SSPD-172 JAV 6 marils.sum of convenient propertiesGet Matrix Eigen values and Eigen VectorThe procedure to find the the eigenvalues and eigenvectors of a matrix involves a few key steps. Here's a concise guide: ### To find the Eigenvalues: 1. **Set up the characteristic equation**: - Start with the matrix ( A ) of size ( n imes n ). - Compute the determinant of ( A - lambda I ), where ( I ) is the identity matrix and ( lambda ) is a scalar. - This determinant gives you the characteristic polynomial of ( A ) : ( det(A - lambda I) = 0 ). 2. **Solve for ( lambda )**: <br> - Solve the characteristic polynomial to get the possible values of ( ). These are the eigenvalues of ( A ). ### To find the Eigenvectors: 1. **For each eigenvalue**: - Take the eigenvalue ( lambda ) and substitute it into the equation ( (A - lambda I) v = 0 ). - Solve the system of linear equations to get the vector ( v ). This vector is the eigenvector associated with the eigenvalue ( ). ### Example: Let's find the eigenvalues and eigenvectors of the matrix ( A = egin{pmatrix} 1 & 1 \ 4 & 1 end{pmatrix} ): 1. **Set up the characteristic equation**: - ( det(A - lambda I) = det egin{pmatrix} 1 - lambda & 1 \ 4 & 1 - lambda end{pmatrix} = (1 - lambda)*(1 - lambda) - 4 = 0 ) - ( (1 - lambda)^2 - 4 = 0 ) - ( lambda^2 - 2lambda -3 = 0 ) 2. **Solve for ( lambda )**: - ( lambda = frac{2 pm sqrt{ left(-2 ight)^2 - 4*(1)(-3) } }{2} ) - ( lambda = frac{2 pm sqrt{ 4 + 12 } }{2} ) - ( lambda = frac{2 pm sqrt{ 16 } }{2} ) - ( lambda = frac{2 pm 4 }{2} ) - ( lambda = 3 ) or ( lambda = -1 ) - Therefore, the eigenvalues are ( 3 ) and ( -1 ). 3. **Find the eigenvectors**: - For ( lambda = 3 ): - ( (A - 3I) egin{pmatrix} v_1 \ v_2 end{pmatrix} = 0 ) - ( egin{pmatrix} 1 - 3 & 1 \ 4 & 1 - 3 end{pmatrix} egin{pmatrix} v_1 \ v_2 end} = 0 ) - ( egin{pmatrix} -2 & 1 \ 4 & -2 end{pmatrix} egin{pmatrix} v_1 \ v_2 end} = 0 ) - ( -2v_1 + v_2 = 0 ) - ( v_2 = 2v_1 ) - Therefore, the eigenvector is ( v = egin{pmatrix} 1 \ 2 end{pmatrix} ). - For ( lambda = -1 ): - ( (A - (-1)I) egin{pmatrix} v_1 \ v_2 end{pmatrix} = 0 ) - ( egin{pmatrix} 1 + 1 1 \ 4 & 1 + 1 end{pmatrix} egin{pmatrix} v_1 \ v_2 end} = 0 ) - ( egin{pmatrix} 2 & 1 \ 4 & 2 end{pmatrix egin{pmatrix} v_1 \ v₂ end{pmatrix} = 0 ) - ( 2v_1 + v_2 = 0 ) - ( v_2 = -2v_1 ) - Therefore, the eigenvector is ( v = egin{pmatrix} 1 \ -2 end{pmatrix} ). ### Final Answer: The eigenvalues are ( 3 ) and ( -1 ). The corresponding eigenvectors are ( egin{pmatrix} 1 \ 2 end{pmatrix} ) and ( egin{pmatrix} 1 \ -2 end{pmatrix} ) . - Cuplikan Gratis dan Subtitle Bahasa Indonesia srt.

239 minit4534 tontonanPopuler!

Unduh Subtitle SSPD-172