Answer on Question #46091-Engineering-Other
Let
A = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) A = \left( \begin{array}{ccc} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{array} \right) A = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞
a) Find the adjoint of A. Find the inverse of A from the adjoint of A. (4)
b) Find the characteristic and minimal polynomials of A. Hence find its eigenvalues and eigenvectors. (6)
c) Why is A diagonalisable? Find a matrix P such that P1AP is diagonal. (2)
d) Verify Cayley-Hamilton theorem for A. Hence, find the inverse of A
Solution
a) Find the adjoint of A.
First find the cofactor of each element.
A 11 = ∣ 7 − 6 12 − 11 ∣ = − 5 , A 12 = − ∣ 6 − 6 12 − 11 ∣ = − 6 , A 13 = ∣ 6 7 12 12 ∣ = − 12 , A 21 = − ∣ 4 − 4 12 − 11 ∣ = − 4 , A 22 = ∣ 5 − 4 12 − 11 ∣ = − 7 , A 23 = − ∣ 5 4 12 12 ∣ = − 12 , A 31 = ∣ 4 − 4 7 − 6 ∣ = 4 , A 32 = − ∣ 5 − 4 6 − 6 ∣ = 6 , A 33 = ∣ 5 4 6 7 ∣ = 11. \begin{array}{l}
A_{11} = \left| \begin{array}{cc} 7 & -6 \\ 12 & -11 \end{array} \right| = -5, A_{12} = -\left| \begin{array}{cc} 6 & -6 \\ 12 & -11 \end{array} \right| = -6, A_{13} = \left| \begin{array}{cc} 6 & 7 \\ 12 & 12 \end{array} \right| = -12, A_{21} = \\
-\left| \begin{array}{cc} 4 & -4 \\ 12 & -11 \end{array} \right| = -4, A_{22} = \left| \begin{array}{cc} 5 & -4 \\ 12 & -11 \end{array} \right| = -7, A_{23} = -\left| \begin{array}{cc} 5 & 4 \\ 12 & 12 \end{array} \right| = -12, A_{31} = \left| \begin{array}{cc} 4 & -4 \\ 7 & -6 \end{array} \right| = \\
4, A_{32} = -\left| \begin{array}{cc} 5 & -4 \\ 6 & -6 \end{array} \right| = 6, A_{33} = \left| \begin{array}{cc} 5 & 4 \\ 6 & 7 \end{array} \right| = 11.
\end{array} A 11 = ∣ ∣ 7 12 − 6 − 11 ∣ ∣ = − 5 , A 12 = − ∣ ∣ 6 12 − 6 − 11 ∣ ∣ = − 6 , A 13 = ∣ ∣ 6 12 7 12 ∣ ∣ = − 12 , A 21 = − ∣ ∣ 4 12 − 4 − 11 ∣ ∣ = − 4 , A 22 = ∣ ∣ 5 12 − 4 − 11 ∣ ∣ = − 7 , A 23 = − ∣ ∣ 5 12 4 12 ∣ ∣ = − 12 , A 31 = ∣ ∣ 4 7 − 4 − 6 ∣ ∣ = 4 , A 32 = − ∣ ∣ 5 6 − 4 − 6 ∣ ∣ = 6 , A 33 = ∣ ∣ 5 6 4 7 ∣ ∣ = 11.
As a result the cofactor matrix of A is
( − 5 − 6 − 12 − 4 − 7 − 12 4 6 11 ) . \left( \begin{array}{ccc} -5 & -6 & -12 \\ -4 & -7 & -12 \\ 4 & 6 & 11 \end{array} \right). ⎝ ⎛ − 5 − 4 4 − 6 − 7 6 − 12 − 12 11 ⎠ ⎞ .
Finally the adjoint of A is the transpose of the cofactor matrix:
( − 5 − 4 4 − 6 − 7 6 − 12 − 12 11 ) . \left( \begin{array}{ccc} -5 & -4 & 4 \\ -6 & -7 & 6 \\ -12 & -12 & 11 \end{array} \right). ⎝ ⎛ − 5 − 6 − 12 − 4 − 7 − 12 4 6 11 ⎠ ⎞ .
Find the inverse of A from the adjoint of A.
det A = a 11 A 11 + a 21 A 21 + a 31 A 31 = − 1. \det A = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31} = -1. det A = a 11 A 11 + a 21 A 21 + a 31 A 31 = − 1. A − 1 = 1 det A ( − 5 − 4 4 − 6 − 7 6 − 12 − 12 11 ) = 1 − 1 ( − 5 − 4 4 − 6 − 7 6 − 12 − 12 11 ) = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) . A^{-1} = \frac{1}{\det A} \left( \begin{array}{ccc} -5 & -4 & 4 \\ -6 & -7 & 6 \\ -12 & -12 & 11 \end{array} \right) = \frac{1}{-1} \left( \begin{array}{ccc} -5 & -4 & 4 \\ -6 & -7 & 6 \\ -12 & -12 & 11 \end{array} \right) = \left( \begin{array}{ccc} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{array} \right). A − 1 = det A 1 ⎝ ⎛ − 5 − 6 − 12 − 4 − 7 − 12 4 6 11 ⎠ ⎞ = − 1 1 ⎝ ⎛ − 5 − 6 − 12 − 4 − 7 − 12 4 6 11 ⎠ ⎞ = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ .
b) The characteristic equation for Q Q Q λ 3 − D 1 λ 2 + D 2 λ − D 3 = 0 \lambda^3 - D_1\lambda^2 + D_2\lambda - D_3 = 0 λ 3 − D 1 λ 2 + D 2 λ − D 3 = 0
D 1 = 5 + 7 − 11 = 1 ; D_1 = 5 + 7 - 11 = 1; D 1 = 5 + 7 − 11 = 1 ; D 2 = A 11 + A 22 + A 33 = − 5 − 7 + 11 = − 1. D_2 = A_{11} + A_{22} + A_{33} = -5 - 7 + 11 = -1. D 2 = A 11 + A 22 + A 33 = − 5 − 7 + 11 = − 1. D 3 = ∣ A ∣ = − 1. D_3 = |A| = -1. D 3 = ∣ A ∣ = − 1.
