Use Cramer’s rule to solve the linear system
and also solve the system using the inverse
A-1 of the coefficient matrix A.
Use Cramer’s rule to solve the linear system
and verify your answer by substituting it into
the left side of the linear system.
OR use a cofactor expansion down
column 1:
For A1 use a cofactor expansion
along row 3:
For A2 use a cofactor expansion
down column 1:
For A3 use a cofactor expansion
along row 3:
Verification of this solution:
Find the adjugate [the transpose of the matrix of cofactors]
and hence the inverse
(A-1) of the matrix
and use this inverse matrix to verify your
answer to question (2) above.
The sub-matrix Aij is what
remains of matrix A after the deletion
of row i and column j.
The cofactors are
From question 2, det A = 88
Find the eigenvalues and corresponding set of basic
eigenvectors for
Write down the matrix P that diagonalizes
A
and verify by matrix
multiplication that
P -1AP = D.
Hence find A 43.
l = –1:
The (–1)-eigenvector is therefore any non-zero multiple of
l = +1:
The (1)-eigenvector is therefore any non-zero multiple of
Find the eigenvalues and corresponding set of basic
eigenvectors for
Write down the matrix P that diagonalizes
A .
The matrix A is upper triangular
All eigenvalues have multiplicity 1, which guarantees the
existence of a matrix P that diagonalizes
A
l = –1:
The (–1)-eigenvector is therefore any non-zero multiple of
l = +1:
The (1)-eigenvector is therefore any non-zero multiple of
l = +2:
The (2)-eigenvector is therefore any non-zero multiple of
Therefore the matrix P that diagonalizes A is
and the set of basic eigenvectors is just the three columns of P .
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