written 5.7 years ago by | • modified 2.2 years ago |
$ A=$ $ \left[ {\begin{array}{cc} -1 & 2 & 3\ 0 & 3 & 5 \ 0 & 0 & -2\ \end{array} } \right] $ $B=A^3+6A^{-1}-2I$
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Medium
written 5.7 years ago by | • modified 2.2 years ago |
$ A=$ $ \left[ {\begin{array}{cc} -1 & 2 & 3\ 0 & 3 & 5 \ 0 & 0 & -2\ \end{array} } \right] $ $B=A^3+6A^{-1}-2I$
Subject: Applied Mathematics 4
Topic: Matrices
Difficulty: Medium
written 5.7 years ago by | • modified 5.7 years ago |
First we will find eigen values and eigen vectors of A. As A is triangular matrix. Eigen values of A are diagonal entries of A.
Therefore, $\lambda=-1,3,-2$ are eigen values of A.
To find eigen vector $X\neq0$ such that $(A-\lambda I)X=0$
$ \left[ {\begin{array}{cc} -1-\lambda & 2 & 3 \\ 0 & 3-\lambda & 5 \\ 0 & 0 & -2-\lambda \end{array} } \right]$ $ \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$= $ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$ (1)
Put $\lambda=-1$ in (1)
$ \left[ {\begin{array}{cc} 0 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & -1 \end{array} } \right]$ $ \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$= $ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$
Choose any two distinct rows. Let us choose 2nd and 3rd row, we get,
$2y+3z=0$ (1)
$4y+5z=0$ (2)
$-z=0$
$z=0$
From (2), $y=0$
Consider X=1
$ X_1=$ $ \left[ {\begin{array}{cc} 1 \\ 0 \\ 0 \end{array} } \right]$
Put $\lambda=3$ in (1) we get,
$\left[ {\begin{array}{cc} -4 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & -5 \end{array} } \right]$$ \left[ {\begin{array}{cc} x \ y \ z \end{array} } \right]$=$ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$
From 2nd and 3rd row as they are distinct
$-4x+2y+3z=0$ (1)
$z=0$
$x-y+0=0$
$y=2x$
for $x=1$, $y=2$
$ X_2=$ $ \left[ {\begin{array}{cc} 1 \\ 2 \\ 0 \end{array} } \right]$
Put $\lambda=-2$ in (1), we get
$ \left[ {\begin{array}{cc} 1 & 2 & 3\\ 0 & 5 & 5 \\ 0 & 0 & 0 \end{array} } \right] $ $ \left[ {\begin{array}{cc} x\\ y \\ z \end{array} } \right] $= $ \left[ {\begin{array}{cc} 0\\ 0 \\ 0 \end{array} } \right]$
$x+2y+3z=0$
$0+5y+5z=0$
By Cramer's rule,
$\frac{x}{\mid {\begin{array}{cc} 2 & 3 \\ 5 & 5 \\ \end{array} } \mid}$=$ \frac{x}{\mid {\begin{array}{cc} 1 & 3 \\ 0 & 5 \\ \end{array} } \mid}$= $\frac{x}{\mid {\begin{array}{cc} 1 & 2 \\ 0 & 5 \\ \end{array} } \mid}$
$\frac{x}{-5}=\frac{-y}{5}=\frac{z}{5}$
$ X_3=$ $ \left[ {\begin{array}{cc} -5\\ -5 \\ 5 \end{array} } \right]$ Now, let us find eigen values and eigen vectors of B=$A^3+6A^{-1}-2I$
Let, $f(A)=A^3+6^{-1}-2I$ (2)
We know that if $\lambda$ is eigen value of A. Then f(A) is eigen value of f(A) and eigen vector corresponding to $\lambda$ and f(A) is same.
From (2),
$f(\lambda)=\lambda^3+6\lambda^{-1}-2=\lambda^3+\frac{6}{\lambda}-2$
For $\lambda=-1=f(-1)=(-1)^3+\frac{6}{-1}-2=-1-6-2=-9$
For $\lambda=3=f(3)=(3)^3+\frac{6}{3}-2=27+2-2=27$
For $\lambda=-2=f(-2)=(-2)^3+\frac{6}{-2}-2=-8-3-2=-13$
Eigen values of $B=A^3+6^{-1}-27$ are -9,27,-13 and corresponding eigen vectors are
$ X_1=$ $ \left[ {\begin{array}{cc} 1\\ 0 \\ 0 \end{array} } \right]$, $ X_2=$ $ \left[ {\begin{array}{cc} 1\\ 2\\ 0 \end{array} } \right]$ , $ X_3=$ $ \left[ {\begin{array}{cc} -5\\ -5 \\ 5 \end{array} } \right]$