We have matrix: A=
⎝⎛21213221−6⎠⎞
Find the eigenvalues of the matrix:
Associate matrix Vlasna vectors values and introduce a variable.
⎝⎛2−λ1213−λ221−6−λ⎠⎞
To do this, find the determinant of the matrix and equate this expression to zero.
(2−λ)∗((3−λ)∗(−6−λ)−2∗1)−1∗(1∗(−6=λ)−2∗2)+2∗(1∗1−(3−λ)∗2)=0
After transformations, we get:
λ3+λ2 −31λ+40=0
After applying the rules, it is clear that the roots of the equation are not integers. Let's calculate the approximate λ .
λ1 =−6,60
λ2 =4,15
λ3 =1,45
Multiply the matrix A by the matrix of its eigenvalues:
⎝⎛21213221−6⎠⎞ ∗ ⎝⎛−6,64,151,45⎠⎞ = ⎝⎛−6,157,3−13,6⎠⎞
Maybe, if I misunderstood something, I add to the report: the sum and product of the matrix A itself:
The sum of the matrix: A + A = 2A
2⎝⎛21213221−6⎠⎞=⎝⎛42426442−12⎠⎞
And product of the matrix: A∗A=A2
A∗A=⎝⎛21213221−6⎠⎞∗⎝⎛21213221−6⎠⎞=⎝⎛97−6912−4−7−142⎠⎞
Multiplication is performed for each component of the matrix separately:
a121=a11∗a11+a12∗a21+a13∗a31=2∗2+1∗1+2∗2=9
a122=a11∗a12+a12∗a22+a13∗a32=2∗1+1∗3+2∗2=2+3+4=9
a123=a11∗a13+a12∗a23+a13∗a33=2∗2+1∗1+2∗(−6)=−7
a221=a21∗a11+a22∗a21+a23∗a31=1∗2+3∗1+1∗2=7
a222=a21∗a12+a22∗a22+a23∗a32=1∗1+3∗3+1∗2=12
a223=a21∗a13+a22∗a23+a23∗a33=1∗2+3∗1+1∗(−6)=−1
a321=a31∗a11+a32∗a21+a33∗a31=2∗2+2∗1+(−6)∗2=−6
a322=a31∗a12+a32∗a22+a33∗a32=2∗1+2∗3+(−6)∗2=−4
a323=a31∗a13+a32∗a23+a33∗a33=2∗2+2∗1+(−6)∗(−6)=42
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