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Question: For the given matrix, find eigen values and eigen vectors of the the equation B.
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$ 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

m4m(64) • 90 views
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modified 4 months ago  • written 4 months ago by gravatar for manasahegde234 manasahegde23420
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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]$

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modified 4 months ago  • written 4 months ago by gravatar for manasahegde234 manasahegde23420
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