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Question: Show that A is diagonalisable. Hence find transforming matrix and diagonal matrix.
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  1. $A=$ $ \left[ {\begin{array}{cc} 8 & -6 & 2\\ -6 & 7 & -4 \\ 2 & -4 & 3\\ \end{array} } \right] $

    1. $ A=$ $ \left[ {\begin{array}{cc} -9 & 4 & 4\\ -8 & 3 & 4 \\ -16 & 8 & 7\\ \end{array} } \right] $

Subject: Applied Mathematics 4

Topic: Matrices

Difficulty: High

m4m(64) • 84 views
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modified 11 weeks ago  • written 11 weeks ago by gravatar for manasahegde234 manasahegde23420
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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 eigen vector $X\neq0$ such that $[A-\lambda I]=0$

$ \left[ {\begin{array}{cc} 8-\lambda & -6 & 2 \\ -6 & 7-\lambda & -4 \\ 2 & -4 & 3-\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 $\lambda=0$ in (1)

$ \left[ {\begin{array}{cc} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{array} } \right]$ $ \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$= $ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$

$8x-6y+2z=0$

$-6x+7y-4z=0$

By Cramer's rule,

$\frac{x}{\mid {\begin{array}{cc} -6 & 2 \\ 7 & -4 \\ \end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc} 8 & 2 \\ -6 & -4 \\ \end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc} 8 & -6 \\ -6 & 7 \\ \end{array} } \mid}$

$\frac{x}{10}=\frac{-y}{-20}=\frac{z}{20}$

$ X_1=$ $ \left[ {\begin{array}{cc} 1 \\ 2 \\ 2 \end{array} } \right]$

Put $\lambda=3$ in (1) we get,

$\left[ {\begin{array}{cc} 5 & -6 & 2 \\ -6 & 4 & -4 \\ 2 & -4 & 0 \end{array} } \right]$$ \left[ {\begin{array}{cc} x \ y \ z \end{array} } \right]$=$ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$

From 1st and 3rd row as they are distinct

$5x-6y+2z=0$

$2x-4y+0=0$

By Cramer's rule,

$\frac{x}{\mid {\begin{array}{cc} -6 & 2 \\ -4 & 0 \\ \end{array} } \mid}$=$ \frac{x}{\mid {\begin{array}{cc} 5 & 2 \\ 2 & 0 \\ \end{array} } \mid}$= $\frac{x}{\mid {\begin{array}{cc} 5 & -6 \\ 2 & -4 \\ \end{array} } \mid}$

$\frac{x}{8}=\frac{-y}{-4}=\frac{z}{-8}$

$ X_2=$ $ \left[ {\begin{array}{cc} 2 \\ 1 \\ -2 \end{array} } \right]$

Put $\lambda=15$ in (1), we get

$ \left[ {\begin{array}{cc} -7 & -6 & 2\\ -6 & 8 & -4 \\ 2 & -4 & -12 \end{array} } \right] $ $ \left[ {\begin{array}{cc} x\\ y \\ z \end{array} } \right] $= $ \left[ {\begin{array}{cc} 0\\ 0 \\ 0 \end{array} } \right]$

$-7x-6y+2z=0$

$-6x-8y-4z=0$

$\frac{x}{\mid {\begin{array}{cc} -6 & 2 \\ -8 & -4\\ \end{array} } \mid}$ = $\frac{-y}{\mid {\begin{array}{cc} -7 & 2 \\ -6 & -4 \\ \end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc} -7 & -6 \\ -6 & -8 \\ \end{array} } \mid}$

$\frac{x}{2}=\frac{-y}{2}=\frac{z}{1} $

$ X_3=$ $ \left[ {\begin{array}{cc} 2\\ -2 \\ 1 \end{array} } \right]$

Transforming matrix $M=$ $ \left[ {\begin{array}{cc} 1 & 2 & 2\\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array} } \right] $

Diagonal matrix $D=$ $ \left[ {\begin{array}{cc} 0 & 0 & 0\\ 0 & 3 & 0 \\ 0 & 0 & 15 \end{array} } \right] $

  1. Characteristic equation of A is $\mid A-\lambda I \mid = 0$

$\lambda^3-(-9+17+3) \lambda^2 + [(-27+21-63)-(-32-64+32)]\lambda - \mid A \mid =0$

$\lambda^3-\lambda^2-5\lambda-3=0$

$\lambda=3,-1, -1$ are eigen values of A

Now to find eigen vector $X\neq0$ such that $[A-\lambda I]=0$

$ \left[ {\begin{array}{cc} 9-\lambda & 4 & 4 \\ -8 & 3-\lambda & 4 \\ -16 & 8 & 7-\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 $\lambda=3$ in (1)

$ \left[ {\begin{array}{cc} -12 & 4 & 4 \\ -8 & 0 & 4 \\ -16 & 8 & 4 \end{array} } \right]$ $ \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$= $ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$

$-8x-0y+4z=0$

$-16x+8y-4z=0$

By Cramer's rule,

$\frac{x}{\mid {\begin{array}{cc} 0 & 4 \\ 8 & 4 \\ \end{array} } \mid}$= $\frac{-y}{\mid {\begin{array}{cc} -8 & 4 \\ -16 & 4 \\ \end{array} } \mid}$= $\frac{z}{\mid {\begin{array}{cc} -8 & 0 \\ -16 & 8 \\ \end{array} } \mid}$

$\frac{x}{-1}=\frac{-y}{1}=\frac{z}{-2}$

$ X_1=$ $ \left[ {\begin{array}{cc} -1 \\ -1 \\ -2 \end{array} } \right]$

Put $\lambda=-1$ in (1) we get,

$\left[ {\begin{array}{cc} -8 & 4 & 4 \\ -8 & 4 & 4 \\ 16 & 8 & 8 \end{array} } \right]$ $ \left[ {\begin{array}{cc} x \\ y \\ z \end{array} } \right]$=$ \left[ {\begin{array}{cc} 0 \\ 0 \\ 0 \end{array} } \right]$

From 1st and 3rd row as they are distinct

$-8x+4y+4z=0$

$-2x+y+z=0$

$x=0$, $x=1$.

$y=1$, $y=0$

$z=-1$, $z=2$

$ X_2=$ $ \left[ {\begin{array}{cc} 0 \\ 1 \\ -1 \end{array} } \right]$

$ X_3=$ $ \left[ {\begin{array}{cc} 1\\ 0 \\ 2 \end{array} } \right]$

For $\lambda=1$ algebraic multiplicity=geometric multiplicity=2. Therefore, algebraic multiplicity=geometric multiplicity. Hence, A is diagonalisable.

Transforming matrix $M=$ $ \left[ {\begin{array}{cc} -1 & 0 & 1\\ -1 & 1 & 0 \\ -2 & -1 & 2 \end{array} } \right] $

Diagonal matrix $D=$ $ \left[ {\begin{array}{cc} 3 & 0 & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} } \right] $

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modified 11 weeks ago  • written 11 weeks ago by gravatar for manasahegde234 manasahegde23420
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