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Question: Are the following matrices derogatory? If yes, find its minimal polynomial.
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1. $A=$ $\left[ {\begin{array}{cc} 6 & -2 & 2\\ -2 & 3 & -1 \\ 0 & 0 & -2\\ \end{array} } \right]$

2. $A=$ $\left[ {\begin{array}{cc} 8 & -8 & -2\\ 4 & -3 & -2 \\ 3 & -4 & 1\\ \end{array} } \right]$

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Medium

m4m(64) • 89 views
 modified 11 weeks ago  • written 11 weeks ago by
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1. Characteristic equation of A is $\mid A-\lambda I \mid=0$

$\lambda^3-(12)\lambda^2+36\lambda-32=0$

$\lambda=8,2,2.$

$(\lambda-8)(\lambda-2)(\lambda-2)=0$

Consider $(\lambda-8)(\lambda)=\lambda^2-0\lambda+16$

Now to check that $\lambda^2-10A+16I$

Therefore, consider $A^2-10A+16I$

$= \left[ {\begin{array}{cc} 44 & -20 & 20\\ 20 & 14 & -10 \\ 20 & -10 & 14\\ \end{array} } \right]$-$10\left[ {\begin{array}{cc} 6 & -2 & 2\\ -2 & 3 & -1 \\ 2 & -1 & 3\\ \end{array} } \right]$+$16 \left[ {\begin{array}{cc} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\\ \end{array} } \right]$

$= \left[ {\begin{array}{cc} 44 & -20 & 20\\ 20 & 14 & -10 \\ 20 & -10 & 14\\ \end{array} } \right]$-$10\left[ {\begin{array}{cc} 60 & -20 & 20\\ -20 & 30 & -10 \\ 20 & -10 & 30\\ \end{array} } \right]$+$16 \left[ {\begin{array}{cc} 16 & 0 & 0\\ 0 & 16 & 0 \\ 0 & 0 & 16\\ \end{array} } \right]$

=$16 \left[ {\begin{array}{cc} 0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{array} } \right]$

$=\lambda^2-10\lambda+16$ annihilates A

Minimal polynomial is $\lambda^2-10\lambda+16$

A is derogatory.

1. Characteristic equation of A is $\mid A-\lambda I \mid=0$

$\lambda^3-(6)\lambda^2+11\lambda-6=0$

$\lambda=1,2,3$

We have $(\lambda-1)(\lambda-2)(\lambda-3)=0$

All eigen values of A are distinct.

Minimal polynomial of $A=(\lambda-1)(\lambda-2)(\lambda-3)$=$\lambda^3-6\lambda^2+11\lambda-6$= characteristic polynomial.

A is non derogatory.