written 5.8 years ago by | • modified 2.3 years ago |
$A=$ $\left[ {\begin{array}{cc} 0 & -1\ 1 & 0 \end{array} } \right] $
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Medium
written 5.8 years ago by | • modified 2.3 years ago |
$A=$ $\left[ {\begin{array}{cc} 0 & -1\ 1 & 0 \end{array} } \right] $
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Medium
written 5.8 years ago by |
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, $\phi(A)=\alpha_1 A+ \alpha_0 I$
$3tanA=\alpha_1 A+ \alpha_0 I$ (1)
$3tan\lambda=\alpha_1 A+ \alpha_0 $ (2)
for $\lambda=3$ we get
$3tan3=\alpha_1 3 +\alpha_0$ (3)
and for $\lambda=-3$ we get from (2),
$3tan(-3_=\alpha_13+\alpha_0$
$-3tan3=-3\alpha_1+\alpha_0$ (4)
Adding (3) and (4) we get,
$0=0+2\alpha_0$
$\alpha=0$
From (3) we get $3tan3=\alpha_1 3$. So, $\alpha_1=tan3$
From (1), we get $3tanA=tan3A$. So, $Atan3$