = 4

but not

2 \det (T_{2})

.

2.

Yes. Check that

\det (\begin{matrix} a_{1} + k a_{2} & b_{1} + k b_{2} \\ c & d \end{matrix}) = (a_{1} d - b_{1} c) + k (a_{2} d - b_{2} c)

= \det (\begin{matrix} a_{1} & b_{1} \\ c & d \end{matrix}) + k \det (\begin{matrix} a_{2} & b_{2} \\ c & d \end{matrix})

and

\det (\begin{matrix} a & b \\ c_{1} + k c_{2} & d_{1} + k d_{2} \end{matrix}) = (a d_{1} - b c_{1}) + k (a d_{2} - b c_{2})

= \det (\begin{matrix} a & b \\ c_{1} & d_{1} \end{matrix}) + k \det (\begin{matrix} a & b \\ c_{2} & d_{2} \end{matrix})

for every scalar $k$ .

3.

No.

A

is invertible if and only if

\det (A) \neq 0

.

4.

No. The value of the area cannot be negative but the value

\det (\begin{matrix} u \\ v \end{matrix})

could be.

5.

Yes. See Exercise 4.1.12.

jephianlin

2011-06-27 00:00

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