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  1. Obtain the eigen values of the matrix

$$A=\begin {bmatrix} 1 & 2 & 34 & 49 \\ 0 & 2 & 43 & 94 \\ 0 & 0 & -2 & 104 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$

asked in Linear Algebra by Veteran (59.6k points) | 597 views

4 Answers

+17 votes
Best answer
$5(a)$ the eigen value for upper triangular/lower triangular/diagonal matrices are the diagonal elements of the matrix
answered by Active (2.4k points)
edited by
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we know for triangular/diagonal the determeinant is product of principal diagonal elements so in A-(lambda)I every value of lambda as any of of the principal diagonal elements will give result as 0. so all the diagonal principal diagonal elements are eigen values.
+8 votes
$A=\begin{bmatrix} 1& 2 &34 &49\\ 0& 2&43 &94\\ 0& 0 & -2&104\\ 0& 0& 0&-1 \end{bmatrix}$

$|A-\lambda I |=0$

$\begin{vmatrix} 1-\lambda& 2 &34 &49\\ 0& 2-\lambda&43 &94\\ 0& 0 & -2-\lambda&104\\ 0& 0& 0&-1-\lambda \end{vmatrix}=0$

$( 1-\lambda)( 2-\lambda)( -2-\lambda)(-1-\lambda)=0$

$\lambda=1\ ,\ -1\ ,\ 2 \ , \ -2$
answered by Boss (40.7k points)
+1 vote
Eigen value $lambda$=1,-1,2,-2 A-$lambda$I=0 just solve it and for the 2nd question i think building a truth table is a naive way of answering it if any1 has better solution then plz reply
answered by Boss (14.4k points)
0 votes
if the matrix is either upper triangular or lower triangular matrix then the principal diagonal element are called eigen value.
1,2,-1,-2 for this question ...
answered by Active (3.9k points)

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