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Question: Verify Cayley-Hamilton theorem for following matrix and also find the following
0

$A^{-1} and A^4$

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: Medium

m4m(64) • 46 views
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 modified 20 days ago  • written 20 days ago by
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Characteristic equation of A is $\mid A-\lambda I \mid=0$

$\lambda^3-6\lambda^2+[(4+4+4)-(1+1+1)]\lambda-\mid A \mid=0$

$\lambda^3-6\lambda^2+(12-3)\lambda-4=0$

$\lambda^3-6\lambda^2+9\lambda-4=0$

To verify Cayley-Hamilton Theorem, we need to prove that $A^3-6A^2-4I=0$

LHS=$A^3-6A^2+9A-4I$

$=$ $\left[ {\begin{array}{cc} 22 & -21 & 21\ -21 & 22 & -21 \ 21 & -21 & 22\ \end{array} } \right]$-$6 \left[ {\begin{array}{cc} 6 & -5 & 5\\ -5 & 6 & -5 \\ 5 & -5 & 6\\ \end{array} } \right]$+$9 \left[ {\begin{array}{cc} 2 & -1 & 1\\ -1 & 2 & -1 \\ 1 & -1 & 2\\ \end{array} } \right]$ - $\left[ {\begin{array}{cc} 4 & 0 & 0\\ 0 & 4 & 0 \\ 0 & 0 & 4\\ \end{array} } \right]$

$=$ $\left[ {\begin{array}{cc} 22 & -21 & 21\ -21 & 22 & -21 \ 21 & -21 & 22\ \end{array} } \right]$-$6 \left[ {\begin{array}{cc} 36 & -30 & 30\\ -30 & 36 & -30 \\ 30 & -30 & 36\\ \end{array} } \right]$+$9 \left[ {\begin{array}{cc} 18 & -9 & 9\\ -9 & 18 & -9 \\ 9 & -9 & 18\\ \end{array} } \right]$

$=$ $\left[ {\begin{array}{cc} 22 & -21 & 21\ -21 & 22 & -21 \ 21 & -21 & 22\ \end{array} } \right]$-$6 \left[ {\begin{array}{cc} 6 & -5 & 5\\ -5 & 6 & -5 \\ 5 & -5 & 6\\ \end{array} } \right]$+$9 \left[ {\begin{array}{cc} 2 & -1 & 1\\ -1 & 2 & -1 \\ 1 & -1 & 2\\ \end{array} } \right]$ - $\left[ {\begin{array}{cc} 4 & 0 & 0\\ 0 & 4 & 0 \\ 0 & 0 & 4\\ \end{array} } \right]$

$=$ $\left[ {\begin{array}{cc} 0 & 0 & 0\ 0 & 0 & 0 \ 0 & 0 & 0\ \end{array} } \right]$= RHS $A^3-6A^2+9A-4I=0$ (1) Cayley-Hamilton Theorem verified. Now to find $A^{-1}$ and $A^4$. multiply (1) by $A{-1}$ we get, $A^{-1}(A^3-6A^2+9A-4I)=A^{-1}0$ $A^2-6A+9I-4A^{-1}=0$ $4A^{-1}=A^2-6A+9I$ $A^{-1}=\frac{1}{4}(A^2-6A+9I)$ $A^{-1}$=$\left[ {\begin{array}{cc} 6 & -5 & 5\\ -5 & 6 & -5 \\ 5 & -5 & 6\\ \end{array} } \right]$ $-6 \left[ {\begin{array}{cc} 2 & -1 & 1\\ -1 & 2 & -1 \\ 1 & -1 & 2\\ \end{array} } \right]$ + $\left[ {\begin{array}{cc} 9 & 0 & 0\\ 0 & 9 & 0 \\ 0 & 0 & 9\\ \end{array} } \right]]$

$\left[ {\begin{array}{cc} 6 & -5 & 5\\ -5 & 6 & -5 \\ 5 & -5 & 6\\ \end{array} } \right]$ -$\left[ {\begin{array}{cc} 12 & -6 & 6\\ -6 & 12 & -6 \\ 6 & -6 & 12\\ \end{array} } \right]$ + $\left[ {\begin{array}{cc} 9 & 0 & 0\\ 0 & 9 & 0 \\ 0 & 0 & 9\\ \end{array} } \right]]$

=$\left[ {\begin{array}{cc} 3 & 1 & 1\\ 1 & 3 & 1 \\ -1 & 1 & 3\\ \end{array} } \right]$

To find $A^4$ multiply (1) by A

$A(A^3-6A^2+9A-4I)=A(0)$

$A^4-6A^3+9A^2-4A=0$

$A^4=6A^3-9A^2+4A$

=6$\left[ {\begin{array}{cc} 22 & -21 & 21\\ -21 & 22 & -21 \\ 21 & -21 & 22\\ \end{array} } \right]$ $-9 \left[ {\begin{array}{cc} 6 & -5 & 5\\ -5 & 6 & -5 \\ 5 & -5 & 6\\ \end{array} } \right]$ +4 $\left[ {\begin{array}{cc} 2 & -1 & 1\\ -1 & 2 & -1 \\ 1 & -1 & 2\\ \end{array} } \right]$

=$\left[ {\begin{array}{cc} 132 & -126 & 126\\ -126 & 132 & -126 \\ 126 & -126 & 132\\ \end{array} } \right]$ $-\left[ {\begin{array}{cc} 54 & -45 & 45\\ -45 & 54 & -45 \\ 45 & -45 & 54\\ \end{array} } \right]$ + $\left[ {\begin{array}{cc} 8 & -4 & 4\\ -4 & 8 & -4 \\ 4 & -4 & 8\\ \end{array} } \right]$

=$\left[ {\begin{array}{cc} 86 & -85 & 85\\ -85 & 86 & -85 \\ 85 & -85 & 86\\ \end{array} } \right]$

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 written 20 days ago by
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