Answer on question 42307 - Math - Linear Algebra
Let
A = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) A = \left( \begin{array}{c c c} 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.
b) Find the characteristic and minimal polynomials of A. Hence find its eigenvalues and eigenvectors.
c) Why is A diagonalisable? Find a matrix P P P P ∧ ( − 1 ) P^{\wedge}(-1) P ∧ ( − 1 )
d) Verify Cayley-Hamilton theorem for A. Hence, find the inverse of A.
Solution
a) Adjoint matrix is the matrix formed by taking the transpose of the cofactor matrix of a given original matrix. To find it we need to find the following determinants
A 11 = ( − 1 ) 1 + 1 ∣ 7 − 6 12 − 11 ∣ = − 77 + 72 = − 5 ; A _ {1 1} = (- 1) ^ {1 + 1} \left| \begin{array}{c c} 7 & - 6 \\ 1 2 & - 1 1 \end{array} \right| = - 7 7 + 7 2 = - 5; A 11 = ( − 1 ) 1 + 1 ∣ ∣ 7 12 − 6 − 11 ∣ ∣ = − 77 + 72 = − 5 ; A 12 = ( − 1 ) 1 + 2 ∣ 6 − 6 12 − 11 ∣ = 66 − 72 = − 6 ; A _ {1 2} = (- 1) ^ {1 + 2} \left| \begin{array}{c c} 6 & - 6 \\ 1 2 & - 1 1 \end{array} \right| = 6 6 - 7 2 = - 6; A 12 = ( − 1 ) 1 + 2 ∣ ∣ 6 12 − 6 − 11 ∣ ∣ = 66 − 72 = − 6 ; A 13 = ( − 1 ) 1 + 3 ∣ 6 7 12 12 ∣ = 72 − 84 = − 12 ; A _ {1 3} = (- 1) ^ {1 + 3} \left| \begin{array}{c c} 6 & 7 \\ 1 2 & 1 2 \end{array} \right| = 7 2 - 8 4 = - 1 2; A 13 = ( − 1 ) 1 + 3 ∣ ∣ 6 12 7 12 ∣ ∣ = 72 − 84 = − 12 ; A 21 = ( − 1 ) 2 + 1 ∣ 4 − 4 12 − 11 ∣ = 44 − 48 = − 4 ; A _ {2 1} = (- 1) ^ {2 + 1} \left| \begin{array}{c c} 4 & - 4 \\ 1 2 & - 1 1 \end{array} \right| = 4 4 - 4 8 = - 4; A 21 = ( − 1 ) 2 + 1 ∣ ∣ 4 12 − 4 − 11 ∣ ∣ = 44 − 48 = − 4 ; A 22 = ( − 1 ) 2 + 2 ∣ 5 − 4 12 − 11 ∣ = − 55 + 48 = − 7 ; A _ {2 2} = (- 1) ^ {2 + 2} \left| \begin{array}{c c} 5 & - 4 \\ 1 2 & - 1 1 \end{array} \right| = - 5 5 + 4 8 = - 7; A 22 = ( − 1 ) 2 + 2 ∣ ∣ 5 12 − 4 − 11 ∣ ∣ = − 55 + 48 = − 7 ; A 23 = ( − 1 ) 2 + 3 ∣ 5 4 12 12 ∣ = − 60 + 48 = − 12 ; A _ {2 3} = (- 1) ^ {2 + 3} \left| \begin{array}{c c} 5 & 4 \\ 1 2 & 1 2 \end{array} \right| = - 6 0 + 4 8 = - 1 2; A 23 = ( − 1 ) 2 + 3 ∣ ∣ 5 12 4 12 ∣ ∣ = − 60 + 48 = − 12 ; A 31 = ( − 