Hence
λ 3 − λ 2 − λ + 1 = 0 → ( λ − 1 ) 2 ( λ + 1 ) = 0. \lambda^3 - \lambda^2 - \lambda + 1 = 0 \rightarrow (\lambda - 1)^2 (\lambda + 1) = 0. λ 3 − λ 2 − λ + 1 = 0 → ( λ − 1 ) 2 ( λ + 1 ) = 0.
The minimal polynomials of A are
( λ − 1 ) , ( λ − 1 ) ( λ + 1 ) , ( λ + 1 ) , ( λ − 1 ) 2 . (\lambda - 1), (\lambda - 1)(\lambda + 1), (\lambda + 1), (\lambda - 1)^2. ( λ − 1 ) , ( λ − 1 ) ( λ + 1 ) , ( λ + 1 ) , ( λ − 1 ) 2 .
The eigenvalues of A are λ 1 = − 1 \lambda_1 = -1 λ 1 = − 1 λ 2 = + 1 \lambda_2 = +1 λ 2 = + 1
Find eigenvector corresponding λ 1 = − 1 \lambda_1 = -1 λ 1 = − 1
( 5 + 1 4 − 4 6 7 + 1 − 6 12 12 − 11 + 1 ) ( x y z ) = ( 0 0 0 ) → v 1 = 1 7 ( 2 3 6 ) . \left( \begin{array}{ccc} 5 + 1 & 4 & -4 \\ 6 & 7 + 1 & -6 \\ 12 & 12 & -11 + 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to v_1 = \frac{1}{7} \left( \begin{array}{c} 2 \\ 3 \\ 6 \end{array} \right). ⎝ ⎛ 5 + 1 6 12 4 7 + 1 12 − 4 − 6 − 11 + 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ → v 1 = 7 1 ⎝ ⎛ 2 3 6 ⎠ ⎞ .
Find eigenvectors corresponding λ 2 = 1 \lambda_2 = 1 λ 2 = 1
( 5 − 1 4 − 4 6 7 − 1 − 6 12 12 − 11 − 1 ) ( x y z ) = ( 0 0 0 ) → x + y − z = 0 → v 2 = 1 2 ( 1 0 1 ) , v 3 = 1 2 ( 0 1 1 ) . \left( \begin{array}{ccc} 5 - 1 & 4 & -4 \\ 6 & 7 - 1 & -6 \\ 12 & 12 & -11 - 1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \to x + y - z = 0 \to v_2 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right), v_3 = \frac{1}{\sqrt{2}} \left( \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right). ⎝ ⎛ 5 − 1 6 12 4 7 − 1 12 − 4 − 6 − 11 − 1 ⎠ ⎞ ⎝ ⎛ x y z ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ → x + y − z = 0 → v 2 = 2 1 ⎝ ⎛ 1 0 1 ⎠ ⎞ , v 3 = 2 1 ⎝ ⎛ 0 1 1 ⎠ ⎞ .
c) Because A have eigenvector basis.
Find a matrix P P P P − 1 A P P^{-1}AP P − 1 A P
P = ( 2 7 1 2 0 3 7 0 1 2 6 7 1 2 1 2 ) . P = \left( \begin{array}{ccc} \frac{2}{7} & \frac{1}{\sqrt{2}} & 0 \\ \frac{3}{7} & 0 & \frac{1}{\sqrt{2}} \\ \frac{6}{7} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array} \right). P = ⎝ ⎛ 7 2 7 3 7 6 2 1 0 2 1 0 2 1 2 1 ⎠ ⎞ . P − 1 A P = D = ( − 1 0 0 0 1 0 0 0 1 ) . P^{-1}AP = D = \left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right). P − 1 A P = D = ⎝ ⎛ − 1 0 0 0 1 0 0 0 1 ⎠ ⎞ .
d) Verify Cayley-Hamilton theorem for A. Hence, find the inverse of A.
We have to show that
A 3 − A 2 − A + I = 0. A^3 - A^2 - A + I = 0. A 3 − A 2 − A + I = 0. A 2 = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) ( 5 4 − 4 6 7 − 6 12 12 − 11 ) = ( 1 0 0 0 1 0 0 0 1 ) . A^2 = \left( \begin{array}{ccc} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{array} \right) \left( \begin{array}{ccc} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right). A 2 = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ = ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ . A 3 = A 2 A = I A = A . A^3 = A^2 A = I A = A. A 3 = A 2 A = I A = A .
So
A − I − A + I = 0. A - I - A + I = 0. A − I − A + I = 0.
This verifies Cayley-Hamilton theorem.
The inverse of A:
A 2 = A A = 1 → A − 1 = A = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) . A^2 = A A = 1 \to A^{-1} = A = \left( \begin{array}{ccc} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{array} \right). A 2 = AA = 1 → A − 1 = A = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ .
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#339914 on Dec 2023