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## User: manasahegde234

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#### Posts by manasahegde234

<prev • 156 results • page 1 of 16 • next >
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... 1. Characteristic equation of A is $\mid A-\lambda I \mid = 0$ $\lambda^3-(8+7+3) \lambda^2 + [(56+21+24)-(36+4+16)]\lambda - \mid A \mid =0$ $\lambda^3-18\lambda^2+45\lambda-0=0$ $\lambda=0,3,15$ are eigen values of A As all eigen values of A are distinct. A is diagonalisable. Now to find ei ...
written 13 days ago by manasahegde23410
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... 1. $A=$ $\left[ {\begin{array}{cc} 8 & -6 & 2\\ -6 & 7 & -4 \\ 2 & -4 & 3\\ \end{array} } \right]$ 2. $A=$ $\left[ {\begin{array}{cc} -9 & 4 & 4\\ -8 & 3 & 4 \\ -16 & 8 & 7\\ \end{array} } \right]$ **Subject:** Applied Mat ...
written 13 days ago by manasahegde23410
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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 $A^2-10A+16I$ $= \left[ ... written 13 days ago by manasahegde23410 1 answer 47 views 1 answer ... 1.$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] $**Subject:** Applied Mathemat ... written 13 days ago by manasahegde23410 1 answer 35 views 1 answers ... Characteristic equation of A is$\mid A-\lambda I \mid=0\lambda^2-(-1+1)\lambda+\mid A \mid=0\lambda^2-0+(-1-8)=0\lambda^2-9=0\lambda^2=9\lambda=\pm3$Let$\phi(A)=3tanA$Consider,$\phi(A)=\alpha_1 A+ \alpha_0 I3tanA=\alpha_1 A+ \alpha_0 I$(1)$3tan\lambda=\alpha_1 A+ \a ...
written 13 days ago by manasahegde23410
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... $A=$ $\left[ {\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array} } \right]$ **Subject:** Applied Mathematics 4 **Topic:** Matrices **Difficulty:** Medium ...
written 13 days ago by manasahegde23410
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... Characteristic equation of $\mid A-\lambda I \mid=0$ Therefore, $\lambda^2-(0)\lambda+\mid A \mid=0$ $\lambda^2+(0-(-1))=0$ $\lambda^2+1=0$ $\lambda=\pm i$ Let, $\phi(A)=e^At$ Consider, $\phi(A)=\alpha_1 A+\alpha_0 I$ $e^{At}=\alpha_1 A+\alpha_0 I$ (1) $e^{\lambda t}=\alpha_1 \lambda+\alpha ... written 13 days ago by manasahegde23410 1 answer 45 views 1 answer ...$A=  \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 1 \\ \end{array} } \right] e^{At}$=$ \left[ {\begin{array}{cc} cost & -sint \\ sint & cost \\ \end{array} } \right] $**Subject:** Applied Mathematics 4 **Topic:** Matrices **Difficulty:** Medium ... written 13 days ago by manasahegde23410 1 answer 41 views 1 answers ... Characteristic equation of A is$\mid A-\lambda I \mid=0\lambda^3-5\lambda^2+[(2+2+4)-(0+1+0)]\lambda-\mid A \mid=0\lambda^3-5\lambda^2+7\lambda-3=0$Therefore, Cayley-Hamilton theorem,$A^3-5A^2+7A-3I=0$from given polynomial in A, i.e$A^8-5\lambda^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$, we ... written 13 days ago by manasahegde23410 1 answer 41 views 1 answer ...$A^8-5A^7+7A^6-3A^5+A^4-5A^3+8A^2-2A+I$where$A= \left[ {\begin{array}{cc} -1 & 2 & 3\\ 0 & 3 & 5 \\ 0 & 0 & -2\\ \end{array} } \right] \$ **Subject:** Applied Mathematics 4 **Topic:** Matrices **Difficulty:** Medium ...