1 ) 3 + 1 ∣ 4 − 4 7 − 6 ∣ = − 24 + 28 = 4 ; A _ {3 1} = (- 1) ^ {3 + 1} \left| \begin{array}{c c} 4 & - 4 \\ 7 & - 6 \end{array} \right| = - 2 4 + 2 8 = 4; A 31 = ( − 1 ) 3 + 1 ∣ ∣ 4 7 − 4 − 6 ∣ ∣ = − 24 + 28 = 4 ; A 32 = ( − 1 ) 3 + 2 ∣ 5 − 4 6 − 6 ∣ = 30 − 24 = 6 ; A _ {3 2} = (- 1) ^ {3 + 2} \left| \begin{array}{c c} 5 & - 4 \\ 6 & - 6 \end{array} \right| = 3 0 - 2 4 = 6; A 32 = ( − 1 ) 3 + 2 ∣ ∣ 5 6 − 4 − 6 ∣ ∣ = 30 − 24 = 6 ; A 12 = ( − 1 ) 3 + 3 ∣ 5 4 6 7 ∣ = 35 − 24 = 11 ; A _ {1 2} = (- 1) ^ {3 + 3} \left| \begin{array}{c c} 5 & 4 \\ 6 & 7 \end{array} \right| = 3 5 - 2 4 = 1 1; A 12 = ( − 1 ) 3 + 3 ∣ ∣ 5 6 4 7 ∣ ∣ = 35 − 24 = 11 ;
Therefore
a d j A = ( − 5 − 4 4 − 6 − 7 6 − 12 − 12 11 ) a d j A = \left( \begin{array}{c c c} - 5 & - 4 & 4 \\ - 6 & - 7 & 6 \\ - 1 2 & - 1 2 & 1 1 \end{array} \right) a d j A = ⎝ ⎛ − 5 − 6 − 12 − 4 − 7 − 12 4 6 11 ⎠ ⎞
Using the triangle rule we find the determinant of A
det ( A ) = ∣ 5 4 − 4 6 7 − 6 12 12 − 11 ∣ = 5 ∗ − 7 ∗ ( − 11 ) + 6 ∗ 12 ∗ ( − 4 ) + 4 ∗ 12 ∗ ( − 6 ) − 12 ∗ 7 ∗ ( − 4 ) + 6 ∗ 4 ∗ ( − 11 ) + 12 ∗ ( − 6 ) ∗ 5 ) = − 1. \det (A) = \left| \begin{array}{c c c} 5 & 4 & - 4 \\ 6 & 7 & - 6 \\ 1 2 & 1 2 & - 1 1 \end{array} \right| = 5 * - 7 * (- 1 1) + 6 * 1 2 * (- 4) + 4 * 1 2 * (- 6) - 1 2 * 7 * (- 4) + 6 * 4 * (- 1 1) + 1 2 * (- 6) * 5) = - 1. det ( A ) = ∣ ∣ 5 6 12 4 7 12 − 4 − 6 − 11 ∣ ∣ = 5 ∗ − 7 ∗ ( − 11 ) + 6 ∗ 12 ∗ ( − 4 ) + 4 ∗ 12 ∗ ( − 6 ) − 12 ∗ 7 ∗ ( − 4 ) + 6 ∗ 4 ∗ ( − 11 ) + 12 ∗ ( − 6 ) ∗ 5 ) = − 1.
Using the formula
A − 1 = 1 det ( A ) ∗ a d j A A ^ {- 1} = \frac {1}{\det (A)} * a d j A A − 1 = det ( A ) 1 ∗ a d j A
We obtain
A − 1 = − 1 ( − 5 − 4 4 − 6 − 7 6 − 12 − 12 11 ) = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) . A ^ {- 1} = - 1 \left( \begin{array}{c c c} - 5 & - 4 & 4 \\ - 6 & - 7 & 6 \\ - 1 2 & - 1 2 & 1 1 \end{array} \right) = \left( \begin{array}{c c c} 5 & 4 & - 4 \\ 6 & 7 & - 6 \\ 1 2 & 1 2 & - 1 1 \end{array} \right). A − 1 = − 1 ⎝ ⎛ − 5 − 6 − 12 − 4 − 7 − 12 4 6 11 ⎠ ⎞ = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ .
b) The characteristic polynomial of the matrix is
P ( x ) = det ( x I − A ) = ∣ x − 5 − 4 4 − 6 x − 7 6 − 12 − 12 x + 11 ∣ = ( x − 5 ) ( x − 7 ) ( x + 11 ) + + 288 + 288 + 48 ( x − 7 ) − 24 ( x + 11 ) + 72 ( x − 5 ) = = x 3 − x 2 − x + 1 = ( x − 1 ) 2 ( x + 1 ) \begin{array}{l} P (x) = \det (x I - A) = \left| \begin{array}{c c c} x - 5 & - 4 & 4 \\ - 6 & x - 7 & 6 \\ - 1 2 & - 1 2 & x + 1 1 \end{array} \right| = (x - 5) (x - 7) (x + 1 1) + \\ + 2 8 8 + 2 8 8 + 4 8 (x - 7) - 2 4 (x + 1 1) + 7 2 (x - 5) = \\ = x ^ {3} - x ^ {2} - x + 1 = (x - 1) ^ {2} (x + 1) \\ \end{array} P ( x ) = det ( x I − A ) = ∣ ∣ x − 5 − 6 − 12 − 4 x − 7 − 12 4 6 x + 11 ∣ ∣ = ( x − 5 ) ( x − 7 ) ( x + 11 ) + + 288 + 288 + 48 ( x − 7 ) − 24 ( x + 11 ) + 72 ( x − 5 ) = = x 3 − x 2 − x + 1 = ( x − 1 ) 2 ( x + 1 )
Thus the distinct eigenvalues of A A A m A m_A m A P ( x ) P(x) P ( x ) A A A m A m_A m A m A ( x ) = x 2 − 1 m_A(x) = x^2 - 1 m A ( x ) = x 2 − 1 m A ( x ) = ( x − 1 ) 2 ( x + 1 ) m_A(x) = (x - 1)^2 (x + 1) m A ( x ) = ( x − 1 ) 2 ( x + 1 )
A 2 − I = ( 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 ) = = ( 25 + 24 − 48 20 + 28 − 48 − 20 − 24 + 44 30 + 42 − 72 24 + 49 − 72 − 24 − 42 + 66 60 + 72 − 132 48 + 84 − 132 − 48 − 72 + 121 ) − ( 1 0 0 0 1 0 0 0 1 ) = ( 0 0 0 0 0 0 0 0 0 ) \begin{array}{l} A ^ {2} - I = \left( \begin{array}{c c c} 5 & 4 & - 4 \\ 6 & 7 & - 6 \\ 1 2 & 1 2 & - 1 1 \end{array} \right) \left( \begin{array}{c c c} 5 & 4 & - 4 \\ 6 & 7 & - 6 \\ 1 2 & 1 2 & - 1 1 \end{array} \right) - \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = \\ = \left( \begin{array}{c c c} 2 5 + 2 4 - 4 8 & 2 0 + 2 8 - 4 8 & - 2 0 - 2 4 + 4 4 \\ 3 0 + 4 2 - 7 2 & 2 4 + 4 9 - 7 2 & - 2 4 - 4 2 + 6 6 \\ 6 0 + 7 2 - 1 3 2 & 4 8 + 8 4 - 1 3 2 & - 4 8 - 7 2 + 1 2 1 \end{array} \right) - \left( \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{c c c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) \\ \end{array} A 2 − I = ⎝ ⎛ 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 ⎠ ⎞ = = ⎝ ⎛ 25 + 24 − 48 30 + 42 − 72 60 + 72 − 132 20 + 28 − 48 24 + 49 − 72 48 + 84 − 132 − 20 − 24 + 44 − 24 − 42 + 66 − 48 − 72 + 121 ⎠ ⎞ − ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 ⎠ ⎞
Hence the minimal polynomial of A A A m A ( x ) = x 2 − 1 m_A(x) = x^2 - 1 m A ( x ) = x 2 − 1
As we said before the eigenvalues of A are -1 and 1 with algebraic multiplicity is 2.
