Question

Question: 20. (M) Use the method of Exercise 19 to guess the determinant of

\(\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}& \cdots &{\bf{1}}\\{\bf{1}}&{\bf{3}}&{\bf{3}}& \cdots &{\bf{3}}\\{\bf{1}}&{\bf{3}}&{\bf{6}}& \cdots &{\bf{6}}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\{\bf{1}}&{\bf{3}}&{\bf{6}}& \cdots &{{\bf{3}}\left( {n - {\bf{1}}} \right)}\end{array}} \right)\)

Justify your conjecture. (Hint: Use Exercise 14(c) and the result of Exercise 19.)

Short answer

\(\det \left( {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\0&2&2& \cdots &2\\0&0&3& \cdots &3\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&0&3& \cdots &{3n}\end{array}} \right) = 2\left( {{3^{n - 2}}} \right)\)

Step-by-step solution

Compute the determinants

Using the matrix program, you can obtain the following:

\(\left| {\begin{array}{*{20}{c}}1&1&1\\1&3&3\\1&3&6\end{array}} \right| = 6 = 2\left( 3 \right)\)

\(\left| {\begin{array}{*{20}{c}}1&1&1&1\\1&3&3&3\\1&3&6&6\\1&3&6&9\end{array}} \right| = 18 = 2\left( {{3^2}} \right)\)

\(\left| {\begin{array}{*{20}{c}}1&1&1&1&1\\1&3&3&3&3\\1&3&6&6&6\\1&3&6&9&9\\1&3&6&9&{12}\end{array}} \right| = 54 = 2\left( {{3^3}} \right)\)

The conjecture is

\(\left| {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\1&3&3& \cdots &3\\1&3&6& \cdots &6\\ \vdots & \vdots & \vdots & \ddots & \vdots \\1&3&6& \cdots &{3\left( {n - 1} \right)}\end{array}} \right| = 2\left( {{3^{n - 2}}} \right)\).

Use the row-replacement operations

The matrix is

\(\left( {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\1&3&3& \cdots &3\\1&3&6& \cdots &6\\ \vdots & \vdots & \vdots & \ddots & \vdots \\1&3&6& \cdots &{3\left( {n - 1} \right)}\end{array}} \right)\).

At row 2, subtract row 1 from row 2. At row 3, subtract row 1 from row 3, and so on.

\(\left( {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\0&2&2& \cdots &2\\0&2&5& \cdots &5\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&2&5& \cdots &{3n - 2}\end{array}} \right)\)

At row 3, subtract row 2 from row 3. At row 4, subtract row 2 from row 4, and so on. Then

\(\left( {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\0&2&2& \cdots &2\\0&0&3& \cdots &3\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&0&3& \cdots &{3n}\end{array}} \right)\)

Use Exercise 14(c) and Exercise 19              

The matrix is of the form \(\left( {\begin{array}{*{20}{c}}A&B\\0&D\end{array}} \right)\), where \(A = \left( {\begin{array}{*{20}{c}}1&1\\0&2\end{array}} \right)\), \(D = \left( {\begin{array}{*{20}{c}}3&3& \cdots &3\\3&6& \cdots &6\\ \vdots & \vdots & \ddots & \vdots \\3&6& \cdots &{3n}\end{array}} \right)\).

Now, using Exercise 14(c), i.e., \(\det \left( {\begin{array}{*{20}{c}}A&B\\0&D\end{array}} \right) = \left( {\det A} \right)\left( {\det D} \right)\), you get the following:

\(\begin{array}{c}\det \left( {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\0&2&2& \cdots &2\\0&0&3& \cdots &3\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&0&3& \cdots &{3n}\end{array}} \right) = \left( {\det \left( {\begin{array}{*{20}{c}}1&1\\0&2\end{array}} \right)} \right)\left( {\det \left( {\begin{array}{*{20}{c}}3&3& \cdots &3\\3&6& \cdots &6\\ \vdots & \vdots & \ddots & \vdots \\3&6& \cdots &{3n}\end{array}} \right)} \right)\\ = \left( 2 \right)\left( {3\det \left[ {\begin{array}{*{20}{c}}1&1& \cdots &1\\3&6& \cdots &6\\ \vdots & \vdots & \ddots & \vdots \\3&6& \cdots &{3n}\end{array}} \right)} \right)\\ = 2\left( {{3^{n - 2}}\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\1&2& \cdots &2\\ \vdots & \vdots & \ddots & \vdots \\1&2& \cdots &n\end{array}} \right)} \right)\end{array}\)

From Exercise 19, \(\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\1&2& \cdots &2\\ \vdots & \vdots & \ddots & \vdots \\1&2& \cdots &n\end{array}} \right) = 1\). Hence, you get

\(\det \left( {\begin{array}{*{20}{c}}1&1&1& \cdots &1\\0&2&2& \cdots &2\\0&0&3& \cdots &3\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&0&3& \cdots &{3n}\end{array}} \right) = 2\left( {{3^{n - 2}}} \right)\).