SOLUTIONS

No problem is ever permanently closed. Any comments, new solutions, or new insights on old problems are always welcome by the editor.

1. Proposed by Robert E. Kennedy and Curtis Cooper, Central Missouri State University, Warrensburg, Missouri.

Let n ≥ 2 and

∑_{j=1}^{i+1} a_{ij} = 0 for i = 1, 2, …, n − 1 .

Find the determinant of the nxn matrix

( \begin{matrix}a_{11} & a_{12} & 0 & 0 & … & 0 & 0 \\ a_{21} & a_{22} & a_{23} & 0 & … & 0 & 0 \\ a_{31} & a_{32} & a_{33} & a_{34} & … & 0 & 0 \\ a_{41} & a_{42} & a_{43} & a_{44} & … & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ a_{n−1,1} & a_{n−1,2} & a_{n−1,3} & a_{n−1,4} & … & a_{n−1,n−1} & a_{n−1,n} \\ 0 & 0 & 0 & 0 & … & 0 & 1 \end{matrix} ) .

I. Solution by Russell Euler, Northwest Missouri State University, Maryville, Missouri.

Using the given summation constraints, one can column reduce

the given matrix, A , to a lower triangular matrix having main

diagonal entries

a_{11}, a_{21} +a_{22}, a_{31} +a_{32} +a_{33}, …, a_{n−1,1} +a_{n−1,2} +⋯ +a_{n−1,n−1} .

Using the constraints again, these diagonal entries can be replaced

by

−a_{12}, −a_{23}, −a_{34}, …, −a_{n−1,n}

respectively. Therefore,

det(A) = (−1)^{n−1}a_{12}a_{23} ⋯ a_{n−1,n} .

II. Solution by Albert Dixon, The School of the Ozarks, Point Lookout, Missouri.

Let

B = ( \begin{matrix}1 & 1 & 1 & 1 & … & 1 & 1 \\ 0 & 1 & 1 & 1 & … & 1 & 1 \\ 0 & 0 & 1 & 1 & … & 1 & 1 \\ 0 & 0 & 0 & 1 & … & 1 & 1 \\ ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & 0 & … & 1 & 1 \\ 0 & 0 & 0 & 0 & … & 0 & 1 \end{matrix} )

and let A be the matrix above. Then the matrix C = AB (using the given constraints twice) is a lower triangular matrix whose entries along the diagonal can be written as

c_{i,i} = −a_{i,i+1} for 1 ≤ i ≤ (n − 1) and c_{n,n} = 1 .

Finally, det(AB) = det(A) det(B) for all matrices A and B and since B is upper triangular while C is lower triangular, their respective determinants are simply the product of their diagonal entries. Therefore, we conclude that

det(A) = det(A) · 1

= det(A) · det(B)

= det(AB) = det(C)

= (−1)^{n−1} ∏_{i=1}^{n−1} a_{i,i+1} .

Also solved by Charles J. Allard, Polo R-VII High School, Polo, Missouri and the proposers.

2. Proposed by Curtis Cooper and Robert E. Kennedy, Central Missouri State University, Warrensburg, Missouri.

A pitcher faces a batter in an at-bat. Let b be the probability the pitch is a ball, w, the probability the batter swings and misses (wiffs), and f, the probability the batter hits a foul ball. What is the probability that the batter strikes out?

Solution by Charles J. Allard, Polo R-VII High School, Polo, Missouri.

The batter will strike out by wiffing after the count reaches two strikes. This ranges from wiffing on an 0-2 count to wiffing with a full count (3-2). This is further complicated by the fact that after two strikes the batter may foul off zero or more pitches without changing the count. Let N represent the event of fouling-off zero or more pitches with two strikes. The sum of the probabilities for event N is

1 + f + f^{2} + ⋯ = \frac{1}{1 − f} .

Let S represent the event of either wiffing or fouling-off a pitch with less than two strikes. The probability that event S can happen is w + f . Furthermore, let B represent the event of taking a ball and W represent the event of wiffing.

A batter can strike out from an 0-2 count with the sequence of events

\begin{matrix}1 & 2 & & 3 \\ S & S & N & W \end{matrix}

The probability that the batter strikes out from an 0-2 count is

(w + f)^{2}w \frac{1}{1 − f} .

A batter can strike out from a 1-2 count with the 3 sequences of events

1 2  3  4
B S  S N W
S B  S N W
S S N B N W

The probability that the batter strikes out from a 1-2 count is

(w + f)^{2}bw (\frac{2}{1 − f} + \frac{1}{(1 − f)^{2}} ) .

A batter can strike out from a 2-2 count with the 6 sequences of events

1 2  3  4  5
B B  S  S N W
B S  B  S N W
S B  B  S N W
B S  S N B N W
S B  S N B N W
S S N B N B N W

The probability that the batter strikes out from a 2-2 count is

(w + f)^{2}b^{2}w (\frac{3}{1 − f} + \frac{2}{(1 − f)^{2}} + \frac{1}{(1 − f)^{3}} ) .

Finally, a batter can strike out from a 3-2 count with the 10 sequences of events

1 2  3  4  5  6
S B  B  B  S N W
B S  B  B  S N W
B B  S  B  S N W
B B  B  S  S N W
S B  B  S N B N W
B S  B  S N B N W
B B  S  S N B N W
S B  S N B N B N W
B S  S N B N B N W
S S N B N B N B N W

The probability that the batter strikes out from a 3-2 count is

(w + f)^{2}b^{3}w (\frac{4}{1 − f} + \frac{3}{(1 − f)^{2}} + \frac{2}{(1 − f)^{3}} + \frac{1}{(1 − f)^{4}} ) .

Adding these probabilities up, we get the probability that the batter strikes out is

∑_{i=0}^{3}(w + f)^{2}b^{i}w ∑_{j=0}^{i} \frac{i + 1 − j}{(1 − f)^{j+1}}

= \frac{(w + f)^{2}w}{(1 − f)^{4}} ∑_{i=0}^{3} ∑_{j=0}^{i} b^{i}(i + 1 − j)(1 − f)^{3−j}

= \frac{(w + f)^{2}w}{(1 − f)^{4}} ∑_{j=0}^{3} ∑_{i=j}^{3} b^{i}(i + 1 − j)(1 − f)^{3−j}

= \frac{(w + f)^{2}w}{(1 − f)^{4}} ∑_{j=0}^{3} ∑_{i=0}^{3−j} b^{i+j}(i + 1)(1 − f)^{3−j}

= \frac{(w + f)^{2}w}{1 − f} ∑_{j=0}^{3} (\frac{b}{1 − f} )^{j} ∑_{i=0}^{3−j}(i + 1)b^{i} .

Also solved by the proposers.