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Consider the following matrix:

$R = \begin{bmatrix} 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \\ 1 & 4 & 16 & 64 \\ 1 & 5 & 25 & 125 \end{bmatrix}$

The absolute value of the product of Eigen values of $R$ is _______
in Linear Algebra by Veteran (425k points)
edited by | 2.7k views
+1
product of eigenvalues is the determinant

$12$ ans
+2
under the pressure, i multiply C2 and C3 and subtracted from C3 ==> C3 is a Zero vector !

Therefore Det will be 0

What a mistake !
+1
still if you didn't get, you have to read det properties again
0
the matrix is in GP, is there any shortcut if you consider it. like if its in AP then Det|A| =0

3 Answers

+16 votes
Best answer

Important properties of Eigen values:

  1. Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements=Trace of the matrix.
  2. Product of all Eigen values$=Det(A)= \mid A \mid$
  3. Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself.

Example:$A=\begin{bmatrix} 1& 0& 0\\ 0&1 &0 \\ 0& 0& 1\end{bmatrix}$

Diagonal matrix

Eigenvalues are $1,1,1$

$B=\begin{bmatrix} 1& 9& 6\\ 0&1 &12 \\ 0& 0& 1\end{bmatrix}$

Upper triangular matrix

Eigenvalues are $1,1,1$

$C=\begin{bmatrix} 1& 0& 0\\ 8&1 &0 \\ 2& 3& 1\end{bmatrix}$

Lower triangular matrix

Eigenvalues are $1,1,1$


Now coming to the actual question

$R=\begin{bmatrix} 1 &2 &4 &8 \\ 1 &3 &9 &27 \\ 1 &4 &16 &64 \\ 1 &5 &25 &125 \end{bmatrix}$

$\mid  R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 1 &3 &9 &27 \\ 1 &4 &16 &64 \\ 1 &5 &25 &125 \end{vmatrix}$

Perform 

  • $R4\rightarrow R_{4}-R_{3}$
  • $R3\rightarrow R_{3}-R_{2}$
  • $R2\rightarrow R_{2}-R_{1}$

$\implies \mid R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &5 &19 \\ 0 &1 &7 &37 \\ 0 &1 &9 &61 \end{vmatrix}$

Perform

  • $R4\rightarrow R_{4}-R_{3}$
  • $R3\rightarrow R_{3}-R_{2}$

$\implies \mid R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &5 &19 \\ 0 &0 &2 &18 \\ 0 &0 &2 &24 \end{vmatrix}$

Perform

  • $R4\rightarrow R_{4}-R_{3}$

$\implies \mid R \mid =\begin{vmatrix} 1 &2 &4 &8 \\ 0 &1 &5 &19 \\ 0 &0 &2 &18 \\ 0 &0 &0 &6 \end{vmatrix}$

 The absolute value of product of  Eigen values$=\text{Det}(A)= \text{Product of diagonal elements }  =12.$

by Veteran (54.9k points)
edited by
+1
Nice approach πŸ™‚
+4 votes

Answer 12

Product of eigen values= |A| (determinant of matrix)

by Active (4.6k points)
0 votes

Product of eigenvalues = determinant of the matrix.

Given matrix:

R2 = R2 - R1; R3 = R3 - R1; R4 = R4 - R1

Expand via first column.

R2 = R2 - 2R1; R3 = R3 - 3R1

Expand via first column

=12

by Active (2.4k points)

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