Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks : 8 M

Year : May 2014, Dec 2014 , May 2015

Characteristic equation is | A- λ I | =0

∴ (8-λ)[(7-λ)(3-λ)-16 ]+6[-6(3-λ)+8 ]+2 [24-2(7- λ)]=0

∴ $(8-λ)(21-10λ+ λ^2-16)+[10+6λ ]+2[10+2λ]=0$

∴ $(8-λ)(λ^2-10λ+5)-60+36 λ+20+4λ=0$

∴ $8λ^2-80λ+40- λ^3 +10 λ^2-5λ-40+40λ=0$

∴ $- λ^3+18 λ^2-45λ+0=0$

∴ $ λ(λ^2-18 λ+45)=0$

∴ λ(λ-3)(λ-15)=0

∴ λ=0 ,3 ,15

Eigen values are 0 , 3, 15 .

Case 1 , when λ=0

| A- λ I | X=0

$ \left[ \begin{array}{ccc} 8 & -6 & 2 \\ -6 & 7 & -4 \\ 2 & -4 & 3 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \\ X_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] $

∴ $8 X_1-6 X_2+2X_3 = 0$ and $- 6 X_1+7 X_(2 )-4X_3 = 0$

Using Cramer’s Rule :

$ \dfrac{X_1}{ \left| \begin{array}{cc} -6 & 2 \\ 7 & -4 \end{array}\right| } \;=\; \dfrac{X_2}{ \left| \begin{array}{cc} 8 & 2 \\ -6 & -4 \end{array}\right| } \;=\; \dfrac{X_3}{ \left| \begin{array}{cc} 8 & -6 \\ -6 & 7 \end{array}\right| } \\ \dfrac{X_1}{ 10} \;=\; \dfrac{-X_2}{ -20} \;=\; \dfrac{X_3}{ 20} \\ \therefore \dfrac{X_1}{ 1} \;=\; \dfrac{-X_2}{ -2} \;=\; \dfrac{X_3}{ 2} $

∴ Eigen vector $X_1=[1 \; 2 \; 2 ]$

If k is non zero scalar then $kX_1$ is also an eigen vector.

Case 2 , when λ=3

| A- λ I | X=0

$ \left[ \begin{array}{ccc} 5 & -6 & 2 \\ -6 & 4 & -4 \\ 2 & -4 & 0 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \\ X_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] $

∴ $5 X_1-6 X_2+2X_3 = 0$ and $- 6 X_1+4 X_(2 )-4X_3 = 0$

Using Cramer’s Rule :

$ \dfrac{X_1}{ \left| \begin{array}{cc} -6 & 2 \\ 4 & -4 \end{array}\right| } \;=\; \dfrac{X_2}{ \left| \begin{array}{cc} 5 & 2 \\ -6 & -4 \end{array}\right| } \;=\; \dfrac{X_3}{ \left| \begin{array}{cc} 5 & -6 \\ -6 & 4 \end{array}\right| } \\ \dfrac{X_1}{ 16} \;=\; \dfrac{-X_2}{ -8} \;=\; \dfrac{X_3}{ -16} \\ \therefore \dfrac{X_1}{ 2} \;=\; \dfrac{-X_2}{ -1} \;=\; \dfrac{X_3}{ -2} $

∴ Eigen vector $X_1=[2 \; 1 \; -2]$

If k is non zero scalar then $kX_2$ is also an eigen vector.

Case 3 , when λ=15

| A- λ I | X=0

$ \left[ \begin{array}{ccc} 7 & -6 & 2 \\ -6 & -8 & -4 \\ 2 & -4 & -12 \end{array}\right] \left[ \begin{array}{ccc} X_1 \\ X_2 \\ X_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] $

∴ $-7 X_1-6 X_2+2X_3 = 0$ and $- 6 X_1-8 X_(2 )-4X_3 = 0$

Using Cramer’s Rule :

$ \dfrac{X_1}{ \left| \begin{array}{cc} -6 & 2 \\ -8 & -4 \end{array}\right| } \;=\; \dfrac{X_2}{ \left| \begin{array}{cc} -7 & 2 \\ -6 & -4 \end{array}\right| } \;=\; \dfrac{X_3}{ \left| \begin{array}{cc} -7 & -6 \\ -6 & -8 \end{array}\right| } \\ \dfrac{X_1}{ 40} \;=\; \dfrac{-X_2}{ 40} \;=\; \dfrac{X_3}{ 20} \\ \therefore \dfrac{X_1}{ 2} \;=\; \dfrac{-X_2}{ -2} \;=\; \dfrac{X_3}{ -1} $

∴ Eigen vector $X_1=[2 \; -2 \; 1]$

If k is non zero scalar then $kX_3$ is also an eigen vector.

Since all the eigen values are distinct , matrix A is diagonalizable.

$M^{-1} A M=D$

So the given matrix A is diagonalized to diagonal matrix D by the transforming matrix M

Where D = $ \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 15 \end{array}\right] \\ \; \\ \; \\ $

And M = $ \left[ \begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & -1 \end{array}\right] $