Question

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\).

1. Reduce \(A\) to echelon form \(U\) with no row scaling, and use \(U\) in formula (1) (before Example 2) to compute \(\det A\). (If \(A\) happens to be singular, start over with a new random matrix.)
2. Compute the eigenvalues of \(A\) and the product of these eigenvalues (as accurately as possible).
3. List the matrix \(A\), and, to four decimal places, list the pivots in \(U\) and the eigenvalues of \(A\). Compute \(\det A\) with your matrix program, and compare it with the products you found in (a) and (b).

Short answer

1. The reduce echelon form of the matrix A is \(\left( {\begin{array}{*{20}{c}}1&0&0&1\\0&1&0&{ - 2}\\0&0&2&1\\0&0&0&{ - 1}\end{array}} \right)\) and \({\rm{det}}A = - 2\).
2. Eigenvalues are \(\left\{ {1,2.81, - 1.34,0.53} \right\}\), and product is 2.
3. It is observed that the determinant of matrix is equal to the product of eigenvalues of the matrix.

Step-by-step solution

(a) Consider a \(4 \times 4\) matrix \(A\) and determine determinant

Consider the matrix\(A = \left( {\begin{array}{*{20}{c}}2&1&0&0\\1&0&0&1\\2&0&0&1\\0&0&2&1\end{array}} \right)\).

Use the following command in MATLAB to find the reduce echelon form of the matrix.

\[\begin{array}{l} > > {\rm{A}} = \left( {\begin{array}{*{20}{c}}2&1&0&{0;}&1&0&0&{1;}\\2&0&0&{1;}&0&0&2&1\end{array}} \right);\\ > > {\rm{R}} = {\rm{rref}}\left( {\rm{A}} \right);\end{array}\]

So, the row reduce echelon form of the matrix A is shown below:

\[\left( {\begin{array}{*{20}{c}}1&0&0&1\\0&1&0&{ - 2}\\0&0&2&1\\0&0&0&{ - 1}\end{array}} \right)\]

Find the determinant by using the formula as shown below:

\[\begin{array}{c}{\rm{det}}A = {\left( { - 1} \right)^4}\left( {1 \cdot 1 \cdot 2 \cdot \left( { - 1} \right)} \right)\\ = - 2\end{array}\]

Thus, \({\rm{det}}A = - 2\).

(b) Find eigenvalues and the product of eigenvalues

Use the following command in MATLAB to find the eigenvalues of the matrix.

\[\begin{array}{l} > > {\rm{A}} = \left( {\begin{array}{*{20}{c}}2&1&0&{0;}&1&0&0&{1;}\\2&0&0&{1;}&0&0&2&1\end{array}} \right);\\ > > {\rm{eig}} = {\rm{eigv}}\left( {\rm{A}} \right);\end{array}\]

So, the eigenvalues of the matrix A are shown as:

\(\left\{ {1,2.81, - 1.34,0.53} \right\}\)

Find the product of eigenvalues.

\[\begin{array}{c}\left( 1 \right) \cdot \left( {2.81} \right) \cdot \left( { - 1.34} \right) \cdot \left( {0.53} \right) = - 1.995662\\ \approx - 2\end{array}\]

Thus, the eigenvalues are \(\left\{ {1,2.81, - 1.34,0.53} \right\}\)and the product of the eigenvalues is \( - 2\).

(c) Compare the result of part (a) and (b)

It is observed that the determinant of matrix is equal to the product of eigenvalues of the matrix.