Let us find the eigenvectors
1) X = − 1 X = -1 X = − 1
( 6 4 − 4 6 8 − 6 12 12 − 10 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) ⇒ ( x 1 x 2 x 3 ) = ( 2 3 6 ) \left( \begin{array}{c c c} 6 & 4 & - 4 \\ 6 & 8 & - 6 \\ 1 2 & 1 2 & - 1 0 \end{array} \right) \left( \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \Rightarrow \left( \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right) = \left( \begin{array}{c} 2 \\ 3 \\ 6 \end{array} \right) ⎝ ⎛ 6 6 12 4 8 12 − 4 − 6 − 10 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ⇒ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 2 3 6 ⎠ ⎞
2) X = 1 X = 1 X = 1
( 4 4 − 4 6 6 − 6 12 12 − 12 ) ( x 1 x 2 x 3 ) = ( 0 0 0 ) ⇒ ( x 1 x 2 x 3 ) = ( c 2 − c 1 c 1 c 2 ) \left( \begin{array}{c c c} 4 & 4 & - 4 \\ 6 & 6 & - 6 \\ 1 2 & 1 2 & - 1 2 \end{array} \right) \left( \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \Rightarrow \left( \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right) = \left( \begin{array}{c} c _ {2} - c _ {1} \\ c _ {1} \\ c _ {2} \end{array} \right) ⎝ ⎛ 4 6 12 4 6 12 − 4 − 6 − 12 ⎠ ⎞ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 0 0 0 ⎠ ⎞ ⇒ ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ c 2 − c 1 c 1 c 2 ⎠ ⎞
Where c 1 c_{1} c 1 c 2 c_{2} c 2
( x 1 x 2 x 3 ) = ( 1 0 1 ) or ( x 1 x 2 x 3 ) = ( − 1 1 0 ) . \left( \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right) = \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right) \text { or } \left( \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right) = \left( \begin{array}{c} - 1 \\ 1 \\ 0 \end{array} \right). ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ 1 0 1 ⎠ ⎞ or ⎝ ⎛ x 1 x 2 x 3 ⎠ ⎞ = ⎝ ⎛ − 1 1 0 ⎠ ⎞ .
c) Use the Theorem: An n × n n \times n n × n A A A n n n
Check whether the eigenvectors of A A A
∣ 2 1 − 1 3 0 1 6 1 0 ∣ = − 3 + 6 − 2 = 1 \left| \begin{array}{c c c} 2 & 1 & - 1 \\ 3 & 0 & 1 \\ 6 & 1 & 0 \end{array} \right| = - 3 + 6 - 2 = 1 ∣ ∣ 2 3 6 1 0 1 − 1 1 0 ∣ ∣ = − 3 + 6 − 2 = 1
Therefore the eigenvectors of A are linearly independent and the matrix A is diagonalizable.
