1 votes 1 votes Shadan Karim asked Jan 4, 2019 Shadan Karim 396 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Lakshman Bhaiya commented Jan 4, 2019 reply Follow Share We can write the polynomial equation $\lambda^{3}-$[Sum of leading diagonal elements]$\lambda^{2}+\left[A_{11}+A_{22}+A_{33}\right]\lambda-det(A)=0--->(1)$ where $A_{11}=$Cofactor of $a_{11}$ $A_{22}=$Cofactor of $a_{22}$ $A_{33}=$Cofactor of $a_{33}$ 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$=Det(A)=|A|$ $(3)$ 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 $1,1,1$ Here in given question $\lambda^{3}-6\lambda^{2}-\lambda+22=0$ compare with equation (1) and get $-det(A)=22$ So$,det(A)=|A|=-22$ 2 votes 2 votes Abhisek Tiwari 4 commented Jan 4, 2019 reply Follow Share @kumar.dilip i did not find all lemda i just find their product which is equal to Det Ax^3 +Bx^2 +Cx +D product of root=-D/A 0 votes 0 votes Lakshman Bhaiya commented Jan 4, 2019 i edited by Lakshman Bhaiya Jan 4, 2019 reply Follow Share We can write like this also Suppose we have $3$ roots$(a,b,c)$ and we want to write the equation. We can write like this. $x^{3}-(a+b+c)x^{2}+(ab+bc+ca)x-abc=0$ 0 votes 0 votes Please log in or register to add a comment.