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Exercise 4.3.27

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1.

A

is not invertible, we have

AC = [\det (C)] I = O

. It’s impossible that

C

is invertible otherwise

A = C^{- 1} O = O

. But the adjoint of the zero matrix

O

is also the zero matrix

O

, which is not invertible. So we know that in this case

C

is not invertible and hence

\det (C) = 0 = {[\det (A)]}^{n - 1}

. Next, if

A

is invertible, we have, by Exercise 4.3.25(c) that

\det (A) \det (C) = \det ([\det (A)] I) = {[\det (A)]}^{n}

So we know that

\det (C) = {[\det (A)]}^{n - 1}

since $\det (A) \neq 0$ .

2.

This is because

\det ({\tilde{A^{t}}}_{ij}) = \det (Ã_{ji}^{t}) = \det (Ã) .

3.

A

is an invertible upper triangular matrix, we claim that

c_{ij} = 0

for all

i

j

with

i > j

. For every

i

j

with

i > j

, we know that

c_{ij} = {(- 1)}^{i + j} \det (Ã_{ij}) .

But $Ã_{ij}$ is an upper triangular matrix with at least one zero diagonal entry if $i > j$ . Since determinant of an upper triangular matrix is the product of all its diagonal entries. We know that for $i > j$ we have $\det (Ã_{ij}) = 0$ and hence $c_{ij} = 0$ . With this, we know that the adjoint of $A$ is also an upper triangular matrix.

jephianlin

2011-06-27 00:00

Exercise 4.3.27

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