P is the matrix that consists of the eigenvectors. Check it
P − 1 A P = ( 2 1 − 1 3 0 1 6 1 0 ) − 1 ( 5 4 − 4 6 7 − 6 12 12 − 11 ) ( 2 1 − 1 3 0 1 6 1 0 ) = = ( − 1 − 1 1 6 6 − 5 3 4 − 3 ) ( 5 4 − 4 6 7 − 6 12 12 − 11 ) ( 2 1 − 1 3 0 1 6 1 0 ) = = ( 1 1 1 6 6 − 5 3 4 − 3 ) ( 2 1 − 1 3 0 1 6 1 0 ) = ( − 1 0 0 0 1 0 0 0 1 ) \begin{array}{l} P ^ {- 1} A P = \left( \begin{array}{c c c} 2 & 1 & - 1 \\ 3 & 0 & 1 \\ 6 & 1 & 0 \end{array} \right) ^ {- 1} \left( \begin{array}{c c c} 5 & 4 & - 4 \\ 6 & 7 & - 6 \\ 1 2 & 1 2 & - 1 1 \end{array} \right) \left( \begin{array}{c c c} 2 & 1 & - 1 \\ 3 & 0 & 1 \\ 6 & 1 & 0 \end{array} \right) = \\ = \left( \begin{array}{c c c} - 1 & - 1 & 1 \\ 6 & 6 & - 5 \\ 3 & 4 & - 3 \end{array} \right) \left( \begin{array}{c c c} 5 & 4 & - 4 \\ 6 & 7 & - 6 \\ 1 2 & 1 2 & - 1 1 \end{array} \right) \left( \begin{array}{c c c} 2 & 1 & - 1 \\ 3 & 0 & 1 \\ 6 & 1 & 0 \end{array} \right) = \\ = \left( \begin{array}{c c c} 1 & 1 & 1 \\ 6 & 6 & - 5 \\ 3 & 4 & - 3 \end{array} \right) \left( \begin{array}{c c c} 2 & 1 & - 1 \\ 3 & 0 & 1 \\ 6 & 1 & 0 \end{array} \right) = \left( \begin{array}{c c c} - 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \end{array} P − 1 A P = ⎝ ⎛ 2 3 6 1 0 1 − 1 1 0 ⎠ ⎞ − 1 ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ ⎝ ⎛ 2 3 6 1 0 1 − 1 1 0 ⎠ ⎞ = = ⎝ ⎛ − 1 6 3 − 1 6 4 1 − 5 − 3 ⎠ ⎞ ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ ⎝ ⎛ 2 3 6 1 0 1 − 1 1 0 ⎠ ⎞ = = ⎝ ⎛ 1 6 3 1 6 4 1 − 5 − 3 ⎠ ⎞ ⎝ ⎛ 2 3 6 1 0 1 − 1 1 0 ⎠ ⎞ = ⎝ ⎛ − 1 0 0 0 1 0 0 0 1 ⎠ ⎞
d) The Cayley-Hamilton theorem states that "substituting" the matrix A A A λ \lambda λ
P ( A ) = A 3 − A 2 − A + 1 = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) 3 − ( 5 4 − 4 6 7 − 6 12 12 − 11 ) 2 − P(A) = A^3 - A^2 - A + 1 = \begin{pmatrix} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{pmatrix}^3 - \begin{pmatrix} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{pmatrix}^2 - P ( A ) = A 3 − A 2 − A + 1 = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ 3 − ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ 2 − − ( 5 4 − 4 6 7 − 6 12 12 − 11 ) + ( 1 0 0 0 1 0 0 0 1 ) = ( 5 4 − 4 6 7 − 6 12 12 − 11 ) − ( 1 0 0 0 1 0 0 0 1 ) − - \begin{pmatrix} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} - − ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ + ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ = ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ − ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ − − ( 5 4 − 4 6 7 − 6 12 12 − 11 ) + ( 1 0 0 0 1 0 0 0 1 ) = ( 0 0 0 0 0 0 0 0 0 ) . - \begin{pmatrix} 5 & 4 & -4 \\ 6 & 7 & -6 \\ 12 & 12 & -11 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. − ⎝ ⎛ 5 6 12 4 7 12 − 4 − 6 − 11 ⎠ ⎞ + ⎝ ⎛ 1 0 0 0 1 0 0 0 1 ⎠ ⎞ = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 ⎠ ⎞ .
So the Theorem holds.
As we know
A A − 1 = I A A^{-1} = I A A − 1 = I
As we could notice A 2 = I A^2 = I A 2 = I A = A − 1 A = A^{-1} A = A − 1
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#339914 on Dec 2023