These summarize the notation used in this book for the
- and
- permutations.
- This exercise is recommended for all readers.
- This exercise is recommended for all readers.
- This exercise is recommended for all readers.
- Problem 3
Use the permutation expansion formula to derive
the formula for
determinants.
- Answer
Following Example 3.6 gives this.

- Problem 4
List all of the
-permutations.
- Answer
This is all of the permutations where

the ones where

the ones where

and the ones where
.

- Problem 6
Prove that
is multilinear if and only if for all
and
, this holds.

- Answer
For the "if" half, the first condition of
Definition 3.2 follows from taking
and the second condition follows from taking
.
The "only if" half also routine.
From
the first condition of Definition 3.2 gives
and the second condition, applied twice, gives the result.
- Problem 7
Find the only nonzero term in the permutation expansion of
this matrix.

Compute that determinant by finding the signum of the associated
permutation.
- Answer
To get a nonzero term in the permutation expansion we must use
the
entry and the
entry.
Having fixed on those two we must also use the
entry and
the
entry.
The signum of
is
because from

the two row swaps
and
will produce the identity matrix.
- Problem 8
How would determinants change if we changed property (4) of the
definition to read that
?
- Answer
They would all double.
- Problem 9
Verify the second and third
statements in Corollary 3.13.
- Answer
For the second statement, given a matrix, transpose it, swap rows, and transpose back. The result is swapped columns, and the determinant changes by a factor of
. The third statement is similar: given a matrix, transpose it, apply multilinearity to what are now rows, and then transpose back the resulting matrices.
- This exercise is recommended for all readers.
- Problem 11
True or false: a matrix whose entries are only zeros or ones has a determinant equal to zero, one, or negative one. (Strang 1980)
- Answer
False.

- Problem 12
- Show that there are
terms in the permutation
expansion formula of a
matrix.
-
How many are sure to be zero if the
entry is zero?
- Answer
- For the column index of the entry in the first row there are
five choices.
Then, for the column index of the entry in the second row there
are four choices (the column index used in the first row cannot
be used here).
Continuing, we get
.
(See also the next question.)
- Once we choose the second column in the first row,
we can choose the other entries in
ways.
- Problem 13
How many
-permutations are there?
- Answer
- This exercise is recommended for all readers.
- Problem 15
What is the smallest number of zeros, and the placement of
those zeros, needed to ensure that a
matrix has a
determinant of zero?
- Answer
Showing that no placement of three zeros suffices is routine. Four zeroes does suffice; put them all in the same row or column.
- This exercise is recommended for all readers.
- Problem 16
If we have
data points
and want to find a
polynomial
passing through those points
then we can plug in the points to get an
equation/
unknown
linear system.
The matrix of coefficients for that system is called the
Vandermonde matrix. Prove that the determinant of the transpose
of that matrix of coefficients

equals the product, over all indices
with
, of terms of the form
.
(This shows that
the determinant is zero, and the linear system has no solution,
if and only if the
's in the data are not distinct.)
- Answer
The
case shows what to do.
The pivot operations of
and
give this.

Then the pivot operation of
gives
the desired result.

- Problem 17
A matrix can be divided into blocks, as here,

which shows four blocks, the square
and
ones
in the upper left and lower right, and the zero blocks in the
upper right and lower left.
Show that if a matrix can be partitioned as

where
and
are square, and
and
are all zeroes,
then
.
- Answer
Let
be
,
let
be
,
and let
be
.
Apply the permutation expansion formula


Because the upper right of
is all zeroes, if a
has at least one of
among its first
column numbers
then the term arising
from
is
(e.g., if
then
is
).
So the above formula reduces to a sum over all permutations
with two halves:
first
are rearranged, and after that comes
a permutation of
.
To see this gives
, distribute.
![{\displaystyle {\bigg [}\sum _{\begin{array}{c}\\[-19pt]\scriptstyle {\text{perms }}\phi _{1}\\[-5pt]\scriptstyle {\text{of }}1,\dots ,p\end{array}}t_{1,\phi _{1}(1)}\cdots t_{p,\phi _{1}(p)}\left|P_{\phi _{1}}\right|{\bigg ]}\cdot {\bigg [}\sum _{\begin{array}{c}\\[-19pt]\scriptstyle {\text{perms }}\phi _{2}\\[-5pt]\scriptstyle {\text{of }}p+1,\dots ,p+q\end{array}}t_{p+1,\phi _{2}(p+1)}\cdots t_{p+q,\phi _{2}(p+q)}\left|P_{\phi _{2}}\right|{\bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b0d9b75cdd834e9d7eb1b2431ba8461ef3e5819)
- This exercise is recommended for all readers.
- Problem 18
Prove that for any
matrix
there are at most
distinct reals
such that the matrix
has
determinant zero
(we shall use this result in Chapter Five).
- Answer
The
case shows what happens.

Each term in the permutation expansion has three factors drawn from
entries in the matrix (e.g.,
and
), and so the determinant is
expressible as a polynomial in
of degree
.
Such a polynomial has at most
roots.
In general, the permutation expansion shows that the determinant can be written as a sum of terms, each with
factors, giving a polynomial of degree
. A polynomial of degree
has at most
roots.
- Problem 20
Show that

(Silverman & Trigg 1963)
- Answer
This is how the answer was given in the cited source.
When the elements of any column are subtracted from the elements of
each of the other two, the elements in two of the columns of the derived
determinant are proportional, so the determinant vanishes.
That is,

- ? Problem 22
Show that the determinant of the
elements in the upper left
corner of the Pascal triangle

has the value unity.
(Rupp & Aude 1931)
- Answer
This is how the answer was given in the cited source.
Denote by
the determinant in question and by
the element in the
-th row and
-th column.
Then from the law of formation of the elements we have

Subtract each row of
from the row following it, beginning the
process with the last pair of rows.
After the
subtractions the above equality shows that
the element
is replaced by the element
, and all the
elements in the first column, except
, become zeroes.
Now subtract each column from the one following it, beginning with the
last pair.
After this process the element
is replaced by
, as shown in the above relation.
The result of the two operations is to replace
by
, and to reduce each element in the first row and in
the first column to zero.
Hence
and consequently

- Silverman, D. L. (proposer); Trigg, C. W. (solver) (Jan. 1963), "Quickie 237", Mathematics Magazine (American Mathematical Society) 36 (1) .
- Strang, Gilbert (1980), Linear Algebra and its Applications (2nd ed.), Hartcourt Brace Javanovich
- Trigg, C. W. (proposer) (Jan. 1963), "Quickie 307", Mathematics Magazine (American Mathematical Society) 36 (1): 77 .
- Trigg, C. W. (proposer); Walker, R. J. (solver) (Jan. 1949), "Elementary Problem 813", American Mathematical Monthly (American Mathematical Society) 56 (1) .
- Rupp, C. A. (proposer); Aude, H. T. R. (solver) (Jun.-July 1931), "Problem 3468", American Mathematical Monthly (American Mathematical Society) 37 (6): 355 .